Class 12 Maths

Class 12 Maths Solutions Vector Algebra

In our daily life, we generally come across two types of quantities, namely scalars and vectors.

Scalars A quantity that has magnitude only is known as a scalar.

Examples Each of the quantities mass, length, time, temperature, density, speed, etc., is a scalar.

Vectors A quantity that has magnitude as well as direction is called a vector.

Examples Each of the quantities force, velocity, acceleration and momentum is a vector.

However, we define a vector as given below.

‘A directed line segment is called a vector’.

A directed line segment with initial point A and the terminal point B, is the vector denoted by \(\overrightarrow{A B}\).

The magnitude of \(\overrightarrow{A B}\) is denoted by |\(\overrightarrow{A B}\)|.

Remark We usually denote a vector by a single letter with an arrow on it and its magnitude is denoted by this letter only.

Thus, \(\overrightarrow{A B}\) = \(\vec{a}\) and |\(\overrightarrow{A B}\)| = |\(\vec{a}\)| = a.

Unit Vector A vector \(\vec{a}\) is called a unit vector if |\(\vec{a}\)| = 1 and it is denoted by \(\hat{a}\) .

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Equal Vectors Two vectors \(\vec{a}\) and \(\vec{b}\) are said to be equal if they have the same magnitude and the same direction regardless of the positions of their initial points.

Negative Of A Vector A vector having the same magnitude as that of a given vector \(\vec{a}\) and the direction opposite to that of \(\vec{a}\) is called the negative of \(\vec{a}\), to be denoted by –\(\vec{a}\).

Thus, if \(\overrightarrow{A B}\) = \(\vec{a}\), then \(\overrightarrow{B A}\) = –\(\vec{a}\).

Zero Or Null Vector A vector whose initial and terminal points coincide is called a zero vector, denoted by \(\overrightarrow{0}\).

Clearly, the magnitude of a zero vector is 0 but it cannot be assigned a definite direction.

Thus, \(\overrightarrow{A A}\) = \(\overrightarrow{0}\).

Coinitial Vectors Two or more vectors having the same initial point are called coinitial vectors.

\(\overrightarrow{O A}\) and \(\overrightarrow{O B}\) are the two coinitial vectors having the same initial point O.

Collinear Vectors Vectors having the same or parallel supports are known as collinear vectors.

In the given figure \(\overrightarrow{A B}\), \(\overrightarrow{B C}\) and \(\overrightarrow{A C}\) are collinear vectors.

Like Vectors Collinear vectors having the same direction are called like vectors.

Thus, \(\overrightarrow{A B}\), \(\overrightarrow{B C}\) and \(\overrightarrow{A C}\) shown above are like vectors.

Unlike Vectors Collinear vectors having opposite directions are known as unlike vectors.

In the given figure PQ ∥ RS.

∴ \(\overrightarrow{P Q}\) and \(\overrightarrow{S R}\) are unlike vectors.

Free Vectors If the initial point of a vector is not specified then it is said to be a free vector.

Localised Vectors A vector drawn parallel to a given vector through a specified point as the initial point is called a localised vector.

Coplanar Vectors Three or more nonzero vectors lying in the same plane or parallel to the same plane are said to be coplanar, otherwise they are called noncoplanar.

Position Vector Of A Point Let O be the origin and let A be a point such that \(\overrightarrow{O A}\) = \(\vec{a}\), then we say that the position vector of A is \(\vec{a}\).

Solved Examples

Example 1 Classify the following measures as scalars and vectors:

(1) 30 seconds

(2) 100 m2

(3) 50 km/hr

(4) 20 gm/cm3

(5) 26 m/s towards south

Solution

(1) 30 seconds represents time, which is scalar.

(2) 100 m2 represents an area, which is scalar.

(3) 50 km/hr represents speed, which is scalar.

(4) 20 gm/cm3 represents density, which is scalar.

(5) 26 m/s towards south represents velocity, which is a vector.

Example 2 Represent graphically a displacement of 50 km, 60° west of north.

Solution

The vector \(\overrightarrow{O A}\) given below represents a displacement of 50 km, 60° west of north.

Vector Addition

Let \(\vec{a}\) and \(\vec{b}\) be any two vectors. Take any point O and draw segments \(\overrightarrow{O A}\) and \(\overrightarrow{A B}\) such that \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{A B}\) = \(\vec{b}\). Join OB. Then, \(\overrightarrow{O B}\) is called the sum or resultant of \(\vec{a}\) and \(\vec{b}\).

∴ \((\vec{a}+\vec{b})=\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B}\)

Triangle Law Of Addition Of Forces

In a △OAB, if \(\overrightarrow{O A}\) and \(\overrightarrow{A B}\) represent \(\vec{a}\) and \(\vec{b}\) respectively, then \(\overrightarrow{O B}\) represents \((\vec{a}+\vec{b})\).

This is known as Triangle Law of Addition of Forces.

Parallelogram Law Of Addition Of Forces In a ∥gm OABC, if \(\overrightarrow{O A}\) and \(\overrightarrow{A B}\) represent \(\vec{a}\) and \(\vec{b}\) respectively, then \(\overrightarrow{O B}\) represents \((\vec{a}+\vec{b})\).

This is known as Parallelogram Law of Addition of Forces.

Laws of Addition of Vectors

Theorem 1 (Commutative Law) Vector addition is commutative, i.e., \(\vec{a}+\vec{b}=\vec{b}+\vec{a} .\)

Proof

Let \(\vec{a}\) and \(\vec{b}\) be the given vectors represented by \(\overrightarrow{O A}\) and \(\overrightarrow{A B}\) respectively. Complete the parallelogram OABC.

Then, \(\overrightarrow{O C}=\overrightarrow{A B}=\vec{b}\)

and \(\overrightarrow{C B}=\overrightarrow{O A}=\vec{a}\)

∴ \(\vec{a}+\vec{b}=\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B}\)

and \(\vec{b}+\vec{a}=\overrightarrow{O C}+\overrightarrow{C B}=\overrightarrow{O B}\).

Hence, \(\vec{a}+\vec{b}=\vec{b}+\vec{a} .\)

WBBSE Class 12 Vector Algebra Solutions

Theorem 2 (Associative Law) Vector addition is associative, i.e., \((\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})\).

Proof

Let \(\overrightarrow{O A}\) = \(\vec{a}\), \(\overrightarrow{A B}\) = \(\vec{b}\) and \(\overrightarrow{B C}\) = \(\vec{c}\).

Join OB, OC and AC.

\((\vec{a}+\vec{b})+\vec{c}=(\overrightarrow{O A}+\overrightarrow{A B})+\overrightarrow{B C}\)

= \((\overrightarrow{O B}+\overrightarrow{B C})\) [∵ \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) = \(\overrightarrow{O B}\)]

= \(\overrightarrow{O C}\)

\(\vec{a}+(\vec{b}+\vec{c})=\overrightarrow{O A}+(\overrightarrow{A B}+\overrightarrow{B C})\)

= \((\overrightarrow{O A}+\overrightarrow{A C})\) [∵ \(\overrightarrow{A B}\) + \(\overrightarrow{B C}\) = \(\overrightarrow{A C}\)]

= \(\overrightarrow{O C}\)

∴ \((\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})\).

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Theorem 3 (Existence of Additive Identity) For any vector \(\vec{a}\), prove that \(\vec{a}+\overrightarrow{0}=\overrightarrow{0}+\vec{a}=\vec{a} .\)

Proof

Let \(\overrightarrow{O A}\) = \(\vec{a}\). Then, \(\vec{a}+\overrightarrow{0}=\overrightarrow{O A}+\overrightarrow{A A}=\overrightarrow{O A}=\vec{a}\)

and, \(\overrightarrow{0}+\vec{a}=\overrightarrow{O O}+\overrightarrow{O A}=\overrightarrow{O A}=\vec{a} .\)

∴ \(\vec{a}+\overrightarrow{0}=\overrightarrow{0}+\vec{a}=\vec{a} .\)

Remark The vector \(\overrightarrow{0}\) is called the additive identity for vectors.

Theorem 4 (Existence of Additive Inverse) For any vector \(\vec{a}\), prove that \(\vec{a}+(-\vec{a})=(-\vec{a})+\vec{a}=\overrightarrow{0}\).

Proof

Let \(\overrightarrow{O A}\) = \(\vec{a}\). Then, \(\overrightarrow{A o}\) = –\(\vec{a}\).

∴ \(\vec{a}+(-\vec{a})=\overrightarrow{O A}+\overrightarrow{A O}=\overrightarrow{O O}=\overrightarrow{0}\)

and, \((-\vec{a})+\vec{a}=\overrightarrow{A O}+\overrightarrow{O A}=\overrightarrow{A A}=\overrightarrow{0}\)

Hence, \(\vec{a}+(-\vec{a})=(-\vec{a})+\vec{a}=\overrightarrow{0}\).

Remark The vector –\(\vec{a}\) is called the additive inverse of \(\vec{a}\).

Difference Of Two Vectors For any two vectors \(\vec{a}\) and \(\vec{b}\), we define \(\vec{a}-\vec{b}=\vec{a}+(-\vec{b})\).

Let \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{O B}\) = \(\vec{b}\). Then, \(\overrightarrow{B O}\) = –\(\vec{b}\).

∴ \((\vec{a}-\vec{b})=\vec{a}+(-\vec{b})\)

= \(\overrightarrow{O A}+\overrightarrow{B O}=\overrightarrow{B O}+\overrightarrow{O A}=\overrightarrow{B A}\)

Thus, (\(\overrightarrow{O A}\) – \(\overrightarrow{O B}\)) = \(\overrightarrow{B A}\).

Similarly, (\(\overrightarrow{O B}\) – \(\overrightarrow{O A}\)) = \(\overrightarrow{A B}\).

Scalar Multiplication of a Vector

The scalar multiple of \(\vec{a}\) by a scalar k is the vector k \(\vec{a}\) such that

(1) \(|k \vec{a}|=|k||\vec{a}|\)

(2) direction of k \(\vec{a}\) is the same as that of \(\vec{a}\), when k > 0 and opposite to that of \(\vec{a}\) when k < 0.

Examples (1) 5 \(\vec{a}\) is the vector whose magnitude is 5 times the magnitude of \(\vec{a}\) and whose direction is the same as that of \(\vec{a}\).

(2) -2 \(\vec{a}\) is the vector whose magnitude is 2 times the magnitude of \(\vec{a}\) and whose direction is opposite to that of \(\vec{a}\).

Components of a Vector

Let O be the origin and let P(x,y,z) be any point in space. Let \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) be unit vectors along the x-axis, y-axis and z-axis respectively. Let the position vector of P be \(\vec{r}\).

Then, \(\vec{r}\) = \((x \hat{i}+y \hat{j}+z \hat{k}) .\)

This form of a vector is called its component form.

Here x, y, z are called the scalar components of \(\vec{r}\) and \(x \hat{i}, y \hat{j}, z \hat{k}\) are called its vector components.

Also, \(|\vec{r}\ = |x \hat{i}+y \hat{j}+z \hat{k}| = \sqrt{x^2+y^2+z^2}\).

Direction Ratios and Direction Cosines of a vector

Consider a vector \(\vec{r} = a \hat{i}+b \hat{j}+c \hat{k}\).

Then, the numbers a, b, c are called the direction ratios of \(\vec{r}\).

Direction cosines of \(\vec{r}\) are given by

\(\frac{a}{\sqrt{a^2+b^2+c^2}} \frac{b}{\sqrt{a^2+b^2+c^2}} \text { and } \frac{c}{\sqrt{a^2+b^2+c^2}} \text {. }\)

Note If l, m, n are the direction cosines of a vector then we always have (l2 + m2 + n2) = 1.

Remark If A(x1, y1, z1) and B(x2, y2, z2) be any two points in space, then direction ratios of \(\overrightarrow{A B}\) are (x2 – x1), (y2 – y1), (z2 – z1) and direction cosines of \(\overrightarrow{A B}\) are:

\(\frac{\left(x_2-x_1\right)}{r}, \frac{\left(y_2-y_1\right)}{r}, \frac{\left(z_2-z_1\right)}{r}\),

where r = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2} .\)

Solved Examples

Example 1 Let \(\vec{a} = a_1 \hat{i}+3 \hat{j}+a_3 \hat{k}\) and \(\vec{b} = 2 \hat{i}+b_2 \hat{j}+\hat{k}\). If \(\vec{a}\) = \(\vec{b}\), find the values of a1, b2 and a3.

Solution

\(\vec{a}\) = \(\vec{b}\) ⇔ \(a_1 \hat{i}+3 \hat{j}+a_3 \hat{k}=2 \hat{i}+b_2 \hat{j}+\hat{k}\)

⇔ a1 = 2, b2 = 3, a3 = 1.

The values of a1, b2 and a3 are 2,3,1.

Example 2 Let \(\vec{a}\) = \(3 \hat{i}+2 \hat{j}\) and \(\vec{b}\) = \(2 \hat{i}+3 \hat{j}\). Is |\(\vec{a}\)| = |\(\vec{b}\)|? Is \(\vec{a}\) = \(\vec{b}\)?

Solution

We have

\(|\vec{a}|=\sqrt{3^2+2^2}=\sqrt{13} \text { and }|\vec{b}|=\sqrt{2^2+3^2}=\sqrt{13} \text {. }\)

∴ |\(\vec{a}\)| = |\(\vec{b}\)|.

But, \(3 \hat{i}+2 \hat{j}\) ≠ \(2 \hat{i}+3 \hat{j}\) and therefore, \(\vec{a}\) ≠ \(\vec{b}\).

Example 3 Find a unit vector in the direction of the vector \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\)

Solution

\(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\) ⇒ \(|\vec{a}|=\sqrt{1^2+2^2+3^2}=\sqrt{14} \text {. }\)

Unit vector in the direction of \(\vec{a}\) is given by

\(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{(\hat{i}+2 \hat{j}+3 \hat{k})}{\sqrt{14}}=\left(\frac{1}{\sqrt{14}} \hat{i}+\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}\right) .\)

Example 4 Find a unit vector in the direction of \(\overrightarrow{A B}\), where A(1,2,3) and B(4, 5, 6) are the given points.

Solution

We have

p.v. of A = \((\hat{i}+2 \hat{j}+3 \hat{k})\) and p.v. of B = \((4 \hat{i}+5 \hat{j}+6 \hat{k})\)

∴ \(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)

= \((4 \hat{i}+5 \hat{j}+6 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})=(3 \hat{i}+3 \hat{j}+3 \hat{k})\), and

\(|\overrightarrow{A B}| = \sqrt{3^2+3^2+3^2}=\sqrt{27}\)

∴ unit vector in the direction of \(\overrightarrow{A B}=\frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}\)

= \(\frac{(3 \hat{i}+3 \hat{j}+3 \hat{k})}{\sqrt{27}}=\frac{3(\hat{i}+\hat{j}+\hat{k})}{3 \sqrt{3}}=\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}\)

= \(\left(\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\right) .\)

Example 5 Find a vector in the direction of the vector \(\vec{a}=(3 \hat{i}+\hat{j})\) that has magnitude 5 units.

Solution

\(\vec{a}=(3 \hat{i}+\hat{j})\) ⇒ \(|\vec{a}| = \sqrt{3^2+1^2}=\sqrt{10}\)

Unit vector in the direction of \vec{a} is given by

\(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{(3 \hat{i}+\hat{j})}{\sqrt{10}}=\left(\frac{3}{\sqrt{10}} \hat{i}+\frac{1}{\sqrt{10}} \hat{j}\right) .\)

Hence, the required vector, 5 \(\hat{a}\) = \(\left(\frac{15}{\sqrt{10}} \hat{i}+\frac{5}{\sqrt{10}} \hat{j}\right)\)

Example 6 Find a vector in the direction of the vector \(3 \hat{i}-\hat{j}+4 \hat{k}\), which has magnitude 6 units.

Solution

Let \(\vec{a}\) = \((3 \hat{i}-\hat{j}+4 \hat{k})\). Then,

|\(\vec{a}\)| = \(\sqrt{3^2+(-1)^2+4^2}=\sqrt{26}\)

∴ unit vector, \(\hat{a}=\left\{\frac{3}{\sqrt{26}} \hat{i}-\frac{1}{\sqrt{26}} \hat{j}+\frac{4}{\sqrt{26}} \hat{k}\right\}\)

Hence, the required vector, \(6 \hat{a}=\left\{\frac{18}{\sqrt{26}} \hat{i}-\frac{6}{\sqrt{26}} \hat{j}+\frac{24}{\sqrt{26}} \hat{k}\right\}\).

Step-by-Step Solutions to Vector Algebra Problems

Example 7 If \(\vec{a}=(\hat{i}+2 \hat{j}-3 \hat{k}) \text { and } \vec{b}=(2 \hat{i}+3 \hat{j}+\hat{k})\), find a unit vector in the direction of \((\vec{a}+\vec{b})\).

Solution

Let \(\vec{c}=(\vec{a}+\vec{b})\). Then,

\(\vec{c}=(\vec{a}+\vec{b})\)

= \((\hat{i}+2 \hat{j}-3 \hat{k})+(2 \hat{i}+3 \hat{j}+\hat{k})\)

= \((3 \hat{i}+5 \hat{j}-2 \hat{k})\)

∴ \(|\vec{c}|=\sqrt{3^2+5^2+(-2)^2}=\sqrt{38} .\)

Hence, the required vector is given by

\(\hat{c}=\frac{\vec{c}}{|\vec{c}|}=\frac{(3 \hat{i}+5 \hat{j}-2 \hat{k})}{\sqrt{38}}=\left(\frac{3}{\sqrt{38}} \hat{i}+\frac{5}{\sqrt{38}} \hat{j}-\frac{2}{\sqrt{38}} \hat{k}\right) .\)

Example 8 Write two different vectors having same magnitude.

Solution

Consider the vectors \(\vec{a}=2 \hat{i}+3 \hat{j}-4 \hat{k} \text { and } \vec{b}=4 \hat{i}+2 \hat{j}+3 \hat{k}\)

Clearly, \(\vec{a}\) ≠ \(\vec{b}\).

But, \(|\vec{a}|=\sqrt{2^2+3^2+(-4)^2}=\sqrt{29} \text { and }|\vec{b}|=\sqrt{4^2+2^2+3^2}=\sqrt{29} \text {. }\)

Thus, |\(\vec{a}\)| = |\(\vec{b}\)| and \(\vec{a}\) ≠ \(\vec{b}\).

Example 9 Write two different vectors having same direction.

Solution

Clearly, 3 \(\overrightarrow{A B}\) and 5 \(\overrightarrow{A B}\) are two different vectors having the same direction.

Example 10 Find the scalar and vector components of the vector with initial point A(3,2) and terminal point B(-5,7).

Solution

Given vector is \(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)

= \((-5 \hat{i}+7 \hat{j})-(3 \hat{i}+2 \hat{j})=(-8 \hat{i}+5 \hat{j})\)

Vector components of \(\overrightarrow{A B}\) are \(-8 \hat{i} and 5 \hat{j}\).

Scalar components of \(\overrightarrow{A B}\) are -8 and 5.

Example 11 If A(1,2,-3) and B(-1,-2,1) are two given points in space then find (1) the direction ratios of \(\overrightarrow{A B}\) and (2) the direction cosines of \(\overrightarrow{A B}\).

Solution

(1) DRs of \(\overrightarrow{A B}\) are (-1-1), (-2-2), (1+3), i.e., -1, -4, 4.

(2) r2 = {(-2)2 + (-4)2 + 42} = 36 ⇒ r = √36 = 6.

Hence, DCs of \(\overrightarrow{A B}\) are \(\frac{-2}{6}, \frac{-4}{6}, \frac{4}{6} \text {, i.e., } \frac{-1}{3}, \frac{-2}{3}, \frac{2}{3} \text {. }\)

Example 12 Find the direction ratios and the direction cosines of the vector \(\vec{r}=2 \hat{i}+3 \hat{j}+\hat{k}\).

Solution

Given vector is \(\vec{r}=2 \hat{i}+3 \hat{j}+\hat{k}\).

DRs of \(\vec{r}\) are 2, 3, 1.

Also, \(|\vec{r}|=\sqrt{2^2+3^2+1^2}=\sqrt{14} \text {. }\)

So, DCs of \(\vec{r}\) are \(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}\)

Secion Formulae

Theorem 1 (Section Formula for Internal Division) Let A and B be two points with position vectors \(\vec{a}\) and \(\vec{b}\) respectively and let P be a point dividing AB internally in the ratio m:n. Let \(\overrightarrow{O P}\) = \(\vec{r}\). Then, prove that

\(\vec{r}=\frac{(m \vec{b}+n \vec{a})}{(m+n)}\)

Proof

Let O be the origin. Then, \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{O B}\) = \(\vec{b}\).

Let P be a point on AB such that \(\frac{A P}{P B}=\frac{m}{n}\). Then,

\(\frac{A P}{P B}=\frac{m}{n}\)

⇒ n.AP = m.PB

⇒ \(n(\overrightarrow{A P})=m(\overrightarrow{P B})\)

⇒ \(n(\overrightarrow{O P}-\overrightarrow{O A})=m(\overrightarrow{O B}-\overrightarrow{O P})\)

⇒ \((m+n) \overrightarrow{O P}=m \overrightarrow{O B}+n \overrightarrow{O A}=m \vec{b}+n \vec{a}\)

⇒ \(\overrightarrow{O P}=\frac{m \vec{b}+n \vec{a}}{(m+n)} ⇒ \vec{r}=\frac{(m \vec{b}+n \vec{a})}{(m+n)}\)

Corollary The position vector of the midpoint of the join of two points with position vectors \(\vec{a}\) and \(\vec{b}\) is \(\frac{1}{2}(\vec{a}+\vec{b}) .\)

Proof

Let A and B be two points with position vectors \(\vec{a}\) and \(\vec{b}\) respectively.

Let P be the midpoint of AB.

Then, P divides AB in the ratio 1:1.

∴ \(\overrightarrow{O P}=\frac{(1 \cdot \vec{b}+1 \cdot \vec{a})}{(1+1)}=\frac{1}{2}(\vec{a}+\vec{b})\)

Theorem 2 (Section Formula for External Division) Let A and B be two points with position vectors \(\vec{a}\) and \(\vec{b}\) respectively and let P be a point dividing AB externally in the ratio m:n. Let \(\overrightarrow{O P}\) = \(\vec{r}\). Then, prove that

\(\vec{r}=\frac{(m \vec{b}-n \vec{a})}{(m-n)}\)

Proof

Let O be the origin. Then, \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{O B}\) = \(\vec{b}\).

Let AB be produced to P such that AP:BP = m:n.

Now, \(\frac{A P}{B P}=\frac{m}{n}\)

⇒ n.AP = m.BP ⇒ n.\(\overrightarrow{A P}\) = m.\(\overrightarrow{B P}\)

⇒ \(n(\overrightarrow{O P}-\overrightarrow{O A})=m(\overrightarrow{O P}-\overrightarrow{O B})\)

⇒ \((m-n) \overrightarrow{O P}=(m \overrightarrow{O B}-n \overrightarrow{O A})=(m \vec{b}-n \vec{a})\)

⇒ \(\overrightarrow{O P}=\frac{(m \vec{b}-n \vec{a})}{(m-n)} ⇒ \vec{r}=\frac{(m \vec{b}-n \vec{a})}{(m-n)}\)

Real-Life Applications of Vector Algebra Concepts

Example 13 Find the position vector of a point R which divides the line joining the points \(P(\hat{i}+2 \hat{j}-\hat{k}) \text { and } Q(-\hat{i}+\hat{j}+\hat{k})\) in the ratio 2:1 (1) internally and (2) externally.

Solution

Here \(\vec{a}=(\hat{i}+2 \hat{j}-\hat{k}) \text { and } \vec{b}=(-\hat{i}+\hat{j}+\hat{k})\). Also, m = 2, n = 1.

(1) When R divides PQ internally in the ratio 2:1

Then, p.v. of R = \(\frac{(m \vec{b}+n \vec{a})}{(m+n)}\)

= \(\frac{2(-\hat{i}+\hat{j}+\hat{k})+1 \cdot(\hat{i}+2 \hat{j}-\hat{k})}{(2+1)}=\frac{(-\hat{i}+4 \hat{j}+\hat{k})}{3} .\)

(2) When R divides PQ externally in the ratio 2:1

Then, p.v. of R = \(\frac{(m \vec{b}-n \vec{a})}{(m-n)}\)

= \(\frac{2(-\hat{i}+\hat{j}+\hat{k})-1 \cdot(\hat{i}+2 \hat{j}-\hat{k})}{(2-1)}=(-3 \hat{i}+3 \hat{k})\)

Example 14 Find the position vector of the midpoint of the vector joining the point \(A(2 \hat{i}+3 \hat{j}+4 \hat{k}) \text { and } B(4 \hat{i}+\hat{j}-2 \hat{k})\)

Solution

The position vectors of A and B are given by

\(\vec{a}=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \text { and } \vec{b}=(4 \hat{i}+\hat{j}-2 \hat{k})\)

∴ p.v. of midpoint of AB

= \(\frac{1}{2}(\vec{a}+\vec{b})\)

= \(\frac{1}{2}\{(2 \hat{i}+3 \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}-2 \hat{k})\}\)

= \((3 \hat{i}+2 \hat{j}+\hat{k})\).

The position vector of the midpoint of the vector joining the point = \((3 \hat{i}+2 \hat{j}+\hat{k})\).

Example 15 Show that the points A, B, C with position vectors \(\vec{a}=(3 \hat{i}-4 \hat{j}-4 \hat{k}), \vec{b}=(2 \hat{i}-\hat{j}+\hat{k}) \text { and } \vec{c}=(\hat{i}-3 \hat{j}-5 \hat{k})\) respectively, from the vertices of a right-angled triangle.

Solution

We have

\(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A) = \((\vec{b}-\vec{a})\)

= \((2 \hat{i}-\hat{j}+\hat{k})-(3 \hat{i}-4 \hat{j}-4 \hat{k})=(-\hat{i}+3 \hat{j}+5 \hat{k})\);

\(\overrightarrow{B C}\) = (p.v. of C)-(p.v. of B) = \((\vec{c}-\vec{b})\)

= \((\hat{i}-3 \hat{j}-5 \hat{k})-(2 \hat{i}-\hat{j}+\hat{k})=(-\hat{i}-2 \hat{j}-6 \hat{k})\);

\(\overrightarrow{C A}\) = (p.v. of A)-(p.v. of C) = \((\hat{a}-\hat{c})\)

= \((3 \hat{i}-4 \hat{j}-4 \hat{k})-(\hat{i}-3 \hat{j}-5 \hat{k})=(2 \hat{i}-\hat{j}+\hat{k})\)

∴ \(|\overrightarrow{A B}|^2=\left\{(-1)^2+3^2+5^2\right\}=35\)

\(|\overrightarrow{A B}|^2=\left\{(-1)^2+3^2+5^2\right\}=35\)

and \(|\overrightarrow{C A}|^2=\left\{2^2+(-1)^2+1^2\right\}=6\)

Thus, \(|\overrightarrow{A B}|^2+|\overrightarrow{C A}|^2=|\overrightarrow{B C}|^2\), i.e., AB2 + CA2 = BC2.

Hence, △ABC is right angled at A.

Scalar Product of Vectors

Angle Between Two Vectors Let \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\) be two given vectors. Take any point O, draw OA ∥ PQ and OB ∥ RS. Then, ∠AOB = θ is called the angle between \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\), provided 0 ≤ θ ≤ π.

If θ = 0 or θ = π then \(\overrightarrow{P Q}\) ∥ \(\overrightarrow{R S}\).

If θ = \(\frac{\pi}{2}\) then \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\) are called perpendicular or orthogonal vectors.

Scalar Product Or Dot Product Of Two Vectors Let \(\vec{a}\) and \(\vec{b}\) be two vectors and let θ be the angle between them. Then, the scalar product, or dot product, of \(\vec{a}\) and \(\vec{b}\), denoted by \(\vec{a}\).\(\vec{b}\), is defined as \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\) cos θ = ab cos θ.

Clearly, the scalar product of two vectors is a scalar.

Types of Vectors Explained

Angle Between Two Vectors In Terms Of Scalar Product Let θ be the angle between two nonzero vectors \(\vec{a}\) and \(\vec{b}\). Then,

\(\vec{a}\).\(\vec{b}\) = \(|\vec{a}||\vec{b}| \cos \theta \text {. }\)

∴ \(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \Rightarrow \theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right) .\)

Length Of A Vector Let \(\vec{a}\) be any vector.

Then, \(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \Rightarrow \theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right) .\)

∴ \(|\vec{a}|=\sqrt{\vec{a} \cdot \vec{a}} .\)

Remarks (1) If \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\), we define \(\vec{a}\).\(\vec{b}\) = 0.

(2) If \(\vec{a}\) and \(\vec{b}\) are like vectors, we have θ = 0.

∴ \(\vec{a} \cdot \vec{b}=a b \cos 0^{\circ}=a b .\)

(3) If \(\vec{a}\) and \(\vec{b}\) are unlike vectors, we have θ = π.

∴ \(\vec{a} \cdot \vec{b}=a b \cos \pi=-a b .\)

(4) If \(\vec{a}\) and \(\vec{b}\) are orthogonal vectors, we have θ = \(\frac{\pi}{2}\).

∴ \(\vec{a} \cdot \vec{b}=a b \cos \frac{\pi}{2}=0 .\)

Example 1 Let \(\vec{a}\) and \(\vec{b}\) be two given vectors such that |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 4 and the angle between them is 60°. Find \(\vec{a}\).\(\vec{b}\).

Solution

Clearly, we have

\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos 60^{\circ}=\left(3 \times 4 \times \frac{1}{2}\right)=6 \text {. }\)

Example 2 Let \(\vec{a}\) and \(\vec{b}\) be two given vectors such that |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 1 and \(\vec{a}\).\(\vec{b}\) = 1. Find the angle between \(\vec{a}\) and \(\vec{b}\).

Solution

Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,

\(\vec{a} \cdot \vec{b}=1 \Rightarrow|\vec{a}||\vec{b}| \cos \theta=1\)

⇒ (2 x 1)cos θ = 1 [∵ |\(\vec{a}\)| = 2 and |\(\vec{b}\)| = 1]

⇒ \(\cos \theta=\frac{1}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)

Hence, the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{3}\).

Orthonormal Vector Triad Let \(\hat{i}, \hat{j}, \hat{k}\) be unit vectors along three mutually perpendicular coordinate axes, the x-axis, y-axis and z-axis respectively. Then, these vectors are said to form an orthonormal triad of vectors.

Clearly, we have

(1) \(\hat{i} \cdot \hat{i}=|\hat{i}||\hat{i}| \cos 0^{\circ}=1\)

Similarly, \(\hat{j} \cdot \hat{j}=1 \text { and } \hat{k} \cdot \hat{k}=1 \text {. }\)

Thus, \(\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1\).

(2) \(\hat{i} \cdot \hat{j}=|\hat{i}||\hat{j}| \cos \frac{\pi}{2}=0 \text {. }\)

Similarly, \(\hat{i} \cdot \hat{k}=0, \hat{j} \cdot \hat{k}=0, \hat{j} \cdot \hat{i}=0, \hat{k} \cdot \hat{i}=0 \text { and } \hat{k} \cdot \hat{j}=0\)

∴ \(\hat{i} \cdot \hat{j}=\hat{i} \cdot \hat{k}=\hat{j} \cdot \hat{i}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=\hat{k} \cdot \hat{j}=0\)

Projection Of \(\vec{b}\) on \(\vec{a}\) Let \(\overrightarrow{O A}=\vec{a}\), \(\overrightarrow{O B}=\vec{b}\) and ∠AOB = θ. Draw BM ⊥ OA. Then, OM is projection of \(\vec{b}\) on \(\vec{a}\).

OM = projection of \(\vec{b}\) on \(\vec{a}\) = \(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|}\)

Properties of Scalar Product

Theorem 1 (Commutative Law) Prove that \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)

Proof

If \(\vec{a}\) or \(\vec{b}\) is a zero vector then \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{b}\).\(\vec{a}\) = 0.

So, in this case, \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)

Now, let \(\vec{a}\) and \(\vec{b}\) be any two nonzero vectors, and let θ be the angle between them. Then,

\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta=a b \cos \theta \text {, }\) and

\(\vec{b} \cdot \vec{a}=|\vec{b}||\vec{a}| \cos (-\theta)=b a \cos \theta=a b \cos \theta\)

So, in this case also, \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)

Hence, \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)

Theorem 2 Prove that \(\vec{a}\).\(\vec{b}\) = 0 ⇔ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\).

Proof

Let \(\vec{a}\).\(\vec{b}\) = 0.

Then, \(\vec{a}\).\(\vec{b}\) = 0 ⇒ ab cos θ = 0

⇒ a = 0 or b = 0 or cos θ = 0

⇒ a = 0 or b = 0 or θ = \(\frac{\pi}{2}\)

⇒ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\).

Thus, \(\vec{a}\).\(\vec{b}\) = 0 ⇒ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\)

Conversely, let \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\).

If \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) then by definition, \(\vec{a}\).\(\vec{b}\) = 0

If \(\vec{a}\) ⊥ \(\vec{b}\) then \(\vec{a}\).\(\vec{b}\) = ab cos \(\frac{\pi}{2}\) = 0.

∴ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) ⇔ \(\vec{a}\) ⊥ \(\vec{b}\) ⇒ \(\vec{a}\).\(\vec{b}\) = 0.

Hence, \(\vec{a}\).\(\vec{b}\) = 0. ⇔ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\).

Theorem 3 (Cauch Schwartz Inequality) Prove that \(|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}|\).

Proof

When \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\), then

\(|\vec{a} \cdot \vec{b}|=0=|\vec{a}||\vec{b}| \text {. }\)

So, let us assume that \(\vec{a}\) ≠ \(\overrightarrow{0}\) and \(\vec{b}\) ≠ \(\overrightarrow{0}\). Then,

\((\vec{a} \cdot \vec{b})=|\vec{a}||\vec{b}| \cos \theta\)

⇒ \(|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}||\cos \theta|\)

⇒ \(\frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|}=|\cos \theta| \leq 1\)

⇒ \(|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}| .\)

Hence, \(|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}| .\)

Theorem 4 (Triangle Inequality) Prove that \(|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}| \text {. }\)

Proof

When \(\vec{a}\) = \(\overrightarrow{0}\), then |\(\vec{a}\)| = 0.

∴ \(|\vec{a}+\vec{b}|=|\overrightarrow{0}+\vec{b}|=|\vec{b}| \text { and }|\vec{a}|+|\vec{b}|=0+|\vec{b}|=|\vec{b}| \text {. }\)

So, in this case, \(|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|\)

Similarly, when \(\vec{b}\) = \(\overrightarrow{0}\), we have \(|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|\)

Let us consider the case when \(\vec{a}\) ≠ \(\overrightarrow{0}\) and \(\vec{b}\) ≠ \(\overrightarrow{0}\).

Now, \(|\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b})^2\)

= \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})\)

= \(\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)

= \(|\vec{a}|^2+2(\vec{a} \cdot \vec{b})+|\vec{b}|^2\) [∵ \(\vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b}\)]

≤ \(|\vec{a}|^2+2|\vec{a} \cdot \vec{b}|+|\vec{b}|^2 \) [∵ α ≤ |α| ∀ α ∈ R]

≤ \(|\vec{a}|^2+2|\vec{a}| \cdot|\vec{b}|+|\vec{b}|^2\) [∵ \(|\vec{a} \cdot \vec{b}| \leq|\vec{a}| \cdot|\vec{b}|\)]

= \(\{|\vec{a}|+|\vec{b}|\}^2\)

∴ \(|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}| \text {. }\)

Common Questions on Vector Algebra and Their Solutions

Theorem 5 (Distributive Law) Prove that \(\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c} .\)

Proof

Let O be the origin, \(\overrightarrow{O A}\) = \(\vec{a}\), \(\overrightarrow{O B}\) = \(\vec{b}\) and \(\overrightarrow{B C}\) = \(\vec{c}\). Then,

\(\overrightarrow{O C}=(\overrightarrow{O B}+\overrightarrow{B C})=(\vec{b}+\vec{c})\)

Draw BM ⊥ OA and CN ⊥ OA. Then,

\(\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \overrightarrow{O C}\)

= \(\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \overrightarrow{O C}\), where ∠AOC = θ.

= a x ON [∵ (OC)cos θ = ON]

= a(OM + MN)

= a(OM) + a(MN)

= a(projection of \(\vec{v}\) on \(\vec{a}\)) + a(projection of \(\vec{c}\) on \(\vec{a}\))

= \((\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c})\)

Hence, \(\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c} .\)

Theorem 6 For any two vectors \(\vec{a}\) and \(\vec{b}\),

(1) \(\vec{a} \cdot(-\vec{b})=-(\vec{a} \cdot \vec{b})=(-\vec{a}) \cdot \vec{b}\)

(2) \((-\vec{a}) \cdot(-\vec{b})=\vec{a} \cdot \vec{b}\)

Proof

Let \(\overrightarrow{O A}\) = \(\vec{a}\) and \(\overrightarrow{O B}\) = \(\vec{b}\). Produce AO and BO to A’ and B’ respectively, such that OA’ = OA and OB’ = OB.

Then, \(\overrightarrow{O A^{\prime}}=-\vec{a} \text { and } \overrightarrow{O B^{\prime}}=-\vec{b}\)

Let ∠AOB = θ.

Then, ∠AOB’ = π – θ.

(1) \(\vec{a} \cdot(-\vec{b})=|\vec{a}||-\vec{b}| \cos (\pi-\theta)\)

= \(|\vec{a}||\vec{b}| \cdot(-\cos \theta)\)

= \(-a b \cos \theta=-(\vec{a} \cdot \vec{b}) .\)

∴ \(\vec{a} \cdot(-\vec{b})=-(\vec{a} \cdot \vec{b}) .\)

Similarly, \((-\vec{a}) \cdot \vec{b}=-(\vec{a} \cdot \vec{b}) .\)

Hence, \(\vec{a} \cdot(-\vec{b})=(-\vec{a}) \cdot \vec{b}=-(\vec{a} \cdot \vec{b})\)

(2) \((-\vec{a}) \cdot(-\vec{b})=|-\vec{a}||-\vec{b}| \cos \angle A^{\prime} O B^{\prime}\)

= \(|\vec{a}||\vec{b}| \cos \theta\) [∵ ∠A’OB’ = θ]

= \(a b \cos \theta=\vec{a} \cdot \vec{b} .\)

Hence, \((-\vec{a}) \cdot(-\vec{b})=\vec{a} \cdot \vec{b}\).

Theorem 7 If \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \quad \text { and } \quad \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\), prove that \((\vec{a} \cdot \vec{b})=\left(a_1 b_1+a_2 b_2+a_3 b_3\right) .\)

Proof

We have

\(\vec{a} \cdot \vec{b}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\)

= \(a_1 \hat{i} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)+a_2 \hat{j} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)+a_3 \hat{k} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\)

= \(\left(a_1 b_1\right)(\hat{i} \cdot \hat{i})+\left(a_1 b_2\right)(\hat{i} \cdot \hat{j})+\left(a_1 b_3\right)(\hat{i} \cdot \hat{k})+\left(a_2 b_1\right)(\hat{j} \cdot \hat{i})\)
\(+\left(a_2 b_2\right)(\hat{j} \cdot \hat{j})+\left(a_2 b_3\right)(\hat{j} \cdot \hat{k})+\left(a_3 b_1\right)(\hat{k} \cdot \hat{i})+\left(a_3 b_2\right)(\hat{k} \cdot \hat{j})+\left(a_3 b_3\right)(\hat{k} \cdot \hat{k})\)

= \(\left(a_1 b_1+a_2 b_2+a_3 b_3\right)\)

[∵ \(\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1, \hat{i} \cdot \hat{j}=\hat{i} \cdot \hat{k}=\hat{j} \cdot \hat{i}\) = … = 0].

Hence, \(\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)=\left(a_1 b_1+a_2 b_2+a_3 b_3\right)\)

Condition Of Perpendicularity

Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \quad \text { and } \quad \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\)

Then, \(\vec{a}\) ⊥ \(\vec{b}\) ⇔ \(\vec{a}\).\(\vec{b}\) = 0 ⇔ \(a_1 b_1+a_2 b_2+a_3 b_3=0 .\)

Thus, \(\vec{a}\) ⊥ \(\vec{b}\) ⇔ \(a_1 b_1+a_2 b_2+a_3 b_3=0 .\)

Angle Between Two Vectors

Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \quad \text { and } \quad \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\), let θ be the angle between them. Then,

\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)

⇔ \(a_1 b_1+a_2 b_2+a_3 b_3=|\vec{a}||\vec{b}| \cos \theta\)

⇔ \(\cos \theta=\frac{a_1 b_1+a_2 b_2+a_3 b_3}{|\vec{a}||\vec{b}|}\)

⇔ \(\theta=\cos ^{-1}\left\{\frac{a_1 b_1+a_2 b_2+a_3 b_3}{\left(\sqrt{a_1^2+a_2^2+a_3^2}\right)\left(\sqrt{b_1^2+b_2^2+b_3^2}\right)}\right\}\)

Solved Examples

Example 1 Let \(\vec{a}\) and \(\vec{b}\) be two given vectors such that \(|\vec{a}|=\sqrt{3},|\vec{b}|=2\) and \(\vec{a} \cdot \vec{b}=\sqrt{6}\). Find the angle between \(\vec{a}\) and \(\vec{b}\).

Solution

Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,

\(\vec{a} \cdot \vec{b}=\sqrt{6}\) ⇒ \(|\vec{a}||\vec{b}| \cos \theta=\sqrt{6}\)

⇒ \((\sqrt{3})(2) \cos \theta=\sqrt{6}\)

⇒ \(\cos \theta=\frac{\sqrt{6}}{2 \sqrt{3}}=\frac{1}{\sqrt{2}} ⇒ \theta=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\)

Hence, the required angle is \(\frac{\pi}{4}\).

Example 2 If \(\vec{a}\) and \(\vec{b}\) are two vectors such that \(|\vec{a}|=|\vec{b}|=\sqrt{2}\) and \(\vec{a}\).\(\vec{b}\) = -1, find the angle between \(\vec{a}\) and \(\vec{b}\).

Solution

Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,

\(\vec{a}\).\(\vec{b}\) = -1

⇔ \(|\vec{a}| \cdot|\vec{b}| \cdot \cos \theta=-1\)

⇔ \(\sqrt{2} \times \sqrt{2} \times \cos \theta=-1\)

⇔ \(\cos \theta=\frac{-1}{2}\)

⇔ \(\theta=\frac{2 \pi}{3}\) [∵ 0 ≤ θ ≤ 2π].

Hence, the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{2 \pi}{3}\).

Example 3 Find the angle between the vectors \(\vec{a}=(3 \hat{i}-2 \hat{j}+\hat{k}) \text { and } \vec{b}=(\hat{i}-2 \hat{j}-3 \hat{k}) \text {. }\)

Solution

Let θ be the angle between \(\vec{a}\) and \(\vec{b}\).

Now, \(|\vec{a}|=\sqrt{3^2+(-2)^2+1^2}=\sqrt{14} \text {. }\)

And, \(|\vec{b}|=\sqrt{1^2+(-2)^2+(-3)^2}=\sqrt{14}\)

Now, \(\vec{a} \cdot \vec{b}=(3 \hat{i}-2 \hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+3 \hat{k})\)

= [(3×1)+(-2)(-2)+(1×3)] = (3+4+3) = 10

⇒ \(|\vec{a}||\vec{b}| \cos \theta=10\)

⇒ \((\sqrt{14} \cdot \sqrt{14}) \cos \theta=10\) ⇒ \(\cos \theta=\frac{10}{14}=\frac{5}{7}\)

⇒ \(\theta=\cos ^{-1}\left(\frac{5}{7}\right) .\)

Hence, the required angle is cos-1(5/7).

Applications of Vector Algebra in Real Life

Example 4 If \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k} \text { and } \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}\), find the value of \((\vec{a}+3 \vec{b}) \cdot(2 \vec{a}-\vec{b}) .\)

Solution

We have

\((\vec{a}+3 \vec{b})=(\hat{i}+\hat{j}+2 \hat{k})+3(3 \hat{i}+2 \hat{j}-\hat{k})\)

= \((\vec{a}+3 \vec{b})=(\hat{i}+\hat{j}+2 \hat{k})+3(3 \hat{i}+2 \hat{j}-\hat{k})\)

= \((10 \hat{i}+7 \hat{j}-\hat{k})\), and

\((2 \vec{a}-\vec{b})=2(\hat{i}+\hat{j}+2 \hat{k})-(3 \hat{i}+2 \hat{j}-\hat{k})\)

= \((2 \hat{i}+2 \hat{j}+4 \hat{k})-(3 \hat{i}+2 \hat{j}-\hat{k})=(-\hat{i}+5 \hat{k})\)

∴ \((\vec{a}+3 \vec{b}) \cdot(2 \vec{a}-\vec{b})=(10 \hat{i}+7 \hat{j}-\hat{k}) \cdot(-\hat{i}+0 \hat{j}+5 \hat{k})\)

= [10 x (-1) + 7 x 0 + (-1) x 5] = -15.

Hence, \((\vec{a}+3 \vec{b}) \cdot(2 \vec{a}-\vec{b})=-15\).

Example 5 Find the value of λ for which the vectors \(\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k} \text { and } \vec{b}=\hat{i}+\lambda \hat{j}-3 \hat{k}\) are perpendicular to each other.

Solution

\(\vec{a}\) ⊥ \(\vec{b}\) ⇔ \(\vec{a}\).\(\vec{b}\) = 0

⇔ \((3 \hat{i}+\hat{j}-2 \hat{k}) \cdot(\hat{i}+\lambda \hat{j}-3 \hat{k})=0\)

⇔ 3 x 1 + 1 x λ + (-2) x (-3) = 0

⇔ λ = -9.

Hence, the required value of λ is -9.

Example 5 If \(\vec{a}=(5 \hat{i}-\hat{j}-3 \hat{k}) \text { and } \vec{b}=(\hat{i}+3 \hat{j}-5 \hat{k})\), then show that \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) are perpendicular to each other.

Solution

We have

\((\vec{a}+\vec{b})=(5 \hat{i}-\hat{j}-3 \hat{k})+(\hat{i}+3 \hat{j}-5 \hat{k})=(6 \hat{i}+2 \hat{j}-8 \hat{k})\), and

\((\vec{a}-\vec{b})=(5 \hat{i}-\hat{j}-3 \hat{k})-(\hat{i}+3 \hat{j}-5 \hat{k})=(4 \hat{i}-4 \hat{j}+2 \hat{k}) .\)

∴ \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(6 \hat{i}+2 \hat{j}-8 \hat{k}) \cdot(4 \hat{i}-4 \hat{j}+2 \hat{k})\)

= (24 – 8 – 16) = 0.

Hence, \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) are perpendicular to each other.

Example 7 If \(\overrightarrow{A B}\) = \((2 \hat{i}+\hat{j}-3 \hat{k})\), A(1,2,-1) is the given point then find the coordinates of B.

Solution

Let O be the origin. Then, \(\overrightarrow{O A}\) = \((\hat{i}+2 \hat{j}-\hat{k})\)

∴ \(\overrightarrow{A B}=(\overrightarrow{O B}-\overrightarrow{O A})\)

⇒ \(\overrightarrow{O B}=\overrightarrow{A B}+\overrightarrow{O A}\)

⇒ \(\overrightarrow{O B}=(2 \hat{i}+\hat{j}-3 \hat{k})+(\hat{i}+2 \hat{j}-\hat{k})=(3 \hat{i}+3 \hat{j}-4 \hat{k})\)

Hence, the coordinates of B are (3,3,-4).

Example 8 Find the projection of \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k} \text { on } \vec{b}=\hat{i}-2 \hat{j}+\hat{k}\)

Solution

Projection of \(\vec{a}\) on \(\vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)

= \(\frac{(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+\hat{k})}{\sqrt{1^2+(-2)^2+1^2}}=\frac{(2+2+1)}{\sqrt{6}}=\frac{5}{\sqrt{6}} .\)

Example 9 If \(\hat{a}\) is a unit vector such that \((\vec{x}-\hat{a}) \cdot(\vec{x}+\hat{a})\) = 3, then find |\(\vec{x}\)|.

Solution

Since \(\hat{a}\) is a unit vector, we have |\(\hat{a}\)| = 1.

Now, \((\vec{x}-\hat{a}) \cdot(\vec{x}+\hat{a})\) = 3

⇒ \(\vec{x} \cdot \vec{x}+\vec{x} \cdot \hat{a}-\hat{a} \cdot \vec{x}-\hat{a} \cdot \hat{a}=3\)

⇒ \(|\vec{x}|^2+\vec{x} \cdot \hat{a}-\vec{x} \cdot \hat{a}-|\hat{a}|^2=3\) [∵ \(\hat{a}\).\(\vec{x}\) = \(\vec{x}\).\(\hat{a}\)]

⇒ \(|\vec{x}|^2-1=3\) [∵ \(|\hat{a} \cdot|^2=1\)]

⇒ \(|\vec{x}|^2=4\)

⇒ \(|\vec{x}|=2\) [∵ |\(\vec{x}\)| ≥ 0].

Example 10 If \(\vec{a}\) and \(\vec{b}\) are vectors such that |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 3 and \(\vec{a}\).\(\vec{b}\) = 4, find \(|\vec{a}-\vec{b}|\).

Solution

We have

\(|\vec{a}-\vec{b}|^2=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})\)

= \(\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\) (by distributive law)

= \(|\vec{a}|^2-2 \vec{a} \cdot \vec{b}+|\vec{b}|^2\) [∵ \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{a}\)]

= (22 – 2 x 4 + 32) = 5.

Hence, \(|\vec{a}-\vec{b}|\) = √5.

Example 11 Find a vector whose magnitude is 3 units and which is perpendicular to each of the vectors \(\vec{a}=3 \hat{i}+\hat{j}-4 \hat{k} \text { and } \vec{b}=6 \hat{i}+5 \hat{j}-2 \hat{k} \text {. }\)

Solution

Let the required vector be \(\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\)

Then, \(|\vec{c}|=3 \Leftrightarrow \sqrt{c_1^2+c_2^2+c_3^2}=3 \Leftrightarrow c_1^2+c_2^2+c_3^2=9\) …(1)

Also, \(\vec{c}\) ⊥ \(\vec{a}\) ⇒ \(\vec{c}\).\(\vec{a}\) = 0

⇒ \(\left(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\right) \cdot(6 \hat{i}+5 \hat{j}-2 \hat{k})=0\)

⇒ 3c1 + c2 – 4c3 = 0 …(2)

And, \(\vec{c}\) ⊥ \(\vec{b}\) ⇒ \(\vec{c}\).\(\vec{b}\) = 0

⇒ \(\left(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\right) \cdot(6 \hat{i}+5 \hat{j}-2 \hat{k})=0\)

⇒ 6c1 + 5c2 – 2c3 = 0 …(3)

From (2) and (3), by cross multiplication, we get

\(\frac{c_1}{(-2+20)}=\frac{c_2}{(-24+6)}=\frac{c_3}{(15-6)}=k \text { (say) }\)

⇒ \(\frac{c_1}{18}=\frac{c_2}{-18}=\frac{c_3}{9}=k \Rightarrow \frac{c_1}{2}=\frac{c_2}{-2}=\frac{c_3}{1}=k\)

⇒ c1 = 2k, c2 = -2 and c3 = 1.

Hence, \(\vec{c}=(2 \hat{i}-2 \hat{j}+\hat{k})\) is the required vector.

Example 12 If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) and |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 5, |\(\vec{c}\)| = 7, find the angle between \(\vec{a}\) and \(\vec{b}\).

Solution

Let θ be the angle between \(\vec{a}\) and \(\vec{b}\).

Now, \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)

⇒ \((\vec{a}+\vec{b})=-\vec{c}\)

⇒ \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=(-\vec{c}) \cdot(-\vec{c})\)

⇒ \(\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+2 \vec{a} \cdot \vec{b}=\vec{c} \cdot \vec{c}\) [∵ \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\)]

⇒ \(|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}||\vec{b}| \cos \theta=|\vec{c}|^2\)

⇒ 9 + 25 + 2 x 3 x 5 x cos θ = 49 [∵ |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 5, |\(\vec{c}\)| = 7]

⇒ cos θ = \(\frac{1}{2}\)

⇒ θ = 60°.

Hence, the angle between \(\vec{a}\) and \(\vec{b}\) is 60°.

Example 13 If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are vectors such that |\(\vec{a}\)| = 5, |\(\vec{b}\)| = 4, |\(\vec{c}\)| = 3 and each is perpendicular to the sum of the other two, find \(|\vec{a}+\vec{b}+\vec{c}|\).

Solution

As given, we have

\(\vec{a} \perp(\vec{b}+\vec{c}), \vec{b} \perp(\vec{c}+\vec{a}) \text { and } \vec{c} \perp(\vec{a}+\vec{b})\)

⇒ \(\vec{a} \cdot(\vec{b}+\vec{c})=0, \vec{b} \cdot(\vec{c}+\vec{a})=0 \text { and } \vec{c} \cdot(\vec{a}+\vec{b})=0\)

⇒ \(\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=0, \vec{b} \cdot \vec{c}+\vec{b} \cdot \vec{a}=0 \text { and } \vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}=0\) [by the distributive law]

⇒ \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\) …(1)

[using \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\), etc., and adding]

∴ \(|\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})\)

= \((\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c})+(\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{b} \cdot \vec{c})+(\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+\vec{c} \cdot \vec{c})\)

= \(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\) [using \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\), etc.]

= (25 + 16 + 9 + 0) = 50

[using (1), and |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 4 and |\(\vec{c}\)| = 5]

⇒ \(|\vec{a}+\vec{b}+\vec{c}|=\sqrt{50}=5 \sqrt{2}\)

Hence, \(|\vec{a}+\vec{b}+\vec{c}|=5 \sqrt{2}\).

Example 14 If \(\hat{a}\) and \(\hat{b}\) are unit vectors and θ is the angle between them, prove that \(\sin \frac{\theta}{2}=\frac{1}{2}|\hat{a}-\hat{b}| .\)

Solution

We have

\(|\hat{a}-\hat{b}|^2=(\hat{a}-\hat{b}) \cdot(\hat{a}-\hat{b})\)

= \(\hat{a} \cdot \hat{a}+\hat{b} \cdot \hat{b}-2 \hat{a} \cdot \hat{b}\) [∵ \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\)]

= (1 + 1 – 2 x 1 x 1 x cosθ) [∵ \(\hat{a} \cdot \hat{a}=1 \text { and } \hat{b} \cdot \hat{b}=1\)]

= \(2(1-\cos \theta)=4 \sin ^2 \frac{\theta}{2}\)

∴ \(\sin ^2 \frac{\theta}{2}=\frac{1}{4} \cdot|\hat{a}-\hat{b}|^2 \Rightarrow \sin \frac{\theta}{2}=\frac{1}{2}|\hat{a}-\hat{b}| .\)

Example 15 Let \(\vec{a}\) and \(\vec{b}\) be two nonzero vectors. Prove that \(\vec{a} \perp \vec{b} \Leftrightarrow|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| .\)

Solution

Let \(\vec{a}\) ⊥ \(\vec{b}\). Then, \((\vec{a} \cdot \vec{b})=0\) …(1)

Now, \(|\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})\)

= \(\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)

= \(|\vec{a}|^2+|\vec{b}|^2\) {∵ \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\) = 0}.

Also, \(|\vec{a}-\vec{b}|^2=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})\)

= \(\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)

= \(|\vec{a}|^2+|\vec{b}|^2\) {∵ \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\) = 0}.

Thus, \(|\vec{a}+\vec{b}|^2=|\vec{a}-\vec{b}|^2\), and therefore, \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \text {. }\)

∴ \(\vec{a} \perp \vec{b} \Rightarrow|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| .\)

Conversely, suppose that \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| .\) Then,

\(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \Rightarrow|\vec{a}+\vec{b}|^2=|\vec{a}-\vec{b}|^2\)

⇒ \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})\)

⇒ \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})\)

⇒ \(\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)

⇒ \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a})=0\)

⇒ \(4(\vec{a} \cdot \vec{b})=0\) [∵ \(\vec{b}\).\(\vec{a}\) = \(\vec{a}\).\(\vec{b}\)]

⇒ \(\vec{a}\).\(\vec{b}\) = 0

⇒ \(\vec{a}\) ⊥ \(\vec{b}\).

Thus, \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \Rightarrow \vec{a} \perp \vec{b} .\)

Hence, \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| ⇔ \vec{a} \perp \vec{b} .\)

Example 16 If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are unit vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) then find the value of \((\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\).

Solution

Since \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are unit vectors, we have

|\(\vec{a}\)| = 1, |\(\vec{b}\)| = 1 and |\(\vec{c}\)| = 1.

Now, \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)

⇒ \((\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=0\) [∵ \(\overrightarrow{0}\).\(\overrightarrow{0}\) = 0]

⇒ \(\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\)

⇒ \(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\)

⇒ \((\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-\frac{3}{2}\) [∵ \(|\vec{a}|^2=1,|\vec{b}|^2=1,|\vec{c}|^2=1\)].

Hence, \((\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-\frac{3}{2}\)

Example 17 If \(\vec{a}\).\(\vec{b}\) = \(\vec{a}\).\(\vec{c}\), show that \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\vec{c}\) or \(\vec{a}\) ⊥ \((\vec{b}-\vec{c})\).

Solution

\(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c} \Rightarrow \vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}=0\)

⇒ \(\vec{a} \cdot(\vec{b}-\vec{c})=0\)

⇒ \(\vec{a}=\overrightarrow{0} \text { or }(\vec{b}-\vec{c})=\overrightarrow{0} \text { or } \vec{a} \perp(\vec{b}-\vec{c})\)

⇒ \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\vec{c} \text { or } \vec{a} \perp(\vec{b}-\vec{c}) \text {. }\)

Hence, \(\vec{a}\).\(\vec{b}\) = \(\vec{a}\).\(\vec{c}\) ⇒ \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{a}\) = \(\vec{c}\) or \(\vec{a}\) ⊥ \((\vec{b}-\vec{c}).\)

Example 18 For any vector \(\vec{a}\) in space, show that \((\vec{a} \cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}=\vec{a}\)

Solution

Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\). Then,

\(\vec{a} \cdot \hat{i}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \cdot \hat{i}=a_1(\hat{i} \cdot \hat{i})+a_2(\hat{j} \cdot \hat{i})+a_3(\hat{k} \cdot \hat{i})\)

= a1 [∵ \(\hat{i} \cdot \hat{i}=1, \hat{j} \cdot \hat{i}=0 \text { and } \hat{k} \cdot \hat{i}=0\)].

Thus, \(\vec{a}\).\(\hat{i}\) = a1.

Similarly, \(\vec{a}\).\(\hat{i}\) = a2 and \(\vec{a}\).\(\hat{k}\) = a3.

∴ \((\vec{a} \cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right)=\vec{a} .\)

Hence, \((\vec{a} \cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}=\vec{a} .\)

Example 19 If \(\vec{a}=(2 \hat{i}+2 \hat{j}+3 \hat{k}), \quad \vec{b}=(-\hat{i}+2 \hat{j}+\hat{k}) \text { and } \vec{c}=(3 \hat{i}+\hat{j})\) such that \((\vec{a}+\lambda \vec{b}) \perp \vec{c}\), then find the value of λ.

Solution

We have

\((\vec{a}+\lambda \vec{b})=(2 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})\)

= \((2-\lambda) \hat{i}+2(1+\lambda) \hat{j}+(3+\lambda) \hat{k}\)

Now,

\((\vec{a}+\lambda \vec{b}) \perp \vec{c} \Rightarrow(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0\)

⇒ \(\{(2-\lambda) \hat{i}+2(1+\lambda) \hat{j}+(3+\lambda) \hat{k}\} \cdot(3 \hat{i}+\hat{j}+0 \cdot \hat{k})=0\)

⇒ 3(2-λ) + 2(1+λ) + 0 = 0 ⇒ λ = 8.

Hence, λ = 8.

Example 20 For any two nonzero vectors \(\vec{a}\) and \(\vec{b}\), show that \((a \vec{b}+b \vec{a}) \perp(a \vec{b}-b \vec{a})\)

Solution

We have

\((a \vec{b}+b \vec{a}) \cdot(a \vec{b}-b \vec{a})\)

= \(a \vec{b} \cdot a \vec{b}-a \vec{b} \cdot b \vec{a}+b \vec{a} \cdot a \vec{b}-b^2 \vec{a} \cdot \vec{a}\)

= \(a^2(\vec{b} \cdot \vec{b})-a b(\vec{b} \cdot \vec{a})+b a(\vec{a} \cdot \vec{b})-b^2(\vec{a} \cdot \vec{a})\)

= (a2b2 – b2a2) = 0 [∵ ab = ba and (\(\vec{b}\).\(\vec{a}\)) = (\(\vec{a}\),.\(\vec{b}\))

Hence, \((a \vec{b}+b \vec{a}) \perp(a \vec{b}-b \vec{a})\).

Example 21 If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three mutually perpendicular vectors of the same magnitude, prove that \((\vec{a}+\vec{b}+\vec{c})\) is equally inclined to the vectors, \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).

Solution

\(|\vec{a}|=|\vec{b}|=|\vec{c}|=a\) (say) [given) …(1)

Since \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors, we have

\(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{c}\) = \(\vec{c}\).\(\vec{a}\) = 0 …(2)

Now, \(|\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})\)

= \(\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)

= \(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\) [using(2)]

= 3a2 [using(1)]

∴ \(\vec{a}+\vec{b}+\vec{c}|=\sqrt{3} a\) …(3)

Suppose \((\vec{a}+\vec{b}+\vec{c})\) makes angles θ1, θ2, θ3 with \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) respectively.

Then, \((\vec{a}+\vec{b}+\vec{c}) \cdot \vec{a}=|\vec{a}+\vec{b}+\vec{c}||\vec{a}| \cos \theta_1\)

= \(\left(\sqrt{3} a \cdot a \cdot \cos \theta_1\right)=\sqrt{3} a^2 \cos \theta_1\)

⇔ \((\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{a}+\vec{c} \cdot \vec{a})=\sqrt{3} a^2 \cos \theta_1\)

⇔ \(|\vec{a}|^2=\sqrt{3} a^2 \cos \theta_1 \Leftrightarrow a^2=\sqrt{3} a^2 \cos \theta_1\)

⇔ \(\cos \theta_1=\frac{1}{\sqrt{3}} \Leftrightarrow \theta_1=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) .\)

Similarly, \(\theta_2=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \text { and } \theta_3=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \text {. }\)

∴ \(\theta_1=\theta_2=\theta_3=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \text {. }\)

Hence, \((\vec{a}+\vec{b}+\vec{c})\) is equally inclined to \(\vec{a}\), \(\vec{b}\), \(\vec{c}\).

Example 22 If A(0,1,1), B(3,1,5) and C(0,3,3) be the vertices of a △ABC, show, using vectors, that △ABC is right angled at C.

Solution

Let O be the origin. Then,

\(\overrightarrow{O A}=0 \hat{i}+\hat{j}+\hat{k}\), \(\overrightarrow{O B}=3 \hat{i}+\hat{j}+5 \hat{k} \text { and } \overrightarrow{O C}=0 \hat{i}+3 \hat{j}+3 \hat{k}\)

∴ \(\overrightarrow{C A}=(\overrightarrow{O A}-\overrightarrow{O C})\)

= \((0 \hat{i}+\hat{j}+\hat{k})-(0 \hat{i}+3 \hat{j}+3 \hat{k})\)

= \((0-0) \hat{i}+(1-3) \hat{j}+(1-3) \hat{k}\)

= \((0 \hat{i}-2 \hat{j}-2 \hat{k})\)

∴ \(\overrightarrow{C B}=(\overrightarrow{O B}-\overrightarrow{O C})\)

= \((3 \hat{i}+\hat{j}+5 \hat{k})-(0 \hat{i}+3 \hat{j}+3 \hat{k})=(3 \hat{i}-2 \hat{j}+2 \hat{k})\)

∴ \(\overrightarrow{C A} \cdot \overrightarrow{C B}=(0 \hat{i}-2 \hat{j}-2 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+2 \hat{k})\)

= [(0.3) + (-2)x(-2) + (-2)x2] = 0

⇒ \(\overrightarrow{C A} \perp \overrightarrow{C B}\)

Hence, △ABC is the right angled at C.

Example 23 Show that the point A(2,-1,1), B(1,-3,-5) and C(3,-4,-4) are the vertices of a right-angled triangle. Also, find the remaining angles of the triangle.

Solution

Let O be the origin. Then,

\(\overrightarrow{O A}=(2 \hat{i}-\hat{j}+\hat{k})\), \(\overrightarrow{O B}=(\hat{i}-3 \hat{j}-5 \hat{k}) \text { and } \overrightarrow{O C}=(3 \hat{i}-4 \hat{j}-4 \hat{k})\)

∴ \(\overrightarrow{A B}=(\overrightarrow{O B}-\overrightarrow{O A})=(-\hat{i}-2 \hat{j}-6 \hat{k})\)

\(\overrightarrow{B C}=(\overrightarrow{O C}-\overrightarrow{O B})=(2 \hat{i}-\hat{j}+\hat{k})\) \(\overrightarrow{C A}=(\overrightarrow{O A}-\overrightarrow{O C})=(-\hat{i}+3 \hat{j}+5 \hat{k})\)

∴ \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{0}\)

This shows that A, B, C are the vertices of a triangle.

\(\overrightarrow{B C} \cdot \overrightarrow{C A}=(2 \hat{i}-\hat{j}+\hat{k}) \cdot(-\hat{i}+3 \hat{j}+5 \hat{k})\)

= [2 x (-1) + (-1) x 3 + 1 x 5] = 0.

∴ \(\overrightarrow{B C} \perp \overrightarrow{C A}\) and therefore, ∠C = 90°.

Hence, △ABC is right angled at C.

Now, ∠A us the angle between \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\).

∴ \(\overrightarrow{A B} \cdot \overrightarrow{A C}=|\overrightarrow{A B}||\overrightarrow{A C}| \cos A\)

⇒ \(\cos A=\frac{\overrightarrow{A B} \cdot \overrightarrow{A C}}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}\)

= \(\frac{(-\hat{i}-2 \hat{j}-6 \hat{k}) \cdot(\hat{i}-3 \hat{j}-5 \hat{k})}{\left\{\sqrt{(-1)^2+(-2)^2+(-6)^2}\right\} \cdot\left\{\sqrt{1^2+(-3)^2+(-5)^2}\right\}}\)

[∵ \(\overrightarrow{A C}=-\overrightarrow{C A}=(\hat{i}-3 \hat{j}-5 \hat{k})\)]

= \(\frac{\{(-1) \times 1+(-2) \times(-3)+(-6) \times(-5)\}}{\sqrt{41} \times \sqrt{35}}=\sqrt{\frac{35}{41}}\)

⇒ A = \(\cos ^{-1}\left(\sqrt{\frac{35}{41}}\right)\)

Further, ∠B is the angle between \(\overrightarrow{B C}\) and \(\overrightarrow{B A}\).

∴ \(\overrightarrow{B C} \cdot \overrightarrow{B A}=|\overrightarrow{B C}||\overrightarrow{B A}| \cos B\)

⇒ \(\cos B=\frac{\overrightarrow{B C} \cdot \overrightarrow{B A}}{|\overrightarrow{B C}||\overrightarrow{B A}|}\)

= \(\frac{(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}+6 \hat{k})}{\left\{\sqrt{2^2+(-1)^2+1^2}\right\}\left\{\sqrt{1^2+2^2+6^2}\right\}}\)

[∵ \(\overrightarrow{B A}=-\overrightarrow{A B}=(\hat{i}+2 \hat{j}+6 \hat{k})\)]

= \(\frac{\{2 \times 1+(-1) \times 2+1 \times 6\}}{\sqrt{6} \times \sqrt{41}}=\sqrt{\frac{6}{41}}\)

⇒ B = \(\cos ^{-1}\left(\sqrt{\frac{6}{41}}\right)\)

Hence, A = \(\cos ^{-1}\left(\sqrt{\frac{35}{41}}\right)\) and B = \(\cos ^{-1}\left(\sqrt{\frac{6}{41}}\right)\).

Example 24 Let \((\hat{i}+\hat{j}+\hat{k})\),\((2 \hat{i}+5 \hat{j})\),\((3 \hat{i}+2 \hat{j}-3 \hat{k})\) and \((\hat{i}-6 \hat{j}-\hat{k})\) be the position vectors of points A, B, C, D respectively. Find the angle between AB and CD. Hence, show that AB ∥ CD.

Solution

Let the angle between \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\) be θ.

Now, \(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)

= \((2 \hat{i}+5 \hat{j})-(\hat{i}+\hat{j}+\hat{k})=(\hat{i}+4 \hat{j}-\hat{k})\)

\(\overrightarrow{C D}\) = (p.v. of D) – (p.v. of C)

= \((\hat{i}-6 \hat{j}-\hat{k})-(3 \hat{i}+2 \hat{j}-3 \hat{k})=(-2 \hat{i}-8 \hat{j}+2 \hat{k})\)

∴ \(|\overrightarrow{A B}|=\sqrt{1^2+4^2+(-1)^2}=\sqrt{18}=3 \sqrt{2}\)

\(|\overrightarrow{C D}|=\sqrt{(-2)^2+(-8)^2+2^2}=\sqrt{72}=6 \sqrt{2}\)

Now, \(\overrightarrow{A B} \cdot \overrightarrow{C D}=(\hat{i}+4 \hat{j}-\hat{k}) \cdot(-2 \hat{i}-8 \hat{j}+2 \hat{k})=(-2-32-2)=-36\)

∴ \(\cos \theta=\frac{\overrightarrow{A B} \cdot \overrightarrow{C D}}{|\overrightarrow{A B}||\overrightarrow{C D}|}=\frac{-36}{(3 \sqrt{2} \times 6 \sqrt{2})}=\frac{-36}{36}=-1\)

⇒ θ = π.

Hence, AB ∥ CD.

Examples of Scalar and Vector Products

Example 15 Express the vector \(\vec{a}=(5 \hat{i}-2 \hat{j}+5 \hat{k})\) as sum of two vectors such that one is parallel to the vector \(\vec{b}=(3 \hat{i}+\hat{k})\) and the other is perpendicular to \(\vec{b}\).

Solution

Any vector parallel to \(\vec{b}\) is the form λ\(\vec{b}\) for some scalar λ.

Let \(\vec{a}\) = λ\(\vec{b}\) + \(\vec{c}\), where \(\vec{c}\) ⊥ \(\vec{b}\). Then,

\(\vec{c}=(\vec{a}-\lambda \vec{b}) \perp \vec{b}\)

⇔ \((\vec{a}-\lambda \vec{b}) \cdot \vec{b}=0 \Leftrightarrow(\vec{a} \cdot \vec{b})-\lambda(\vec{b} \cdot \vec{b})=0\)

⇔ \((5 \hat{i}-2 \hat{j}+5 \hat{k}) \cdot(3 \hat{i}+\hat{k})-\lambda(3 \hat{i}+\hat{k}) \cdot(3 \hat{i}+\hat{k})=0\)

⇔ (15-0+5) – λ(9+1) = 0 ⇔ 10λ = 20 ⇔ λ = 2.

∴ \(\lambda \vec{b}=2 \vec{b}=(6 \hat{i}+2 \hat{k})\)

And, \(\vec{c}=(\vec{a}-2 \vec{b})=(5 \hat{i}-2 \hat{j}+5 \hat{k})-2(3 \hat{i}+\hat{k})=(-\hat{i}-2 \hat{j}+3 \hat{k})\)

Hence, the required vectors are \((6 \hat{i}+2 \hat{k}) \text { and }(-\hat{i}-2 \hat{j}+3 \hat{k}) \text {. }\)

Vector (or Cross) Product of Vectors

Vector Product Of Two Vectors Let \(\vec{a}\) and \(\vec{b}\) be two nonzero, nonparallel vectors, and let θ be the angle between them such that 0 < θ < π.

Then, the vector product of \(\vec{a}\) and \(\vec{b}\) is defined as \(\vec{a} \times \vec{b}=(|\vec{a}||\vec{b}| \sin \theta) \hat{n}\),

where \(\hat{n}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\), such that \(\vec{a}\), \(\vec{b}\), \(\hat{n}\) form a right-handed system.

When a right-handed screw is rotated from \(\vec{a}\) to \(\vec{b}\) and it advances along \(\hat{n}\) then the system is said to be right-handed.

Remark 1 If \(\vec{a}\) and \(\vec{b}\) are parallel or collinear, i.e., when θ = 0 or θ = π then we define, \(\vec{a}\) x \(\vec{b}\) = \(\overrightarrow{0}\).

In particular, \(\vec{a}\) x \(\vec{a}\) = \(\overrightarrow{0}\).

Remark 2 If \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\), we define, \(\vec{a}\) x \(\vec{b}\) = \(\overrightarrow{0}\).

Remark 3 For any vector \(\vec{a}\), we have \(\vec{a} \times \vec{a}=(|\vec{a}||\vec{a}| \sin 0) \hat{n}=0 \hat{n}=\overrightarrow{0}\)

Angle Between Two Vectors Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,

\(\vec{a} \times \vec{b}=(a b \sin \theta) \hat{n} \text {, where }|\vec{a}|=a \text { and }|\vec{b}|=b\)

⇒ \(|\vec{a} \times \vec{b}|=a b \sin \theta\)

⇒ \(\sin \theta=\frac{|\vec{a} \times \vec{b}|}{a b}=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\)

⇒ \(\theta=\sin ^{-1}\left\{\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\right\} .\)

Unit Vector Perpendicular To Two Given Vectors A unit vector \(\hat{n}\) perpendicular to each one of \(\vec{a}\) and \(\vec{b}\) is given by

\(\hat{n}=\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}\)

Properties of Vector Product

Result 1 Vector product is not commutative.

In fact, we have \((\vec{b} \times \vec{a})=-(\vec{a} \times \vec{b})\)

Proof Let \(\vec{a}\), \(\vec{b}\), \(\hat{n}\) from a right-handed system. Then,

\((\vec{a} \times \vec{b})=(a b \sin \theta) \hat{n}\) …(1)

Then, \(\vec{b}\), \(\vec{a}\) , –\(\hat{n}\) form a right-handed system.

∴ \((\vec{b} \times \vec{a})=(b a \sin \theta)(-\hat{n})\)

⇒ \(-(\vec{b} \times \vec{a})=(a b \sin \theta) \hat{n}\) …(2)

From (1) and (2), we get \((\vec{a} \times \vec{b})=-(\vec{b} \times \vec{a})\)

Result 2 For any two vectors \(\vec{a}\) and \(\vec{b}\), prove that \((-\vec{b}) \times \vec{a}=(\vec{a} \times \vec{b})=\vec{b} \times(-\vec{a})\).

Proof

Let θ be the angle between \(\vec{a}\) and \(\vec{b}\).

Then, the angle between (-\(\vec{b}\)) is \(\vec{a}\) is (π – θ).

And, (-\(\vec{b}\)), \(\vec{a}\), \(\hat{n}\) form a right-handed system.

∴ \((-\vec{b}) \times \vec{a}=|-\vec{b}||\vec{a}| \sin (\pi-\theta) \hat{n}\)

= \((|\vec{b}||\vec{a}| \sin \theta) \hat{n}=(a b \sin \theta) \hat{n}=(\vec{a} \times \vec{b})\)

∴ \((-\vec{b}) \times \vec{a}=(\vec{a} \times \vec{b}) .\)

Similarly, \(\vec{b} \times(-\vec{a})=(\vec{a} \times \vec{b})\)

Hence, \((-\vec{b}) \times \vec{a}=(\vec{a} \times \vec{b})=\vec{b} \times(-\vec{a})\)

Result 3 For any scalar m, we have \((m \vec{a} \times \vec{b})=m(\vec{a} \times \vec{b})=\vec{a} \times(m \vec{b})\)

An Important Note At a later stage, we shall prove that for any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), we have \(\vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{b} \times \vec{c}\), i.e., the position of dot and cross can be interchanged. However, we shall use this fact in proving the following theorem.

Result 4 (Distributive law) For any vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), prove that \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)

Proof

Let \(\vec{p}=[\vec{a} \times(\vec{b}+\vec{c})-(\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})]\), and let \(\vec{r}\) be any arbitraty vector.

Then,

\(\vec{r} \cdot \vec{p}=\vec{r} \cdot[\vec{a} \times(\vec{b}+\vec{c})-(\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})]\)

= \(\vec{r} \cdot[\vec{a} \times(\vec{b}+\vec{c})]-\vec{r} \cdot(\vec{a} \times \vec{b})-\vec{r} \cdot(\vec{a} \times \vec{c})\)

= \((\vec{r} \times \vec{a}) \cdot(\vec{b}+\vec{c})-(\vec{r} \times \vec{a}) \cdot \vec{b}-(\vec{r} \times \vec{a}) \cdot \vec{c}\)

= \((\vec{r} \times \vec{a}) \cdot \vec{b}+(\vec{r} \times \vec{a}) \cdot \vec{c}-(\vec{r} \times \vec{a}) \cdot \vec{b}-(\vec{r} \times \vec{a}) \cdot \vec{c}\)

= 0.

Now, \(\vec{r}\).\(\vec{p}\) = 0 ⇒ \(\vec{r}\) = \(\overrightarrow{0}\) or \(\vec{r}\) = \(\overrightarrow{0}\) or \((\vec{r} ⊥ \vec{p})\).

Since \(\vec{r}\) is an arbitraty vector, we may choose it in such a way that \(\vec{r}\) ≠ \(\overrightarrow{0}\) and \(\vec{r}\) is not perpendicular to \(\vec{p}\),.

Then, \(\vec{p}\) = \(\overrightarrow{0}\)

⇒ \(\vec{a} \times(\vec{b}+\vec{c})-(\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})=\overrightarrow{0}\)

⇒ \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)

Hence, \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\).

Result 5 For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), prove that \(\vec{a} \times(\vec{b}-\vec{c})=(\vec{a} \times \vec{b})-(\vec{a} \times \vec{c}) .\)

Proof

We have

\(\vec{a} \times(\vec{b}-\vec{c})=\vec{a} \times[\vec{b}+(-\vec{c})]\)

= \((\vec{a} \times \vec{b})+\vec{a} \times(-\vec{c})\) [by the distributive law]

= \((\vec{a} \times \vec{b})+[-(\vec{a} \times \vec{c})]\) [∵ \(\vec{a} \times(-\vec{c})=-(\vec{a} \times \vec{c})\)]

= \((\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})\)

Hence, \(\vec{a} \times(\vec{b}-\vec{c})=(\vec{a} \times \vec{b})-(\vec{a} \times \vec{c}) .\)

Result Prove that two nonzero vectors \(\vec{a}\)and \(\vec{b}\) are parallel or collinear if and only if \((\vec{a} \times \vec{b})=\overrightarrow{0}\)

Proof

Let \(\vec{a}\) ≠ \(\overrightarrow{0}\) and \(\vec{b}\) ≠ \(\overrightarrow{0}\) and let θ be the angle between them.

Let \(\vec{a}\) and \(\vec{b}\) be parallel or collinear.

Then, θ = 0 or θ = π

⇒ sin θ = 0 ⇒ \((\vec{a} \times \vec{b})=(a b \sin \theta) \hat{n}=0 \hat{n}=\overrightarrow{0} .\)

Thus, when \(\vec{a}\) and \(\vec{b}\) are parallel or collinear, then \((\vec{a} \times \vec{b})=\overrightarrow{0}\).

Again, let \(\vec{a}\) ≠ \(\overrightarrow{0}\), \(\vec{b}\) ≠ \(\overrightarrow{0}\) and \((\vec{a} \times \vec{b})=\overrightarrow{0}\). Then

\((\vec{a} \times \vec{b})=\overrightarrow{0}\) ⇒ \(|\vec{a} \times \vec{b}|=0\)

⇒ ab sinθ = 0

⇒ sin θ = 0 [a ≠ 0 and b ≠ 0]

⇒ θ = 0 or θ = π

⇒ (\(vec{a}\) ∥ \(\vec{b}\)) or (\(\vec{a}\) and \(\vec{b}\) are collinear).

Result 7 Prove that

\((\vec{a} \times \vec{b})=\overrightarrow{0} \Rightarrow \vec{a}=\overrightarrow{0} \text {, or } \vec{b}=\overrightarrow{0} \text {, or } \vec{a} \| \vec{b} \text { or } \vec{a} \text { and } \vec{b}\) are collinear.

Proof

Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,

\(\vec{a} \times \vec{b}=\overrightarrow{0} \Rightarrow(a b \sin \theta) \hat{n}=\overrightarrow{0}\)

⇒ ab sinθ = 0

⇒ a = 0, or b = 0, or sin θ = 0

⇒ \(\vec{a}\) = \(\overrightarrow{0}\), or \(\vec{b}\) = \(\overrightarrow{0}\), or θ = 0 or θ = π

⇒ \(\vec{a}\) = \(\overrightarrow{0}\), or \(\vec{b}\) = \(\overrightarrow{0}\), or \(\vec{a}\) and \(\vec{b}\) are parallel or collinear.

Corollary Show that \(\vec{a}\) x \(\vec{b}\) = \(\vec{a}\) x \(\vec{c}\) does not imply that \(\vec{b}\) = \(\vec{c}\).

Proof

\(\vec{a} \times \vec{b}=\vec{a} \times \vec{c} \Rightarrow \vec{a} \times \vec{b}-\vec{a} \times \vec{c}=\overrightarrow{0}\)

⇒ \(\vec{a} \times(\vec{b}-\vec{c})=\overrightarrow{0}\)

⇒ \(\vec{a}=\overrightarrow{0}, \text { or }(\vec{b}-\vec{c})=0 \text {, or } \vec{a} \|(\vec{b}-\vec{c})\)

⇒ \(\vec{a}=\overrightarrow{0}, \text { or }(\vec{b}-\vec{c})=0 \text {, or } \vec{a} \|(\vec{b}-\vec{c})\)

∴ \(\vec{a} \times \vec{b}=\vec{a} \times \vec{c}\) does not always mean that \(\vec{b}\) = \(\vec{c}\).

Summary

(1) \((\vec{a} \times \vec{b})=-(\vec{b} \times \vec{a})\)

(2) \((-\vec{b}) \times \vec{a}=(\vec{a} \times \vec{b})=\vec{b} \times(-\vec{a})\)

(3) \(m \vec{a} \times \vec{b}=m(\vec{a} \times \vec{b})=\vec{a} \times(m \vec{b})\)

(4) \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)

(5) \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)

(6) \(\vec{a} \times \vec{b}=\overrightarrow{0} \Leftrightarrow(\vec{a} \| \vec{b})\) or (\(\vec{a}\) and \(\vec{b}\) are collinear), when \(\vec{a}\) ≠ \(\overrightarrow{0}\) and \(\vec{b}\) ≠ \(\overrightarrow{0}\)

Area of a Parallelogram

Theorem 1 Two adjacent sides of a ∥gm are represented by \(\vec{a}\) and \(\vec{b}\) respectively. Prove that the area of the ∥gm = \(|\vec{a} \times \vec{b}|\).

Proof

Let ABCD be a ∥gm. Join BD.

Let \(\overrightarrow{A B}\) = \(\vec{a}\), \(\overrightarrow{A D}\) = \(\vec{b}\) and ∠BAD = θ.

Draw DL ⊥ AB. Then,

AB = a and DL = AD sin θ = b sin θ.

∴ ar(△ABD) = \(\frac{1}{2}\) x base x altitude

= \(\left(\frac{1}{2} \times A B \times D L\right)=\frac{1}{2} a b \sin \theta=\frac{1}{2}|\vec{a} \times \vec{b}|\)

∴ ar(∥gm ABCD) = 2 x ar(△ABD) = \(|\vec{a} \times \vec{b}|\).

Remark (\(\vec{a}\) x \(\vec{b}\)) is called the vector area of the ∥gm.

Area of a Triangle

Theorem 2 Prove that the area of △ABC, where \(\overrightarrow{A B}\) = \(\vec{a}\) and \(\overrightarrow{A C}\) = \(\vec{b}\), is \(\frac{1}{2}|\vec{a} \times \vec{b}|\).

Proof

In △ABC, let \(\overrightarrow{A B}\) = \(\vec{a}\) and \(\overrightarrow{A C}\) = \(\vec{b}\) and ∠BAC = θ.

Draw CL ⊥ AB. Then,

AB = a and CL = (AC)sin θ = b sin θ

and ar(△ABC) = \(\frac{1}{2}\) x AB x CL

= \(\frac{1}{2} a b \sin \theta=\frac{1}{2}|\vec{a} \times \vec{b}|\)

∴ ar(△ABC) = \(\frac{1}{2}|\vec{a} \times \vec{b}|\).

Remark \(\frac{1}{2}(\vec{a} \times \vec{b})\) is called the vector area of △ABC.

Area of a Quadrilateral

Theorem 3 Prove that the area of a quadrilateral ABCD with diagonals AC and BD is \(\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}|\).

Proof

Vector area of quad. ABCD

= (vector area of △ABC) + (vector area of △ACD)

= \(\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{A C})+\frac{1}{2}(\overrightarrow{A C} \times \overrightarrow{A D})\)

= \(-\frac{1}{2}(\overrightarrow{A C} \times \overrightarrow{A B})+\frac{1}{2}(\overrightarrow{A C} \times \overrightarrow{A D})\)

= \(\frac{1}{2} \cdot\{\overrightarrow{A C} \times(\overrightarrow{A D}-\overrightarrow{A B})\}=\frac{1}{2}(\overrightarrow{A C} \times \overrightarrow{B D})\)

∴ ar(quad.ABCD) = \(\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}|\).

Summary

(1) ar(∥gm ABCD) = \(|\vec{a} \times \vec{b}|\), where \(\overrightarrow{A B}\) = \(\vec{a}\) and \(\overrightarrow{A D}\) = \(\vec{b}\).

(2) ar(△ABC) = \(\frac{1}{2}|\vec{a} \times \vec{b}|\), where \(\overrightarrow{A B}\) = \(\vec{a}\) and \(\overrightarrow{A C}\) = \(\vec{b}\).

(3) ar(quad. ABCD) = \(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B D}|\), where AC and BD are its diagonals.

Vector Product Of An Orthonormal Vector Triad

For mutually perpendicular unit vectors \(\hat{i}, \hat{j}, \hat{k}\) we have

\(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\hat{0}\) \(\hat{i} \times \hat{j}=\hat{k}=-\hat{j} \times \hat{i}, \hat{j} \times \hat{k}=\hat{i}=-\hat{k} \times \hat{j} \text { and } \hat{k} \times \hat{i}=\hat{j}=-\hat{i} \times \hat{k}\)

Vector Product in Terms of Components

Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \text { and } \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\). Then,

\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|\)

Proof

We have

\(\vec{a} \times \vec{b}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \times\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\)

= \(a_1 b_1(\hat{i} \times \hat{i})+a_1 b_2(\hat{i} \times \hat{j})+a_1 b_3(\hat{i} \times \hat{k})\)

\(+a_2 b_1(\hat{j} \times \hat{i})+a_2 b_2(\hat{j} \times \hat{j})+a_2 b_3(\hat{j} \times \hat{k})\) \(+a_3 b_1(\hat{k} \times \hat{i})+a_3 b_2(\hat{k} \times \hat{j})+a_3 b_3(\hat{k} \times \hat{k})\)

= \(\left(a_2 b_3-a_3 b_2\right) \hat{i}+\left(a_3 b_1-a_1 b_3\right) \hat{j}+\left(a_1 b_2-a_2 b_1\right) \hat{k}\)

[∵ \(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\hat{0}, \hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i}, \hat{k} \times \hat{i}=\hat{j}\text { and }\)

\(\hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{j}=-\hat{i} \text { and } \hat{i} \times \hat{k}=-\hat{j}\)]

= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|\)

Solved Examples

Example 1 If \(\vec{a}=(3 \hat{i}+\hat{j}-4 \hat{k}) \text { and } \vec{b}=(6 \hat{i}+5 \hat{j}-2 \hat{k}) \text {, find }(\vec{a} \times \vec{b}) \text { and }|\vec{a} \times \vec{b}|\)

Solution

We have

\((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & -4 \\
6 & 5 & -2
\end{array}\right|\)

= \((-2+20) \hat{i}-(-6+24) \hat{j}+(15-6) \hat{k}\)

= \((18 \hat{i}-18 \hat{j}+9 \hat{k})\)

\(|\vec{a} \times \vec{b}|^2=\left\{(18)^2+(-18)^2+9^2\right\}=729\)

⇒ \(|\vec{a} \times \vec{b}|=\sqrt{729}=27 \text {. }\)

Example 2 Find the values of λ and μ for which \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}\)

Solution

Let \(\vec{a}=(2 \hat{i}+6 \hat{j}+27 \hat{k}) \text { and } \vec{b}=\hat{i}+\lambda \hat{j}+\mu \hat{k}\). Then,

\((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 27 \\
1 & \lambda & \mu
\end{array}\right|\)

= \((6 \mu-27 \lambda) \hat{i}-(2 \mu-27) \hat{j}+(2 \lambda-6) \hat{k} .\)

Now, \((\vec{a} \times \vec{b})=\overrightarrow{0}\)

⇔ 2λ – 6 = 0 and 2μ – 27 = 0

⇔ λ = 3 and μ = \(\frac{27}{2}\).

Hence, λ = 3 and μ = \(\frac{27}{2}\).

Example 3 If \(\vec{a}=(\hat{i}-2 \hat{j}+3 \hat{k}) \text { and } \vec{b}=(2 \hat{i}+3 \hat{j}-5 \hat{k})\), then find \((\vec{a} \times \vec{b})\) and verify that \((\vec{a} \times \vec{b})\) is perpendicular to each one of \(\vec{a}\) and \(\vec{b}\).

Solution

We have

\((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
2 & 3 & -5
\end{array}\right|\)

= \((10-9) \hat{i}-(-5-6) \hat{j}+(3+4) \hat{k}=(\hat{i}+11 \hat{j}+7 \hat{k})\)

Now, \((\vec{a} \times \vec{b}) \cdot \vec{a}=(\hat{i}+11 \hat{j}+7 \hat{k}) \cdot(\hat{i}-2 \hat{j}+3 \hat{k})\)

∴ \((\vec{a} \times \vec{b}) \cdot \vec{a}\) = (1 – 22 + 21) = 0

∴ \((\vec{a} \times \vec{b}) \perp \vec{a} .\)

And, \((\vec{a} \times \vec{b}) \cdot \vec{b}=(\hat{i}+11 \hat{j}+7 \hat{k}) \cdot(2 \hat{i}+3 \hat{j}-5 \hat{k})\) = (2 + 33 – 35) = 0.

∴ \((\vec{a} \times \vec{b}) \perp \vec{b}\)

Example 4 Find a unit vector perpendicular to each one of the vectors \(\vec{a}=(4 \hat{i}-\hat{j}+3 \hat{k}) \text { and } \vec{b}=(2 \hat{i}+2 \hat{j}-\hat{k})\)

Solution

We know that \((\vec{a} \times \vec{b})\) is a vector perpendicular to each one of \(\vec{a}\) and \(\vec{b}\). So, the required vector is \(\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}\).

Now, \((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
4 & -1 & 3 \\
2 & 2 & -1
\end{array}\right|\)

= \((1-6) \hat{i}-(-4-6) \hat{j}+(8+2) \hat{k}\)

= \((-5 \hat{i}+10 \hat{j}+10 \hat{k})\)

\(|\vec{a} \times \vec{b}|=\sqrt{(-5)^2+(10)^2+(10)^2}=\sqrt{225}=15\)

Hence, the required unit vector = \(\frac{(-5 \hat{i}+10 \hat{j}+10 \hat{k})}{15}\)

= \(\frac{1}{3}(-\hat{i}+2 \hat{j}+2 \hat{k}) .\)

Example 5 Find a vector of magnitude 15, which is perpendicular to both the vectors \((4 \hat{i}-\hat{j}+8 \hat{k}) \text { and }(-\hat{j}+\hat{k})\)

Solution

Let \(\vec{a}=(4 \hat{i}-\hat{j}+8 \hat{k}) \text { and } \vec{b}=(-\hat{j}+\hat{k})\)

A unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) = \(\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}\).

Now, \(\vec{a} \times \vec{b}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
4 & -1 & 8 \\
0 & -1 & 1
\end{array}\right|\)

= \((-1+8) \hat{i}-(4-0) \hat{j}+(-4-0) \hat{k}\)

= \((7 \hat{i}-4 \hat{j}-4 \hat{k})\)

∴ \(|\vec{a} \times \vec{b}|=\sqrt{7^2+(-4)^2+(-4)^2}=\sqrt{81}=9 .\)

So, a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\)

= \(|\vec{a} \times \vec{b}|=\sqrt{7^2+(-4)^2+(-4)^2}=\sqrt{81}=9 .\)

The required vector = \(\frac{15(7 \hat{i}-4 \hat{j}-4 \hat{k})}{9}=\frac{5}{3}(7 \hat{i}-4 \hat{j}-4 \hat{k})\)

Example 6 If \(|\vec{a}|=2,|\vec{b}|=7 \text { and }(\vec{a} \times \vec{b})=(3 \hat{i}+2 \hat{j}+6 \hat{k})\), find the angle between \(\vec{a}\) and \(\vec{b}\).

Solution

Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,

\(\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}\)

⇒ \(|\vec{a} \times \vec{b}|=\sqrt{3^2+2^2+6^2}=\sqrt{49}=7\)

⇒ \(|\vec{a}||\vec{b}| \sin \theta=7\) [∵ \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta\)]

⇒ \(\sin \theta=\frac{7}{|\vec{a}||\vec{b}|}=\frac{7}{(2 \times 7)}=\frac{1}{2}\)

⇒ θ = \(\frac{\pi}{6}\).

Hence, the required angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\).

Example 7 If |\(\vec{a}\)| = √26, |\(\vec{b}\)| = 7 and \(|\vec{a} \times \vec{b}|\) = 35, find \(\vec{a}\).\(\vec{b}\)

Solution

Given that |\(\vec{a}\)| = √26, |\(\vec{b}\)| = 7 and \(|\vec{a} \times \vec{b}|\) = 35.

∴ \(|\vec{a} \times \vec{b}|=35 \Rightarrow|\vec{a}||\vec{b}| \sin \theta=35\)

⇒ \(\sin \theta=\frac{35}{|\vec{a}||\vec{b}|}=\frac{35}{(\sqrt{26}) \times 7}=\frac{5}{\sqrt{26}}\)

Now, \(\cos \theta=\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{25}{26}}=\frac{1}{\sqrt{26}} .\)

∴ \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta=\left(\sqrt{26} \times 7 \times \frac{1}{\sqrt{26}}\right)=7 .\)

Hence, \(\vec{a}\).\(\vec{b}\) = 7.

Example 8 If \(\vec{a}=4 \hat{i}+3 \hat{j}+2 \hat{k} \text { and } \vec{b}=3 \hat{i}+2 \hat{k} \text {, find }|\vec{b} \times 2 \vec{a}| \text {. }\)

Solution

We have

\(\vec{b}=(3 \hat{i}+2 \hat{k}) \text { and } 2 \vec{a}=(8 \hat{i}+6 \hat{j}+4 \hat{k}) \text {. }\)

∴ \((\vec{b} \times 2 \vec{a})=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 0 & 2 \\
8 & 6 & 4
\end{array}\right|\)

= \((0-12) \hat{i}+(16-12) \hat{j}+(18-0) \hat{k}\)

= \((-12 \hat{i}+4 \hat{j}+18 \hat{k})\)

∴ \(|\vec{b} \times 2 \vec{a}|=|-12 \hat{i}+4 \hat{j}+18 \hat{k}|\)

= \(\sqrt{(-12)^2+4^2+(18)^2}=\sqrt{484}=22\)

Hence, \(|\vec{b} \times 2 \vec{a}|\) = 22.

Example 9 Find the sine of the angle between the vectors \(\vec{a}=(2 \hat{i}-\hat{j}+3 \hat{k})\) and \(\vec{b}=(\hat{i}+3 \hat{j}+2 \hat{k})\).

Solution

We have

\((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 3 \\
1 & 3 & 2
\end{array}\right|\)

= \((-2-9) \hat{i}-(4-3) \hat{j}+(6+1) \hat{k}=(-11 \hat{i}-\hat{j}+7 \hat{k})\)

\(|\vec{a} \times \vec{b}|=\sqrt{(-11)^2+(-1)^2+7^2}=\sqrt{171}=3 \sqrt{19}\) \(|\vec{a}|=\sqrt{2^2+(-1)^2+3^2}=\sqrt{14}\) \(|\vec{b}|=\sqrt{1^2+3^2+2^2}=\sqrt{14}\)

Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,

\(\sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}=\frac{3 \sqrt{19}}{(\sqrt{14})(\sqrt{14})}=\frac{3}{14} \sqrt{19} .\)

Example 10 Find the area of the parallelogram whose adjacent sides are represented by the vectors \((3 \hat{i}+\hat{j}-2 \hat{k}) \text { and }(\hat{i}-3 \hat{j}+4 \hat{k})\)

Solution

Let \(\vec{a}=(3 \hat{i}+\hat{j}-2 \hat{k}) \text { and } \vec{b}=(\hat{i}-3 \hat{j}+4 \hat{k})\)

Then, vector area of the ∥gm is \((\vec{a} \times \vec{b})\).

Now, \((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & -2 \\
1 & -3 & 4
\end{array}\right|\)

= \((4-6) \hat{i}-(12+2) \hat{j}+(-9-1) \hat{k}\)

= \((-2 \hat{i}-14 \hat{j}-10 \hat{k})\)

Required area = \(|\vec{a} \times \vec{b}|\)

= \(\sqrt{(-2)^2+(-14)^2+(-10)^2}\) sq units.

= \(\sqrt{300} \text { sq units }=10 \sqrt{3} \text { sq units. }\)

Example 11 Find the area of the parallelogram whose diagonals are represented by the vectors \(\vec{d}_1=(2 \hat{i}-\hat{j}+\hat{k}) \text { and } \vec{d}_2=(3 \hat{i}+4 \hat{j}-\hat{k}) \text {. }\)

Solution

Given that \(\vec{d}_1=(2 \hat{i}-\hat{j}+\hat{k}) \text { and } \vec{d}_2=(3 \hat{i}+4 \hat{j}-\hat{k}) \text {. }\)

Vector area of the ∥gm is \(\frac{1}{2}\left(\vec{d}_1 \times \vec{d}_2\right)\)

Now, \(\left(\vec{d}_1 \times \vec{d}_2\right)=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
3 & 4 & -1
\end{array}\right|\)

= \((1-4) \hat{i}-(-2-3) \hat{j}+(8+3) \hat{k}\)

= \((-3 \hat{i}+5 \hat{j}+11 \hat{k})\)

Required area = \(\frac{1}{2}\left|\overrightarrow{d_1} \times \overrightarrow{d_2}\right|\)

= \(\frac{1}{2} \sqrt{(-3)^2+5^2+(11)^2}\) sq units

= \(\frac{1}{2} \sqrt{155}\) sq units.

Example 12 Find the area of the triangle whose adjacent sides are determined by the vectors \(\vec{a}=(-2 \hat{i}-5 \hat{k}) \text { and } \vec{b}=(\hat{i}-2 \hat{j}-\hat{k})\)

Solution

Two adjacent sides of the given triangle are represented by the vectors \(\vec{a}=(-2 \hat{i}-5 \hat{k}) \text { and } \vec{b}=(\hat{i}-2 \hat{j}-\hat{k})\)

So, the area of the triangle is \(\frac{1}{2}|\vec{a} \times \vec{b}|\).

Now, \((\vec{a} \times \vec{b})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 0 & -5 \\
1 & -2 & -1
\end{array}\right|\)

= \((0-10) \hat{i}-(2+5) \hat{j}+(4-0) \hat{k}\)

= \((-10 \hat{i}-7 \hat{j}+4 \hat{k})\)

∴ required area = \(\frac{1}{2}|\vec{a} \times \vec{b}|\)

= \(\frac{1}{2} \cdot\left\{\sqrt{(-10)^2+(-7)^2+4^2}\right\}\) sq units

= \(\frac{1}{2} \sqrt{165}\) sq units.

Example 13 Using vector method, find the area of the triangle whose vertices are A(1,1,1), B(1,2,3) and C(2,3,1).

Solution

Position vector of A = \((\hat{i}+\hat{j}+\hat{k})\);

position vector of B = \((\hat{i}+2 \hat{j}+3 \hat{k})\); and

position vector of C = \((2 \hat{i}+3 \hat{j}+\hat{k})\).

∴ \(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)

= \((\hat{i}+2 \hat{j}+3 \hat{k})-(\hat{i}+\hat{j}+\hat{k})=(\hat{j}+2 \hat{k}) .\)

And, \(\overrightarrow{A C}\) = (p.v. of C) – (p.v. of A)

= \((2 \hat{i}+3 \hat{j}+\hat{k})-(\hat{i}+\hat{j}+\hat{k})=(\hat{i}+2 \hat{j})\)

∴ area of △ABC = \(\left|\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{A C})\right|\).

Now, \(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 1 & 2 \\
1 & 2 & 0
\end{array}\right|\)

= \((0-4) \hat{i}+(2-0) \hat{j}+(0-1) \hat{k}=(-4 \hat{i}+2 \hat{j}-\hat{k})\)

⇒ \(\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{A C})=\left(-2 \hat{i}+\hat{j}-\frac{1}{2} \hat{k}\right)\)

⇒ area of △ABC = \(\left|-2 \hat{i}+\hat{j}-\frac{1}{2} \hat{k}\right|\)

= \(\sqrt{(-2)^2+1^2+\left(-\frac{1}{2}\right)^2}=\frac{\sqrt{21}}{2}\) sq units.

Hence, the area of the given triangle is \(\frac{\sqrt{21}}{2}\) sq units.

Example 14 Show that the points whose vectors are \((5 \hat{i}+6 \hat{j}+7 \hat{k}), (7 \hat{i}-8 \hat{j}+9 \hat{k}) \text { and }(3 \hat{i}+20 \hat{j}+5 \hat{k})\) are collinear.

Solution

Let the given points be A, B, C. Then,

\(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)

= \((7 \hat{i}-8 \hat{j}+9 \hat{k})-(5 \hat{i}+6 \hat{j}+7 \hat{k})=(2 \hat{i}-14 \hat{j}+2 \hat{k})\).

And, \(\overrightarrow{AC}\) = (p.v. of C) – (p.v. of A)

= \((3 \hat{i}+20 \hat{j}+5 \hat{k})-(5 \hat{i}+6 \hat{j}+7 \hat{k})\)

= \((-2 \hat{i}+14 \hat{j}-2 \hat{k})\).

∴ \(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -14 & 2 \\
-2 & 14 & -2
\end{array}\right|\)

= \(2 \times(-2) \cdot\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -7 & 1 \\
1 & -7 & 1
\end{array}\right|=(-4 \times 0)=\overrightarrow{0}\)

[∵ R2 are R3 are identical]

Thus, \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) are parallel vectors, having a common point, A. Hence, the point A, B, C are collinear.

Example 15 Using vector method, show that the points A(2,-1,3), B(4,3,1) and C(3,1,2) are collinear.

Solution

We know, the points A, B, C are collinear ⇔ area of △ABC = 0.

Now, p.v. of A = \((2 \hat{i}-\hat{j}+3 \hat{k})\), p.v. of B = \((4 \hat{i}+3 \hat{j}+\hat{k})\), and p.v. of C = \((3 \hat{i}+\hat{j}+2 \hat{k})\).

\(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)

= \(4 \hat{i}+3 \hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})=(2 \hat{i}+4 \hat{j}-2 \hat{k})\)

And, \(\overrightarrow{A C}\) = (p.v. of C) – (p.v. of A)

= \((3 \hat{i}+\hat{j}+2 \hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})=(\hat{i}+2 \hat{j}-\hat{k})\)

∴ \(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & 4 & -2 \\
1 & 2 & -1
\end{array}\right|=2 \cdot\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -1 \\
1 & 2 & -1
\end{array}\right|=\overrightarrow{0}\)

[∵ R2 and R3 are identical].

Thus, \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) are parallel vectors, having a common point, A. Hence, the points A, B, C are collinear.

Example 15 Show that the points having position vectors \((\vec{a}-2 \vec{b}+3 \vec{c}),(-2 \vec{a}+3 \vec{b}+2 \vec{c}),(-8 \vec{a}+13 \vec{b})\) are collinear, whatever be \(\vec{a}\), \(\vec{b}\), \(\vec{c}\).

Solution

Let A, B, C be the given points whose position vectors are \((\vec{a}-2 \vec{b}+3 \vec{c}),(-2 \vec{a}+3 \vec{b}+2 \vec{c})\) and \((-8 \vec{a}+13 \vec{b})\) respectively.

Then,

\(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)

= \((-2 \vec{a}+3 \vec{b}+2 \vec{c})-(\vec{a}-2 \vec{b}+3 \vec{c})=(-3 \vec{a}+5 \vec{b}-\vec{c})\),

and \(\overrightarrow{A C}\) = (p.v. of C) – (p.v. of A)

= \((-8 \vec{a}+13 \vec{b})-(\vec{a}-2 \vec{b}+3 \vec{c})=(-9 \vec{a}+15 \vec{b}-3 \vec{c})\)

∴ \(\overrightarrow{A B} \times \overrightarrow{A C}=(-3 \vec{a}+5 \vec{b}-\vec{c}) \times(-9 \vec{a}+15 \vec{b}-3 \vec{c})\)

= \(\vec{d} \times 3 \vec{d} \text {, where } \vec{d}=-3 \vec{a}+5 \vec{b}-\vec{c}\)

= \(\overrightarrow{0}\) [∵ \(\vec{d} \times \vec{d}=\overrightarrow{0}\)].

∴ \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) are parallel vectors, a common point, A. Hence, the points A, B, C are collinear.

Lagrange’s Identity

Example 17 Prove that \(|\vec{a} \times \vec{b}|^2=\left|\begin{array}{ll}
\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\
\vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b}
\end{array}\right| .\)

Solution

Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then,

\(|\vec{a} \times \vec{b}|^2=(\vec{a} \times \vec{b}) \cdot(\vec{a} \times \vec{b})\)

= \((a b \sin \theta) \hat{n} \cdot(a b \sin \theta) \hat{n}\)

= \(\left(a^2 b^2 \sin ^2 \theta\right)(\hat{n} \cdot \hat{n})=a^2 b^2 \sin ^2 \theta\)

= \(a^2 b^2\left(1-\cos ^2 \theta\right)=a^2 b^2-(a b \cos \theta)^2\)

= \((\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b})-(\vec{a} \cdot \vec{b})^2\)

= \(\left|\begin{array}{ll}
\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\
\vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b}
\end{array}\right|\)

Hence, \(|\vec{a} \times \vec{b}|^2=\left|\begin{array}{ll}
\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\
\vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b}
\end{array}\right| .\)

Example 18 Prove that \(\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})=\overrightarrow{0} .\)

Solution

We have

\(\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})\)

= \((\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})+(\vec{b} \times \vec{c})+(\vec{b} \times \vec{a})+(\vec{c} \times \vec{a})+(\vec{c} \times \vec{b})\)

[by the distributive law]

= \((\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})-(\vec{a} \times \vec{b})-(\vec{b} \times \vec{c})-(\vec{c} \times \vec{a})\)

= \(\overrightarrow{0}\)

[∵ \((\vec{b} \times \vec{a})=-(\vec{a} \times \vec{b}),(\vec{c} \times \vec{b})=-(\vec{b} \times \vec{c}) \text { and }(\vec{a} \times \vec{c})=-(\vec{c} \times \vec{a})\)].

Hence, \(\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})=\overrightarrow{0} .\)

Example 19 If \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), prove that \((\vec{a} \times \vec{b})=(\vec{b} \times \vec{c})=(\vec{c} \times \vec{a})\).

Solution

\(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \Rightarrow \vec{a}+\vec{b}=-\vec{c}\)

⇒ \((\vec{a}+\vec{b}) \times \vec{b}=(-\vec{c}) \times \vec{b}\)

⇒ \((\vec{a} \times \vec{b})+(\vec{b} \times \vec{b})=(-\vec{c}) \times \vec{b}\)

[by the distributive law]

⇒ \((\vec{a} \times \vec{b})+\overrightarrow{0}=(\vec{b} \times \vec{c})\)

[∵ \(\vec{b} \times \vec{b}=\overrightarrow{0} \text { and }(-\vec{c}) \times \vec{b}=\vec{b} \times \vec{c}\)]

⇒ \(\vec{a} \times \vec{b}=\vec{b} \times \vec{c}\) …(1)

Also, \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \Rightarrow \vec{b}+\vec{c}=-\vec{a}\)

⇒ \((\vec{b}+\vec{c}) \times \vec{c}=(-\vec{a}) \times \vec{c}\)

⇒ \((\vec{b} \times \vec{c})+(\vec{c} \times \vec{c})=(-\vec{a}) \times \vec{c}\) [by the distributive law]

⇒ \((\vec{b} \times \vec{c})+\overrightarrow{0}=\vec{c} \times \vec{a}\)

[∵ \(\vec{c} \times \vec{c}=\overrightarrow{0} \text { and }(-\vec{a}) \times \vec{c}=\vec{c} \times \vec{a}\)]

⇒ \(\vec{b} \times \vec{c}=\vec{c} \times \vec{a}\) …(2)

From (1) and (2), we get \(\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} .\)

Example 20 Prove that the points A, B, C with position vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are collinear if and only if \((\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})+(\vec{a} \times \vec{b})=\overrightarrow{0}\).

Proof

We have

\(\overrightarrow{A B}\) = (position vector of B) – (position vector of A) = \((\vec{b}-\vec{a})\)

and \(\overrightarrow{B C}\) = (position vector of C) – (position vector of B) = \((\vec{c}-\vec{b})\).

Now, A, B, C are collinear

⇔ \(\overrightarrow{A B}\) and \(\overrightarrow{B C}\) are parallel

⇔ \((\vec{b}-\vec{a}) \times(\vec{c}-\vec{b})=\overrightarrow{0}\)

⇔ \((\vec{b}-\vec{a}) \times \vec{c}-(\vec{b}-\vec{a}) \times \vec{b}=\overrightarrow{0}\) [by the distributive law]

⇔ \(\vec{b} \times \vec{c}-\vec{a} \times \vec{c}-\vec{b} \times \vec{b}+\vec{a} \times \vec{b}=\overrightarrow{0}\) [by the distributive law]

⇔ \((\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})+(\vec{a} \times \vec{b})=\overrightarrow{0}\) [∵ \(\vec{b} \times \vec{b}=\overrightarrow{0} \text { and }-\vec{a} \times \vec{c}=\vec{c} \times \vec{a}\)].

Thus, A, B, C are collinear ⇔ \((\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})+(\vec{a} \times \vec{b})=\overrightarrow{0}\)

Example 21 Prove that \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\)

Proof

We have

\((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})\)

= \(\vec{a} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{a}-\vec{b} \times \vec{b}\) [by the distributive law]

= \(\vec{a} \times \vec{b}-\vec{b} \times \vec{a}\) [∵ \(\vec{a} \times \vec{a}=\overrightarrow{0} \text { and } \vec{b} \times \vec{b}=\overrightarrow{0}\)]

= \((\vec{a} \times \vec{b})+(\vec{a} \times \vec{b})\) [∵ \(-\vec{b} \times \vec{a}=(\vec{a} \times \vec{b})\)]

= \(2(\vec{a} \times \vec{b})\).

Example 22 If \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a} x \vec{b} = \overrightarrow{0}\), prove that \(\vec{a}\) = \(\overrightarrow{0}\) or \(\vec{b}\) = \(\overrightarrow{0}\).

Proof

Let \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a}\) x \(\vec{b}\) = \(\overrightarrow{0}\). Then,

\(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a}\) x \(\vec{b}\) = \(\overrightarrow{0}\)

⇒ \((\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0} \text { or } \vec{a} \perp \vec{b}) \text { and }(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0} \text { or } \vec{a} \| \vec{b})\)

⇒ \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0}\) [∵ \(\vec{a}\) ⊥ \(\vec{b}\) and \(\vec{a}\) ∥ \(\vec{b}\) can never hold simultaneously].

Hence, \((\vec{a} \cdot \vec{b}=0 \text { and } \vec{a} \times \vec{b}=\overrightarrow{0}) \Rightarrow(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0}) \text {. }\)

Example 23 If \(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}, \vec{a} \times \vec{b}=\vec{a} \times \vec{c} \text { and } \vec{a} \neq \overrightarrow{0}\), then prove that \(\vec{b}\) = \(\vec{c}.

Proof

[latex]\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c} \text { and } \vec{a} \neq \overrightarrow{0}\)

⇒ \(\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}=0 \text { and } \vec{a} \neq \overrightarrow{0}\)

⇒ \(\vec{a} \cdot(\vec{b}-\vec{c})=0 \text { and } \vec{a} \neq \overrightarrow{0}\)

⇒ \(\vec{b}-\vec{c}=\overrightarrow{0} \text { or } \vec{a} \perp(\vec{b}-\vec{c})\)

\(\vec{b}=\vec{c} \text { or } \vec{a} \perp(\vec{b}-\vec{c})\) …(1)

Again, \(\vec{a} \times \vec{b}=\vec{a} \times \vec{c} \text { and } \vec{a} \neq \overrightarrow{0}\)

⇒ \((\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})=\overrightarrow{0} \text { and } \vec{a} \neq \overrightarrow{0}\)

⇒ \(\vec{a} \times(\vec{b}-\vec{c})=\overrightarrow{0} \text { and } \vec{a} \neq \overrightarrow{0}\)

⇒ \((\vec{b}-\vec{c})=\overrightarrow{0} \text { or } \vec{a} \|(\vec{b}-\vec{c})\)

⇒ \(\vec{b}=\vec{c} \text { or } \vec{a} \|(\vec{b}-\vec{c})\) …(2)

From (1) and (2), we get \(\vec{b}\) = \(\vec{c}\)

[∵ \(\vec{a} \perp(\vec{b}-\vec{c}) \text { and } \vec{a} \|(\vec{b}-\vec{c})\) both cannot hold simultaneously].

Example 24 If \(\vec{a} \times \vec{b}=\vec{c} \times \vec{d} \text { and } \vec{a} \times \vec{c}=\vec{b} \times \vec{d}\), show that \((\vec{a}-\vec{d})\) is parallel to \((\vec{b}-\vec{c})\), it being given that a ≠ d and b ≠ c.

Solution

\(\vec{a} \times \vec{b}=\vec{c} \times \vec{d} \text {, and } \vec{a} \times \vec{c}=\vec{b} \times \vec{d}\)

⇒ \(\vec{a} \times \vec{b}-\vec{a} \times \vec{c}=\vec{c} \times \vec{d}-\vec{b} \times \vec{d}\)

⇒ \(\vec{a} \times \vec{b}-\vec{a} \times \vec{c}+\vec{b} \times \vec{d}-\vec{c} \times \vec{d}=\overrightarrow{0}\)

⇒ \(\vec{a} \times(\vec{b}-\vec{c})+(\vec{b}-\vec{c}) \times \vec{d}=\overrightarrow{0}\)

⇒ \(\vec{a} \times(\vec{b}-\vec{c})-\vec{d} \times(\vec{b}-\vec{c})=\overrightarrow{0}\)

⇒ \((\vec{a}-\vec{d}) \times(\vec{b}-\vec{c})=\overrightarrow{0}\)

⇒ \((\vec{a}-\vec{d}) \|(\vec{b}-\vec{c})\)

Hence, \((\vec{a}-\vec{d})\) is parallel to \((\vec{b}-\vec{c})\).

Example 25 If \(\vec{a}=\hat{i}+\hat{j}+\hat{k} \text { and } \vec{b}=\hat{j}-\hat{k}\), find a vector \(\vec{c}\) such that \(\vec{a}\) x \(\vec{c}\) = \(\vec{b}\) and \(\vec{a}\).\(\vec{c}\) = 3.

Solution

Let \(\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\). Then,

\((\vec{a} \times \vec{c})=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
c_1 & c_2 & c_3
\end{array}\right|\)

= \(\left(c_3-c_2\right) \hat{i}+\left(c_1-c_3\right) \hat{j}+\left(c_2-c_1\right) \hat{k}\)

∴ \((\vec{a} \times \vec{c})=\vec{b} \Rightarrow\left(c_3-c_2\right) \hat{i}+\left(c_1-c_3\right) \hat{j}+\left(c_2-c_1\right) \hat{k}=\hat{j}-\hat{k}\)

⇒ c3 – c2 = 0, c1 – c3 = 1 and c2 – c1 = -1

⇒ c3 = c2, c1 – c3 = 1 and c2 – c1 = -1 …(1)

\(\vec{a} \cdot \vec{c}=(\hat{i}+\hat{j}+\hat{k}) \cdot\left(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\right)\)

⇒ \(\vec{a} \cdot \vec{c}=c_1+c_2+c_3\)

⇒ c1 + c2 + c3 = 3 …(2) [∵ \(\vec{a}\).\(\vec{c}\) = 3]

⇒ c1 + c2 + c1 – 1 = 3 [∵ c1 – c3 = 1]

⇒ 2c1 + c2 = 4 …(3)

On solving c1 – c2 = 1 and 2c1 + c2 = 4, we get

3c1 = 5 ⇒ c1 = \(\frac{5}{3}\).

∴ \(c_2=\left(c_1-1\right)=\left(\frac{5}{3}-1\right)=\frac{2}{3} \text {. }\)

And, c3 = c2 = \(\frac{2}{3}\).

Hence, \(\vec{c}=\left(\frac{5}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right) .\)

Product Of Three Vectors

Scalar Triple Product The scalar triple product of three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) is defined as \((\vec{a} \times \vec{b}) \cdot \vec{c}\), and it is denoted by [\(\vec{a}\), \(\vec{b}\), \(\vec{c}\)].

Thus \([\vec{a} \vec{b} \vec{c}]=(\vec{a} \times \vec{b}) \cdot \vec{c} .\)

Since \((\vec{a} x \vec{b})\) is a vector, \((\vec{a} \times \vec{b}) \cdot \vec{c}\) is a scalar.

Consequently, \([\vec{a} \vec{b} \vec{c}]\) is a scalar quantity.

Some Theorems on Scalar Triple Product

Theorem 1 (Geometrical interpretation) \((\vec{a} \times \vec{b}) \cdot \vec{c}\) represents the volume of a parallelpiped whose coterminous edges are represented by \(\vec{a}\), \(\vec{b}\), \(\vec{c}\).

Proof

Let us consider a parallelepiped having coterminous edges OA, OB and OC respectively.

Let \(\overrightarrow{O A}\) = \(\vec{a}\), \(\overrightarrow{O B}\) = \(\vec{B}\) and \(\overrightarrow{O C}\) = \(\vec{c}\).

Then, \((\vec{a} \times \vec{b})\) is a vector perpendicular to the plane of \(\vec{a}\) and \(vec{b}\).

Let θ be the angle between \((\vec{a} \times \vec{b}) and [latex]vec{c}\).

∴ \([\vec{a} \vec{b} \vec{c}]=(\vec{a} \times \vec{b}) \cdot \vec{c} .\)

= \(|\vec{a} \times \vec{b}||\vec{c}| \cos \theta\)

= (area of ∥gm OADB) [projection of \(\vec{c}\) along \((\vec{a} \times \vec{b})\)]

= (area of ∥gm OADB).OL

= (area of the base of the parallelepiped with coterminous edges \(\vec{a}\), \(\vec{v}\), \(\vec{c}\).

Hence \([\vec{a} \vec{b} \vec{c}]\) represents the volume of the parallelepiped witn coterminous edges \(\vec{A}\), \(\vec{b}\), \(\vec{c}\) forming a right-handed system.

Theorem 2 For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), \((\vec{a} \times \vec{b}) \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})\).

Proof

Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) form a right-handed system, representing the coterminous edges of a parallelepiped of volume V.

Then, V = \((\vec{a} \times \vec{b}) \cdot \vec{c} .\)

Again, \(\vec{b}\), \(\vec{c}\), \(\vec{a}\) form a right-handed system, representing the coterminous edges of the same parallelepiped.

∴ V = \((\vec{b} \times \vec{c}) \cdot \vec{a}=\vec{a} \cdot(\vec{b} \times \vec{c})\) [∵ \(\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A}\)]

Hence, \((\vec{a} \times \vec{b}) \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})\).

Theorem 3 The scalar triple product of three vectors remains unchanged so long as their cyclic order remains unchanged, i.e., \((\vec{a} \times \vec{b}) \cdot \vec{c}=(\vec{b} \times \vec{c}) \cdot \vec{a}=(\vec{c} \times \vec{a}) \cdot \vec{b} .\)

Proof

Let \(\vec{a}\), \(\vec{b}\), \(\vec{C}\) form a right-handed system, representing the coterminous edges of a rectangular parallelepiped of volume V.

Then, V = \((\vec{a} \times \vec{b}) \cdot \vec{c} .\)

Clearly, \(\vec{b}\), \(\vec{c}\), \(\vec{a}\) as well as \(\vec{c}\), \(\vec{a}\), \(\vec{b}\) form a right-handed system, representing the coterminous edges of the same parallelepiped.

∴ \(V=(\vec{b} \times \vec{c}) \cdot \vec{a} \text { and } V=(\vec{c} \times \vec{a}) \cdot \vec{b} .\)

Thus, \((\vec{a} \times \vec{b}) \cdot \vec{c}=(\vec{b} \times \vec{c}) \cdot \vec{a}=(\vec{c} \times \vec{a}) \cdot \vec{b}\) [each equal to V]

Hence, \([\vec{a} \vec{b} \vec{c}]=[\vec{b} \vec{c} \vec{a}]=[\vec{c} \vec{a} \vec{b}]\)

Theorem 4 The scalar triple product changes in sign but not in magnitude when the cyclic order of vectors is changed.

Proof

For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) we know that

\([\vec{a} \vec{b} \vec{c}]=[\vec{b} \vec{c} \vec{a}]=[\vec{c} \vec{a} \vec{b}]\)

∴ \([\vec{c} \vec{b} \vec{a}]=(\vec{c} \times \vec{b}) \cdot \vec{a}=-(\vec{b} \times \vec{c}) \cdot \vec{a}=-[\vec{b} \vec{c} \vec{a}]=-[\vec{a} \vec{b} \vec{c}]\)

And, \([\vec{a} \vec{c} \vec{b}]=(\vec{a} \times \vec{c}) \cdot \vec{b}=-(\vec{c} \times \vec{a}) \cdot \vec{b}=-[\vec{c} \vec{a} \vec{b}]=-[\vec{a} \vec{b} \vec{c}] .\)

∴ \([\vec{c} \vec{b} \vec{a}]=-[\vec{a} \vec{b} \vec{c}] \text { and }[\vec{a} \vec{c} \vec{b}]=-[\vec{a} \vec{b} \vec{c}] \text {. }\)

Theorem 5 The scalar triple product vanishes if any two of its vectors are equal, i.e., \(\left[\begin{array}{lll}
\vec{a} & \vec{a} & \vec{b}
\end{array}\right]=0, \left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]=0 \text { and }\left[\begin{array}{lll}
\vec{b} & \vec{a} & \vec{a}
\end{array}\right]=0.\)

Proof

We have \([\vec{a} \vec{a} \vec{b}]=(\vec{a} \times \vec{a}) \cdot \vec{b}=\overrightarrow{0} \cdot \vec{b}=0\)

\(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{a} & \vec{b}
\end{array}\right]\) [cyclic-ordered property]

= \((\vec{a} \times \vec{a}) \cdot \vec{b}=\overrightarrow{0} \cdot \vec{b}=0 \text {; }\)

and \(\left[\begin{array}{lll}
\vec{b} & \vec{a} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]\)

= 0. [cyclic-ordered property]

Hence, \(\left[\begin{array}{lll}
\vec{a} & \vec{a} & \vec{b}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{b} & \vec{a} & \vec{a}
\end{array}\right]=0 .\)

Theorem 6 The scalar triple product vanishes if any two of its vectors are parallel or collinear.

Proof

Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three vectors such that \(\vec{a}\) ∥ \(\vec{b}\), or \(\vec{a}\) and \(\vec{b}\) are collinear.

Then, \(\vec{a} = m\vec{b}\) for some scalar m.

∴ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left[\begin{array}{llll}
m & \vec{b} & \vec{b} & \vec{c}
\end{array}\right]=(m \vec{b} \times \vec{b}) \cdot \vec{c}=\overrightarrow{0} \cdot \vec{c}=0 .\)

Theorem 7 The necessary and sufficient condition for three nonzero, noncollinear vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) to be coplanar is that \([\vec{a} \vec{b} \vec{c}]=0\)

Proof

Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three nonzero, noncollinear and coplanar vectors. Then,

\(\vec{b}\) x \(\vec{c}\) is perpendicular to the plane of \(\vec{b}\) and \(\vec{c}\)

⇒ \((\vec{b} \times \vec{c}) \perp \vec{a}\) [∵ \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar]

⇒ \(\vec{a} \cdot(\vec{b} \times \vec{c})=0\)

⇒ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=0\)

Thus, whenever \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar, we have \([\vec{a} \vec{b} \vec{c}]=0\).

Conversely, let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three nonzero, noncollinear vectors such that \([\vec{a} \vec{b} \vec{c}]=0\). Then,

\([\vec{a} \vec{b} \vec{c}]=0 \Rightarrow \vec{a} \cdot(\vec{b} \times \vec{c})=0\)

⇒ \(\vec{a}=\overrightarrow{0}, \text { or }(\vec{b} \times \vec{c})=\overrightarrow{0}, \text { or }(\vec{b} \times \vec{c}) \perp \vec{a}\)

⇒ \((\vec{b} \times \vec{c})=\overrightarrow{0} \text {, or }(\vec{b} \times \vec{c}) \perp \vec{a}\) [∵ \(\vec{a}\) ≠ \(\vec{0}\)]

⇒ \((\vec{b} \times \vec{c}) \perp \vec{a}\)

[∵ \(\vec{b}\) ≠ \(\vec{0}\), \(\vec{c}\) ≠ \(\vec{0}\), and \(\vec{b}\), \(\vec{c}\), are noncollinear ⇒ \(\vec{b} \times \vec{c} \neq \overrightarrow{0}\)].

Thus, \((\vec{b} \times \vec{c})\) is perpendicular to \(\vec{a}\).

But, \((\vec{b} \times \vec{c})\) is perpendicular to the plane of \(\vec{b}\) and \(\vec{c}\).

∴ \(\vec{a}\) must lie in the plane of \(\vec{b}\) and \(\vec{c}\).

Hence \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) must be coplanar.

Theorem 8 (Scalar triple product in terms of components) Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k} \text { and } \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k} \text {. }\)

Then, \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left|\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|\)

Proof

We have \(\vec{a} \times \vec{b}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|\)

= \(\left(a_2 b_3-a_3 b_2\right) \hat{i}+\left(a_3 b_1-a_1 b_3\right) \hat{j}+\left(a_1 b_2-a_2 b_1\right) \hat{k}\)

∴ \([\vec{a} \vec{b} \vec{c}]=(\vec{a} \times \vec{b}) \cdot \vec{c}\)

= \(\left(a_2 b_3-a_3 b_2\right) c_1+\left(a_3 b_1-a_1 b_3\right) c_2+\left(a_1 b_2-a_2 b_1\right) c_3\)

= \(a_1\left(b_2 c_3-b_3 c_2\right)+a_2\left(b_3 c_1-b_1 c_3\right)+a_3\left(b_1 c_2-b_2 c_1\right)\)

= \(\left|\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right| .\)

Solved Examples

Example 1 Prove that \([\hat{i} \hat{j} \hat{k}]=1 \text {, and }[\hat{i} \hat{k} \hat{j}]=-1\)

Solution

\([\hat{i} \hat{j} \hat{k}]=(\hat{i} \times \hat{j}) \cdot \hat{k}=\hat{k} \cdot \hat{k}=1\)

And, \([\hat{i} \hat{k} \hat{j}]=(\hat{i} \times \hat{k}) \cdot \hat{j}=-\hat{j} \cdot \hat{j}==-1\)

Example 2 If \(\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k} and \vec{c}=-3 \hat{i}+\hat{j}+2 \hat{k} \text {, find }[\vec{a} \vec{b} \vec{c}]\)

Solution

We have

\(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left|\begin{array}{rrr}
2 & 1 & 3 \\
-1 & 2 & 1 \\
-3 & 1 & 2
\end{array}\right|\)

= \(\left|\begin{array}{rrr}
0 & 5 & 5 \\
-1 & 2 & 1 \\
0 & -5 & -1
\end{array}\right|\)

\(\left[R_1 \rightarrow R_1+2 R_2, R_3 \rightarrow R_3-3 R_2\right]\)

= -(-1).[-5+25] = 20.

Example 3 Find the volume of the parallelepiped whose coterminous edges are represented by the vectors \(\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k} \text { and } \vec{c}=2 \hat{i}+\hat{j}-\hat{k}\)

Solution

We have

\(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left|\begin{array}{rrr}
2 & -3 & 1 \\
1 & -1 & 2 \\
2 & 1 & -1
\end{array}\right|\)

= \(\left|\begin{array}{rrr}
0 & -1 & -3 \\
1 & -1 & 2 \\
0 & 3 & -5
\end{array}\right|\)

\(\left[R_1 \rightarrow R_1-2 R_2 \text {, and } R_3 \rightarrow R_3-2 R_2\right]\)

= -(1).(5+9) = -14.

∴ volume of the parallelepiped

= \(|[\vec{a} \vec{b} \vec{c}]|=|-14|\) = 14 cubic units.

Example 4 Show that the vectors \(\hat{i}-3 \hat{j}+4 \hat{k}, 2 \hat{i}-\hat{j}+2 \hat{k} \text { and } 4 \hat{i}-7 \hat{j}+10 \hat{k}\) are coplanar.

Solution

Let \(\vec{a}=\hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}+2 \hat{k} \text { and } \vec{c}=4 \hat{i}-7 \hat{j}+10 \hat{k}\)

∴ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=\left|\begin{array}{rrr}
1 & -3 & 4 \\
2 & -1 & 2 \\
4 & -7 & 10
\end{array}\right|=\left|\begin{array}{rrr}
1 & -3 & 4 \\
0 & 5 & -6 \\
0 & 5 & -6
\end{array}\right|\)

= \(\left|\begin{array}{l}
R_2 \rightarrow R_2-2 R_1 \\
R_3 \rightarrow R_3-4 R_1
\end{array}\right|\)

= (-30+30) = 0,

Hence, the given vectors are coplanar.

Example 5 Find the value of λ so that the vectors \(\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-3 \hat{k} \text { and } \vec{c}=\hat{j}+\lambda \hat{k}\) are coplanar.

Solution

The given vectors will be coplanar if \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=0\)

Now, \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=0 ⇔ \left|\begin{array}{rrr}
2 & -3 & 1 \\
1 & 2 & -3 \\
0 & 1 & \lambda
\end{array}\right|=0\)

⇔ \(\left|\begin{array}{rrr}
0 & -7 & 7 \\
1 & 2 & -3 \\
0 & 1 & \lambda
\end{array}\right|=0\)

\(\left[R_1 \rightarrow R_1-2 R_2\right]\)

⇔ (-1)(-7λ-7) = 0 ⇔ 7λ + 7 = 0 ⇔ λ = -1.

Hence, the given vectors are coplanar when λ = -1.

Example 6 Show that the four points with position vectors \((4 \hat{i}+5 \hat{j}+\hat{k}),(-\hat{j}-\hat{k}),(3 \hat{i}+9 \hat{j}+4 \hat{k}) \text { and } 4(-\hat{i}+\hat{j}+\hat{k})\) are coplanar.

Solution

Let the given points be A, B, C, D respectively.

Points A, B, C, D are coplanar ⇔ \(\overrightarrow{A B}, \overrightarrow{A C} \text { and } \overrightarrow{A D}\) are coplanar

⇔ \([\overrightarrow{A B} \overrightarrow{A C} \overrightarrow{A D}]=0\)

Now, \(\overrightarrow{A B}\) = (p.v. of B)-(p.v. of A)

= \((-\hat{j}-\hat{k})-(4 \hat{i}+5 \hat{j}+\hat{k})=(-4 \hat{i}-6 \hat{j}-2 \hat{k}) .\)

\(\overrightarrow{A C}\) = (p.v. of C)-(p.v. of A)

= \((3 \hat{i}+9 \hat{j}+4 \hat{k})-(4 \hat{i}+5 \hat{j}+\hat{k})=(-\hat{i}+4 \hat{j}+3 \hat{k})\)

\(\overrightarrow{A D}\) = (p.v. of D)-(p.v. of A)

= \((-4 \hat{i}+4 \hat{j}+4 \hat{k})-(4 \hat{i}+5 \hat{j}+\hat{k})=(-8 \hat{i}-\hat{j}+3 \hat{k})\)

∴ \(\left[\begin{array}{lll}
\overrightarrow{A B} & \overrightarrow{A C} & \overrightarrow{A D}
\end{array}\right]=\left|\begin{array}{rrr}
-4 & -6 & -2 \\
-1 & 4 & 3 \\
-8 & -1 & 3
\end{array}\right|\)

= \(\left|\begin{array}{rrr}
0 & -22 & -14 \\
-1 & 4 & 3 \\
0 & -21 & -33
\end{array}\right|\)

\(\left\{\begin{array}{l}
R_1 \rightarrow R_1-4 R_2 \\
R_3 \rightarrow R_3-8 R_2
\end{array}\right\}\)

= -(-1)[462-462] = 0.

∴ \(\overrightarrow{A B}\), \(\overrightarrow{A C}\) and \(\overrightarrow{A D}\) are coplanar.

Hence, the points A, B, C, D are coplanar.

Example 7 Find the value of λ so that the four points with position vectors \((-6 \hat{i}+3 \hat{j}+2 \hat{k}),(3 \hat{i}+\lambda \hat{j}+4 \hat{k}),(5 \hat{i}+7 \hat{j}+3 \hat{k})\) and \((-13 \hat{i}+17 \hat{j}-2 \hat{k})\) are coplanar.

Solution

Let the given points be A, B, C, D respectively. Then,

\(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)

= \((3 \hat{i}+\lambda \hat{j}+4 \hat{k})-(-6 \hat{i}+3 \hat{j}+2 \hat{k})=9 \hat{i}+(\lambda-3) \hat{j}+2 \hat{k}\)

\(\overrightarrow{A C}\) = (p.v. of C) – (p.v. of A)

= \((5 \hat{i}+7 \hat{j}+3 \hat{k})-(-6 \hat{i}+3 \hat{j}+2 \hat{k})=(11 \hat{i}+4 \hat{j}+\hat{k})\)

\(\overrightarrow{A D}\) = (p.v. of D) – (p.v. of A)

= \((-13 \hat{i}+17 \hat{j}-\hat{k})-(-6 \hat{i}+3 \hat{j}+2 \hat{k})=(-7 \hat{i}+14 \hat{j}-3 \hat{k})\)

Now, A, B, C, D are coplanar

⇔ \(\left[\begin{array}{lll}
\overrightarrow{A B} & \overrightarrow{A C} & \overrightarrow{A D}
\end{array}\right]=0\)

⇔ \(\left|\begin{array}{rcr}
9 & \lambda-3 & 2 \\
11 & 4 & 1 \\
-7 & 14 & -3
\end{array}\right|=0\)

⇔ 9(-12-14) – (λ-3)(-33+7) + 2(154+28) = 0

⇔ -234 + 26λ – 78 + 364 = 0

⇔ 26λ = -52 ⇔ λ = -2.

Hence, the required value of λ is -2.

Example 8 Show that the points A(-1,4,-3), B(3,2,-5), C(-3,8,-5) and D(-3,2,1) are coplanar.

Solution

Clearly, the position vectors of A, B, C, D are \((-\hat{i}+4 \hat{j}-3 \hat{k}), (3 \hat{i}+2 \hat{j}-5 \hat{k}),(-3 \hat{i}+8 \hat{j}-5 \hat{k}) and (-3 \hat{i}+2 \hat{j}+\hat{k})\) respectively.

∴ \(\overrightarrow{A B}\) = (p.v. of B) – (p.v. of A)

= \((3 \hat{i}+2 \hat{j}-5 \hat{k})-(-\hat{i}+4 \hat{j}-3 \hat{k})=(4 \hat{i}-2 \hat{j}-2 \hat{k})\)

\(\overrightarrow{A C}\) = (p.v. of C) – (p.v. of A)

= \((-3 \hat{i}+8 \hat{j}-5 \hat{k})-(-\hat{i}+4 \hat{j}-3 \hat{k})=(-2 \hat{i}+4 \hat{j}-2 \hat{k})\)

\(\overrightarrow{A D}\) = (p.v. of D) – (p.v. of A)

= \((-3 \hat{i}+2 \hat{j}+\hat{k})-(-\hat{i}+4 \hat{j}-3 \hat{k})=(-2 \hat{i}-2 \hat{j}+4 \hat{k})\)

∴ \(\left[\begin{array}{lll}
\overrightarrow{A B} & \overrightarrow{A C} & \overrightarrow{A D}
\end{array}\right]=\left|\begin{array}{rrr}
4 & -2 & -2 \\
-2 & 4 & -2 \\
-2 & -2 & 4
\end{array}\right|\)

= \(2 \times(-2) \times(-2) \cdot\left|\begin{array}{rrr}
2 & -1 & -1 \\
1 & -2 & 1 \\
1 & 1 & -2
\end{array}\right|\)

= \(8 \cdot\left|\begin{array}{rrr}
0 & -3 & 3 \\
0 & -3 & 3 \\
1 & 1 & -2
\end{array}\right|\)

\(\left[R_1 \rightarrow R_1-2 R_3, R_2 \rightarrow R_2-R_3\right]\)

= (8 x 0) = 0 [∵ R1 and R2 are identical]

⇒ \(\overrightarrow{A B}\), \(\overrightarrow{A C}\), \(\overrightarrow{A D}\) are coplanar

⇒ the points A, B, C, D are coplanar.

Example 9 For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) show that the vectors \((\vec{a}-\vec{b}),(\vec{b}-\vec{c}),(\vec{c}-\vec{a})\) are coplanar.

Solution

We know that \((\vec{a}-\vec{b}),(\vec{b}-\vec{c}),(\vec{c}-\vec{a})\) are coplanar

⇔ \(\left[\begin{array}{lll}
\vec{a}-\vec{b} & \vec{b}-\vec{c} & \vec{c}-\vec{a}
\end{array}\right]=0 .\)

Now, \(\left[\begin{array}{lll}
\vec{a}-\vec{b} & \vec{b}-\vec{c} & \vec{c}-\vec{a}
\end{array}\right]\)

= \((\vec{a}-\vec{b}) \cdot[(\vec{b}-\vec{c}) \times(\vec{c}-\vec{a})]\)

= \((\vec{a}-\vec{b}) \cdot[\vec{b} \times \vec{c}-\vec{b} \times \vec{a}-\vec{c} \times \vec{c}+\vec{c} \times \vec{a}]\) [by distribution law]

= \((\vec{a}-\vec{b}) \cdot[\vec{b} \times \vec{c}+\vec{a} \times \vec{b}+\vec{c} \times \vec{a}] [∵ -\vec{b} \times \vec{a}=\vec{a} \times \vec{b} \text {, and } \vec{c} \times \vec{c}=\overrightarrow{0}]\)

= \(\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{a} \cdot(\vec{a} \times \vec{b})+\vec{a} \cdot[\vec{c} \times \vec{a}]-\vec{b} \cdot(\vec{b} \times \vec{c})-\vec{b} \cdot(\vec{a} \times \vec{b})-\vec{b} \cdot(\vec{c} \times \vec{a})\)

= \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{a} & \vec{b}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{c} & \vec{a}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{b} & \vec{c}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{a} & \vec{b}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]\)

= \([\vec{a} \vec{b} \vec{c}]-[\vec{b} \vec{c} \vec{a}]\)

[∵ scalar triple product with two equal vectors is 0]

= \([\vec{a} \vec{b} \vec{c}]-[\vec{a} \vec{b} \vec{c}]=0\)

{∵ \(\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)}.

Hence, \((\vec{a}-\vec{b}),(\vec{b}-\vec{c}),(\vec{c}-\vec{a})\) are coplanar.

Example 10 Prove that \(\left[\begin{array}{llll}
\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}
\end{array}\right]=2\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] .\)

Solution

We have

\([\vec{a}+\vec{b} \vec{b}+\vec{c} \vec{c}+\vec{a}]\)

= \((\vec{a}+\vec{b}) \cdot[(\vec{b}+\vec{c}) \times(\vec{c}+\vec{a})]\)

= \((\vec{a}+\vec{b}) \cdot[\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+\vec{c} \times \vec{c}+\vec{c} \times \vec{a}]\) [by distribution law]

= \((\vec{a}+\vec{b}) \cdot[(\vec{b} \times \vec{c})-(\vec{a} \times \vec{b})+(\vec{c} \times \vec{a})]\)

[∵ \(\vec{c} \times \vec{c}=\overrightarrow{0}, \text { and } \vec{b} \times \vec{a}=-\vec{a} \times \vec{b}\)]

= \(\vec{a} \cdot(\vec{b} \times \vec{c})-\vec{a} \cdot(\vec{a} \times \vec{b})+\vec{a} \cdot(\vec{c} \times \vec{a})+\vec{b} \cdot(\vec{b} \times \vec{c})-\vec{b} \cdot(\vec{a} \times \vec{b})+\vec{b} \cdot(\vec{c} \times \vec{a})\)

[by distribution law]

= \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]-\left[\begin{array}{lll}
\vec{a} & \vec{a} & \vec{b}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{c} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{b} & \vec{c}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{a} & \vec{b}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]\)

= \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]\)

[∵ scalar triple product with two equal vectors is 0]

= \(2[\vec{a} \vec{b} \vec{c}]\) {∵ \(\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)}

Hence, \(\left[\begin{array}{lll}
\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}
\end{array}\right]=2\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] .\)

Example 11 Show that \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are noncoplanar if and only if \(\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}\) are noncoplanar.

Solution

\(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are noncoplanar

⇔ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] \neq 0 ⇔2 \left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] \neq 0\)

⇔ \(\left[\begin{array}{lll}
\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}
\end{array}\right] \neq 0\)

⇔ \(\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}\) are noncoplanar.

Example 12 If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are the position vectors of points A, B, C prove that \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a})\) is a vector perpendicular tot he plane of triangle ABC.

Solution

In order to prove the required result, we have to show that \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a})\) is perpendicular to each of the vectors \(\overrightarrow{A B}\), \(\overrightarrow{B C}\) and \(\overrightarrow{C A}\).

We have, \(\overrightarrow{A B}=(\vec{b}-\vec{a}), \overrightarrow{B C}=(\vec{c}-\vec{b}) \text { and } \overrightarrow{C A}=(\vec{a}-\vec{c}) \text {. }\)

Now, \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}) \cdot(\vec{b}-\vec{a})\)

= \((\vec{a} \times \vec{b}) \cdot(\vec{b}-\vec{a})+(\vec{b} \times \vec{c}) \cdot(\vec{b}-\vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{b}-\vec{a})\)

= \((\vec{a} \times \vec{b}) \cdot \vec{b}-(\vec{a} \times \vec{b}) \cdot \vec{a}+(\vec{b} \times \vec{c}) \cdot \vec{b}-(\vec{b} \times \vec{c}) \cdot \vec{a}+(\vec{c} \times \vec{a}) \cdot \vec{b}-(\vec{c} \times \vec{a}) \cdot \vec{a}\)

= \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{b}
\end{array}\right]-\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{b}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{c} & \vec{a} & \vec{b}
\end{array}\right]-\left[\begin{array}{lll}
\vec{c} & \vec{a} & \vec{a}
\end{array}\right]\)

= \(\left[\begin{array}{lll}
\vec{c} & \vec{a} & \vec{b}
\end{array}\right]-\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]\)

[∵ scalar triple product with two equal vectors is 0].

= 0 (∵ \(\left[\begin{array}{lll}
\vec{c} & \vec{a} & \vec{b}
\end{array}\right]=\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]\))

Similarly, \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}) \cdot(\vec{c}-\vec{b})=0\)

And, \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}) \cdot(\vec{a}-\vec{c})=0\)

Thus, \((\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a})\) is perpendicular to each one of the vectors \(\overrightarrow{A B}\), \(\overrightarrow{B C}\) and \(\overrightarrow{C A}\), and therefore, it is perpendicular to the plane of △ABC.

Example 13 Show that \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar if and only if \((\vec{a}+\vec{b}),(\vec{b}+\vec{c}),(\vec{c}+\vec{a})\) are coplanar.

Solution \((\vec{a}+\vec{b}),(\vec{b}+\vec{c}),(\vec{c}+\vec{a})\) are coplanar

⇔ \(\left[\begin{array}{lll}
\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}
\end{array}\right]=0\)

⇔ \((\vec{a}+\vec{b}) \cdot\{(\vec{b}+\vec{c}) \times(\vec{c}+\vec{a})\}=0\)

⇔ \((\vec{a}+\vec{b}) \cdot\{\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+\vec{c} \times \vec{c}+\vec{c} \times \vec{a}\}=0\)

⇔ \((\vec{a}+\vec{b}) \cdot\{\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+\vec{c} \times \vec{a}\}=0\)

[∵ \(\vec{c} \times \vec{c}=0\)]

⇔ \(\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{a} \cdot(\vec{b} \times \vec{a})+\vec{a} \cdot(\vec{c} \times \vec{a})\)

\(+\vec{b} \cdot(\vec{b} \times \vec{c})+\vec{b} \cdot(\vec{b} \times \vec{a})+\vec{b} \cdot(\vec{c} \times \vec{a})=0\)

⇔ \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{c} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{b} & \vec{c}
\end{array}\right]\)

\(+\left[\begin{array}{lll}
\vec{b} & \vec{b} & \vec{a}
\end{array}\right]+\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]=0\)

⇔ \([\vec{a} \vec{b} \vec{c}]+\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]=0\)

[∵ scalar triple product with two equal vectors is 0]

⇔ \(2[\vec{a} \vec{b} \vec{c}]=0\) {∵ \(\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{a}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)}

⇔ \([\vec{a} \vec{b} \vec{c}]=0\)

⇔ \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar.

Hence, \((\vec{a}+\vec{b}), \quad(\vec{b}+\vec{c}), \quad(\vec{c}+\vec{a})\) are coplanar ⇔ \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar.

Example 14. For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) prove that \(\left[\begin{array}{lll}
\vec{a} & \vec{b}+\vec{c} & \vec{a}+\vec{b}+\vec{c}
\end{array}\right]=0\)

Solution

We have

\(\left[\begin{array}{lll}
\vec{a} & \vec{b}+\vec{c} & \vec{a}+\vec{b}+\vec{c}
\end{array}\right]\)

= \(\{\vec{a} \times(\vec{b}+\vec{c})\} \cdot(\vec{a}+\vec{b}+\vec{c})\)

= \(\{(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\} \cdot(\vec{a}+\vec{b}+\vec{c})\) [by the distribution law]

= \((\vec{a} \times \vec{b}) \cdot \vec{a}+(\vec{a} \times \vec{b}) \cdot \vec{b}+(\vec{a} \times \vec{b}) \cdot \vec{c}+(\vec{a} \times \vec{c}) \cdot \vec{a}\)

\(+(\vec{a} \times \vec{c}) \cdot \vec{b}+(\vec{a} \times \vec{c}) \cdot \vec{c}\) [by the distribution law]

= \(\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]+\left[\begin{array}{lll}
\vec{a} & \vec{c} & \vec{b}
\end{array}\right]\)

[∵ scalar triple product with two equal vectors is 0]

= \([\vec{a} \vec{b} \vec{c}]-[\vec{a} \vec{b} \vec{c}]=0\)

Hence, \(\left[\begin{array}{lll}
\vec{a} & \vec{b}+\vec{c} & \vec{a}+\vec{b}+\vec{c}
\end{array}\right]=0\)

Class 12 Maths Solutions For Matrices

Matrix A rectangular array of mn numbers in the form of m horizontal lines (called rows) and n vertical lines (called columns) is called a matrix of order m by n, written as an m x n matrix.

Such an array is enclosed by [ ] or ( ).

Each of the mn numbers constituting the matrix is called an element or an entry of the matrix.

Usually, we denote a matrix by a capital letter.

The plural of matrix is matrices.

Examples (1) A = \(\left[\begin{array}{rrr}
3 & 5 & -4 \\
0 & 1 & 9
\end{array}\right]\) is a matrix, having 2 rows and 3 columns.

Its order is 2 x 3 and it has 6 elements.

(2) B = \(\left[\begin{array}{rrrr}
9 & 4 & \sqrt{2} & -1 \\
1 & 8 & -3 & 2 \\
6 & 0 & 5 & 7
\end{array}\right]\) is a matrix, having 3 rows and 4 columns. Its order is 3 x 4 and it has 12 elements.

How to Describe a Matrix

To locate the position of a particular element of a matrix, we have to specify the number of the row and that of the column in which the element occurs.

An element occuring in the ith row and jth column of a matrix A will be called the (i,j)th element of A, to be denoted by air.

Read and Learn More  Class 12 Math Solutions

In general, as m x n matrix A may be written as

A = \(\left[\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & \ldots & a_{1 n} \\
a_{21} & a_{22} & a_{23} & \ldots & a_{2 n} \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
a_{i 1} & a_{i 2} & a_{i 3} & \ldots & a_{i n} \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
a_{m 1} & a_{m 2} & a_{m 3} & \cdots & a_{n n}
\end{array}\right]=\left[a_{i j}\right]_{m \times n} .\)

Example 1 Consider the matrix A = \(\left[\begin{array}{rrr}
3 & -2 & 5 \\
6 & 9 & 1
\end{array}\right]\)

Clearly, the element in the 1st row and 2nd column is -2.

So, we write a12 = -2.

Similarly, a11 = 3; a12 = -2; a13 = 5; a21 = 6; a22 = 9 and a23 = 1.

Example 2 Construct a 3 x 2 matrix whose elements are given by aij = (i+2j).

Solution

A 3 x 2 matrix has 3 rows and 2 columns.

In general, a 3 x 2 matrix is given by

A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array}\right]_{3 \times 2}\)

Thus aij = (i+2j) for i=1, 2,3 and j = 1,2.

∴ a11 = (1+2×1) = 3; a12 = (1+2×2) = 5;

a21 = (2+2×1) = 4; a22=(2+2×2) = 6;

a31 = (3+2×1) = 5; a32 = (3+2×2) = 7.

Hence, A = \(\left[\begin{array}{ll}
3 & 5 \\
4 & 6 \\
5 & 7
\end{array}\right]_{3 \times 2}\)

Example 3 Construct a 2 x 3 matrix whose elements are given by aij = \(\frac{1}{2}|5 i-3 j| .\)

Solution

A 2 x 3 matrix has 2 rows and 3 columns.

In general, a 2 x 3 matrix is given by

A = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right]_{2 \times 3}\)

Thus, \(a_{i j}=\frac{1}{2}|5 i-3 j|\), where i = 1,2 and j = 1,2,3.

∴ \(a_{11}=\frac{1}{2}|5 \times 1-3 \times 1|=\frac{1}{2} \cdot|2|=\frac{1}{2} \times 2=1 ;\)

\(a_{12}=\frac{1}{2}|5 \times 1-3 \times 2|=\frac{1}{2} \cdot|5-6|=\frac{1}{2} \cdot|-1|=\frac{1}{2} \times 1=\frac{1}{2} \text {; }\) \(a_{13}=\frac{1}{2}|5 \times 1-3 \times 3|=\frac{1}{2} \cdot|5-9|=\frac{1}{2} \cdot|-4|=\frac{1}{2} \times 4=2 \text {; }\) \(a_{21}=\frac{1}{2}|5 \times 2-3 \times 1|=\frac{1}{2} \cdot|10-3|=\frac{1}{2} \cdot|7|=\frac{1}{2} \times 7=\frac{7}{2} ;\) \(a_{22}=\frac{1}{2}|5 \times 2-3 \times 2|=\frac{1}{2} \cdot|10-6|=\frac{1}{2} \cdot|4|=\frac{1}{2} \times 4=2\) \(a_{23}=\frac{1}{2}|5 \times 2-3 \times 3|=\frac{1}{2} \cdot|10-9|=\frac{1}{2} \cdot|1|=\frac{1}{2} \times 1=\frac{1}{2}\)

Hence, A = \(\left[\begin{array}{lll}
1 & \frac{1}{2} & 2 \\
\frac{7}{2} & 2 & \frac{1}{2}
\end{array}\right]\).

WBBSE Class 12 Matrices Solutions

Example 4 If a matrix has 12 elements, what are the possible orders it can have?

Solution

We know that a matrix of order m x n has mn elements.

Hence, all possible orders of a matrix having 12 elements are (12×1), (1×12), (6×2), (2×6), (4×3) and (3×4).

Various Types of Matrices

Row Matrix A matrix having only one row is known as a row matrix or a row vector.

Examples (1) A = [5 18] is a row matrix of order 1 x 2.

(2) B = [2 √5 -9 0] is a row matrix of order 1 x 4.

Column Matrix A matrix having only one column is known as a column matrix or a column vector.

Examples (1) A = \(\left[\begin{array}{r}
2 \\
7 \\
-3
\end{array}\right]\) is a column matrix of order 3 x 1.

(2) B = \(\left[\begin{array}{l}
6 \\
4
\end{array}\right]\) is a column matrix of order 2 x 1.

Zero Or Null Matrix A matrix each of whose elements is zero is called a zero matrix or a null matrix.

Example The matrices [0], [0 0], \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\) and \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) are null matrices of order (1×1), (1×2), (2×2) and (2×3) respectively.

Square Matrix A matrix having the same number of rows and columns is called a square matrix.

A matrix of order (n x n) is called a square matrix of order n or an n-rowed square matrix.

A matrix of order m x n, where m ≠ n, is called a rectangular matrix.

Examples (1) The matrix \(\left[\begin{array}{rr}
3 & 2 \\
6 & -5
\end{array}\right]\) is a 2-rowed square matrix.

(2) The matrix \(\left[\begin{array}{ccc}
5 & 3 & 6 \\
7 & \sqrt{2} & -4 \\
-9 & \frac{1}{3} & 0
\end{array}\right]\) is a 3-rowed square matrix.

Diagonal Elements Of A Matrix Let A = [aij]mxn be an m x n matrix. Then, the elements aij for which i = j, are called the diagonal elements of A.

Thus, the diagonal elements of A = [aij]mxn are a11, a22, a33, a44, etc.

The line along which the diagonal elements lie is called the diagonal of the matrix.

Example Let A = \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
\sqrt{5} & \frac{5}{8} & 7 \\
6 & -4 & \sqrt{2}
\end{array}\right]\) Then, the diagonal elements of A are: a11 = 3, a22 = \(\frac{5}{8}\), a33 = √2.

Understanding Matrices Concepts for Class 12

Diagonal Matrix A square matrix in which every nondiagonal element is zero is called a diagonal matrix.

If A = [aij]mxn be a diagonal matrix then aij=0 when i ≠ j and we write it as

A = diag[a11, a22, a33,…, ann].

Example Let A = \(\left[\begin{array}{rrr}
6 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & -2
\end{array}\right]\). Then, A is a diagonal matrix. We may write it as, A = diag[6,4,-2].

Scalar Matrix A square matrix in which every nondiagonal element is zero and all diagonal elements re equal is known as a scalar matrix.

Examples (1) A = \(\left[\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right]\) is a scalar matrix of order 2.

(2) B = \(\left[\begin{array}{rrr}
-3 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & -3
\end{array}\right]\) is a scalar matrix of order 3.

Unit Matrix A square matrix in which every nondiagonal element is 0 and every diagonal element is 1 is called a unit matrix or an identity matrix.

Thus, a square matrix [aij]nxn is a unit matrix if

\(a_{i j}=\left\{\begin{array}{l}
0 \text { when } i \neq j, \\
1 \text { when } i=j .
\end{array}\right.\)

A unit matrix of order n will be denoted by In or simply by I.

Examples (1) I2 = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) is a unit matrix of order 2.

(2) I3 = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) is a unit matrix of order 3.

Comparable Matrices Two matrices A and B are said to be comparable if they are of the same order, i.e., they have the same number of rows and the same umber of columns.

Example A = \(\left[\begin{array}{rrr}
2 & -5 & 1 \\
0 & 3 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
3 & 7 & 0 \\
1 & 4 & -9
\end{array}\right]\) are comparable matrices, each being of order (2×3).

Equal Matrices Two matrices A and B are said to be equal, written as A = B, if they are of the same order and their corresponding elements are equal.

Example 1 Find x, y, z when \(\left[\begin{array}{ll}
5 & 3 \\
x & 7
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 7
\end{array}\right]\)

Solution

Since the corresponding elements of equal matrices are equal, we have

\(\left[\begin{array}{ll}
5 & 3 \\
x & 7
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 7
\end{array}\right]\) ⇔ x = 1, y = 5 and z = 3.

Step-by-Step Solutions to Matrix Problems

Example 2 Find x, y, z w when \(\left[\begin{array}{ll}
x-y & 2 x+z \\
2 x-y & 3 z+w
\end{array}\right]=\left[\begin{array}{rr}
-1 & 5 \\
0 & 13
\end{array}\right]\).

Solution

We know that in equal matrices, the corresponding elements are equal.

∴ \(\left[\begin{array}{rr}
x-y & 2 x+z \\
2 x-y & 3 z+w
\end{array}\right]=\left[\begin{array}{rr}
-1 & 5 \\
0 & 13
\end{array}\right]\)

⇔ x – y = -1, 2x – y = 0, 2x + z = 5 and 3z + w = 13.

Solving the first two equations, we get x = 1 and y = 2.

Putting x = 1 in 2x + z = 5, we get z = 3.

Putting z = 3 in 3z + w = 13, we get w = 4.

∴ x = 1, y = 2, z = 3 and w = 4.

Example 3 \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \neq\left[\begin{array}{ll}
0 & 0 \\
0 & 0 \\
0 & 0
\end{array}\right] \text {. }\) Why?

Solution

Since the given null matrices are not comparable, they are not equal.

Operations On Matrices

Mainly we have four operations on matrices, namely:

Transposition, Matrix Addition, Matrix multiplication and Scalar multiplication.

Out of these operations, transposition is a unary operation while matrix addition and matrix multiplication are both binary operations and scalar multiplication is an external composition.

Transposition

Transpose Of A Matrix Let A be an (mxn) matrix. Then, the matrix obtained by interchanging the rows and columns of A is called the transpose of A, denoted by A’ or A’.

If A = [aij]mxn, then A’ = [aij]nxm.

Remarks (1) If A is an (mxn) matrix, then A’ is an (nxm) matrix.

(2) (i,j)th element of A = (j,i)th element of A’.

Examples (1) If A = \(\left[\begin{array}{rrr}
2 & 3 & -1 \\
4 & -2 & 5
\end{array}\right]\), then \(A^t=\left[\begin{array}{rr}
2 & 4 \\
3 & -2 \\
-1 & 5
\end{array}\right]\)

Here A is a (2×3) matrix and A’ is a (3×2) matrix.

(2) If B = \(\left[\begin{array}{r}
3 \\
-4 \\
6
\end{array}\right]\), then Bt = [3 -4 6].

Here, B is a (3×1) matrix and Bt is a (1×3) matrix.

Theorem (Involution) For any matrix A, prove that (A’)’ = A.

Proof

Let A = [aij]mxn matrix ⇒ A’ is an (nxm) matrix.

⇒ (A’)’ is a (mxn) matrix.

∴ A and (A’)’ are matrices of the same order.

Also, (i,j)th element of A = (j,i)th element of A’

= (i,j)th element of (A’)’.

Thus, A and (A’)’ are comparable matrices having their corresponding elements equal.

Hence, (A’)’ = A.

Symmetric Matrix A square matrix A is said to be symmetric if A’ = A.

Thus, A is symmetric ⇔ aji = aij.

Examples (1) If A = \(\left[\begin{array}{rr}
4 & 2 \\
2 & -3
\end{array}\right]\), then A’ = \(\left[\begin{array}{rr}
4 & 2 \\
2 & -3
\end{array}\right]\) = A.

∴ A is symmetric.

(2) If B = \(\left[\begin{array}{rrr}
6 & 8 & -4 \\
8 & 3 & 0 \\
-4 & 0 & 5
\end{array}\right]\), then B’ = \(\left[\begin{array}{rrr}
6 & 8 & -4 \\
8 & 3 & 0 \\
-4 & 0 & 5
\end{array}\right]\) = B.

∴ B is symmetric.

Skew-Symmetric Matrix A square matrix A is said to be skew-symmetric if A’ = -A, where -A is the matrix obtained by replacing each element of A by its negative.

Remark A is skew-symmetric ⇒ A’ = -A

⇒ aji = -aij, where A = [aij]nxn

⇒ aii = -aii

⇒ 2aii = 0 ⇒ aii = 0

⇒ every diagonal element of A is 0.

Thus, every diagonal element of a skew-symmetric matrix is 0.

Types of Matrices Explained

Examples (1) If A = \(\left[\begin{array}{rr}
0 & 8 \\
-8 & 0
\end{array}\right]\), then A’ = \(\left[\begin{array}{rr}
0 & -8 \\
8 & 0
\end{array}\right]\) = -A.

Hence, A is skew-symmetric.

(2) Let B = \(\left[\begin{array}{ccc}
0 & h & -g \\
-h & 0 & f \\
-g & f & 0
\end{array}\right]\), then B’ = \(\left[\begin{array}{ccc}
0 & -h & -g \\
h & 0 & -f \\
g & f & 0
\end{array}\right]\) = -B.

Hence, B is skew-symmetric.

Addition Of Matrices

Let A and B be two comparable matrices, each of order (mxn). Then, their sum (A+B) is a matrix of order (mxn), obtained by adding the corresponding elements of A and B.

Example 1 If A = \(\left[\begin{array}{ll}
2 & 1 \\
0 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 4 & 5 \\
1 & 2 & 3
\end{array}\right]\) then A and B are matrices order 2 x 2 and 2 x 3 respectively.

So, A and B are not comparable.

Hence, A+B is not defined.

Example 2 Let A = \(\left[\begin{array}{lll}
5 & 0 & -2 \\
3 & 2 & -7
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
4 & -3 & -6 \\
-1 & 0 & 4
\end{array}\right] \text {. }\)

Clearly, each one of A and B is a 2 x 3 matrix.

∴ A + B is defined.

We have: A+B = \(\left[\begin{array}{lll}
5+4 & 0+(-3) & -2+(-6) \\
3+(-1) & 2+0 & -7+4
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
9 & -3 & -8 \\
2 & 2 & -3
\end{array}\right]\)

Some Results on Addition of Matrices

Theorem 1 Matrix addition is commutative, i.e., A + B = B + A for all comparable matrices A and B.

Proof

Let A = [aij]mxn abd B = [bij]mxn. Then,

A + B = [aij]mxn + [bij]mxn

= [aij + bij]mxn [by the definition of addition of matrices]

= [bij + aij]mxn [∵ addition of numbers is commutative]

= [bij]mxn + [aij]mxn = B + A.

Hence, A + B = B + A.

Theorem 2 Matrix addition is associative, i.e., (A + B)+C = A+(B+C) for all comparable matrices A, B and C.

Proof

Let A = [aij]mxn, B = [bij]mxn and C = [cij]mxn. Then,

(A+B)+c = ([aij]mxn + [bij]mxn) + [cij]mxn

= [aij + bij]mxn + [cij]mxn

= [(aij+bij)+cij]mxn

= [aij + (bij + cij)]mxn

[∵ addition of numbers is associative]

= [aij]mxn + [bij+cij]mxn

= [aij]mxn + ([bij]m+n + [cij]mxn) = A + (B+C).

Hence, (A+B)+C = A+(B+C).

Theorem 3 If A is an mxn matrix and O is an mxn null matrix, then A + O = O + A = A.

Proof

Let A = [aij]mxn and O = [bij]mxn,

where bij = 0 for all suffixes i and j.

Then, A + O = [aij]mxn + [bij]mxn = [aij+bij]mxn

= [aij+0]mxn [∵ bij = 0]

= [aij]mxn = A.

∴ A + O = A.

Similarly, O + A = A.

Hence, A + O = O + A = A.

Remark The null matrix O of order m x n is the additive identity in the set of all m x n matrices.

Example 3 Let A = \(\left[\begin{array}{rrr}
3 & 5 & 4 \\
1 & 2 & -3
\end{array}\right]\) and O = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\), then verify that A + O = O + A = A.

Solution

Clearly, each one of A and O is a matrix of order (2×3).

So, (A+O) and (O+A) are both defined.

Now, A + O = \(\left[\begin{array}{rrr}
3 & 5 & 4 \\
1 & 2 & -3
\end{array}\right]+\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
3+0 & 5+0 & 4+0 \\
1+0 & 2+0 & -3+0
\end{array}\right]+\left[\begin{array}{rrr}
3 & 5 & 4 \\
1 & 2 & -3
\end{array}\right]\) = A.

And, O + A = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]+\left[\begin{array}{rrr}
3 & 5 & 4 \\
1 & 2 & -3
\end{array}\right]\)

= \(\left[\begin{array}{lll}
0+3 & 0+5 & 0+4 \\
0+1 & 0+2 & 0+(-3)
\end{array}\right]=\left[\begin{array}{llr}
3 & 5 & 4 \\
1 & 2 & -3
\end{array}\right]\) = A.

Hence, A + O = O + A = A.

Common Questions on Matrices and Their Solutions

Negative Of A Matrix Let A = [aij]mxn. Then, the negative of A is the matrix (-A) = [-aij]mxn, obtained by replacing each element of A with its corresponding additive inverse. (-A) is called the additive inverse of A.

Example 4 If A = \(\left[\begin{array}{rrr}
3 & -2 & 0 \\
-5 & 7 & \sqrt{2}
\end{array}\right]\), find (-A) and verify that A + (-A) = (-A) + A = 0.

Solution

Clearly, we have

(-A) = \(\left[\begin{array}{rrr}
-3 & 2 & 0 \\
5 & -7 & -\sqrt{2}
\end{array}\right]\)

Now, A + (-A) = \(\left[\begin{array}{rrr}
3 & -2 & 0 \\
-5 & 7 & \sqrt{2}
\end{array}\right]+\left[\begin{array}{rrr}
-3 & 2 & 0 \\
5 & -7 & -\sqrt{2}
\end{array}\right]\)

= \(\left[\begin{array}{ccc}
3+(-3) & -2+2 & 0+0 \\
-5+5 & 7+(-7) & \sqrt{2}+(-\sqrt{2})
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=O\)

and, (-A) + A = \(\left[\begin{array}{rrr}
-3 & 2 & 0 \\
5 & -7 & -\sqrt{2}
\end{array}\right]+\left[\begin{array}{rrr}
3 & -2 & 0 \\
-5 & 7 & \sqrt{2}
\end{array}\right]\)

= \(\left[\begin{array}{ccc}
-3+3 & 2+(-2) & 0+0 \\
5+(-5) & -7+7 & -\sqrt{2}+\sqrt{2}
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

Hence, A + (-A) = (-A) + A = O.

Theorem 4 If A and B are two matrices of the same order then prove that (A+B)’ = (A’+B’).

Proof Let A = [aij]mxn and B = [bij]mxn. Then,

A is an (mxn) matrix, B is an (mxn) matrix

⇒ (A+B) is an (mxn) matrix

⇒ (A + B)’ is an (nxm) matrix.

Also, A is an (mxn) matrix and B is an (mxn) matrix

⇒ A’ is an (nxm) matrix and B’ is an (nxm) matrix

⇒ (A’ + B’) is an (nxm) matrix.

Thus, (A+B)’ and (A’+B’) are comparable matrices.

Also, (j,i)th element of (A+B)’

= (i,j)th element of (A+B)

= (i,j)th element of A + (i,j)th element of B

= (j, i)th element of A’ + (j, i)th element of B’

= (j, i)th element of (A’ + B’).

Thus, (A + B)’ and (A’ + B’) are comparable and their corresponding elements are equal.

Hence, (A+B)’ = (A’ + B’).

Solved Examples

Example 1 Let A = \(\left[\begin{array}{rrr}
2 & 3 & -5 \\
0 & -4 & 8
\end{array}\right]\) Verify that (A’)’ = A.

Solution

We have A’ = \(\left[\begin{array}{rrr}
2 & 3 & -5 \\
0 & -4 & 8
\end{array}\right]^t=\left[\begin{array}{rr}
2 & 0 \\
3 & -4 \\
-5 & 8
\end{array}\right]\)

⇒ \(\left(A^t\right)^t=\left[\begin{array}{rr}
2 & 0 \\
3 & -4 \\
-5 & 8
\end{array}\right]^t=\left[\begin{array}{rrr}
2 & 3 & -5 \\
0 & -4 & 8
\end{array}\right]=A\)

Hence, (A’)’ = A.

Example 2 Let A = \(\left[\begin{array}{rrr}
2 & 3 & 5 \\
-1 & 0 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
4 & -2 & 3 \\
2 & 6 & -1
\end{array}\right]\). Verify that A + B = B + A.

Solution

Here, A is a 2 x 3 matrix and B is a 2 x 3 matrix. So, A and B are comparable.

Therefore, (A+B) and (B+A) both exist and each is a 2 x 3 matrix.

Now, A + B = \(\left[\begin{array}{rrr}
2 & 3 & 5 \\
-1 & 0 & 4
\end{array}\right]+\left[\begin{array}{rrr}
4 & -2 & 3 \\
2 & 6 & -1
\end{array}\right]\)

= \(\left[\begin{array}{rll}
2+4 & 3+(-2) & 5+3 \\
-1+2 & 0+6 & 4+(-1)
\end{array}\right]=\left[\begin{array}{lll}
6 & 1 & 8 \\
1 & 6 & 3
\end{array}\right] .\)

And, B + A = \(\left[\begin{array}{rrr}
4 & -2 & 3 \\
2 & 6 & -1
\end{array}\right]+\left[\begin{array}{rrr}
2 & 3 & 5 \\
-1 & 0 & 4
\end{array}\right]\)

= \(\left[\begin{array}{lrr}
4+2 & -2+3 & 3+5 \\
2+(-1) & 6+0 & (-1)+4
\end{array}\right]=\left[\begin{array}{lll}
6 & 1 & 8 \\
1 & 6 & 3
\end{array}\right]\)

Hence, A + B = B + A.

Example 3 Let A = \(\left[\begin{array}{rr}
1 & -2 \\
5 & 4 \\
3 & 0
\end{array}\right]\), B = \(\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-3 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
4 & 3 \\
-2 & 2 \\
1 & 6
\end{array}\right]\). Verify that (A + B) + C = A + (B + C).

Solution

Clearly, each one of the matrices A, B, C is a (3×2) matrix. So, (A + B) + C and A + (B + C) are both defined and each one is a 3 x 2 matrix.

Now, (A+B) = \(\left[\begin{array}{rr}
1 & -2 \\
5 & 4 \\
3 & 0
\end{array}\right]+\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-3 & 5
\end{array}\right]\)

= \(\left[\begin{array}{lr}
1+3 & -2+1 \\
5+0 & 4+2 \\
3+(-3) & 0+5
\end{array}\right]=\left[\begin{array}{rr}
4 & -1 \\
5 & 6 \\
0 & 5
\end{array}\right]\)

∴ (A+B)+C = \(\left[\begin{array}{rr}
4 & -1 \\
5 & 6 \\
0 & 5
\end{array}\right]+\left[\begin{array}{rr}
4 & 3 \\
-2 & 2 \\
1 & 6
\end{array}\right]\)

= \(\left[\begin{array}{lr}
4+4 & -1+3 \\
5+(-2) & 6+2 \\
0+1 & 5+6
\end{array}\right]=\left[\begin{array}{rr}
8 & 2 \\
3 & 8 \\
1 & 11
\end{array}\right]\)

Also, (B + C) = \(\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-3 & 5
\end{array}\right]+\left[\begin{array}{rr}
4 & 3 \\
-2 & 2 \\
1 & 6
\end{array}\right]\)

= \(\left[\begin{array}{ll}
3+4 & 1+3 \\
0+(-2) & 2+2 \\
-3+1 & 5+6
\end{array}\right]=\left[\begin{array}{rr}
7 & 4 \\
-2 & 4 \\
-2 & 11
\end{array}\right] \text {. }\)

∴ A + (B + C) = \(\left[\begin{array}{rr}
1 & -2 \\
5 & 4 \\
3 & 0
\end{array}\right]+\left[\begin{array}{rr}
7 & 4 \\
-2 & 4 \\
-2 & 11
\end{array}\right]\)

= \(\left[\begin{array}{ll}
1+7 & -2+4 \\
5+(-2) & 4+4 \\
3+(-2) & 0+11
\end{array}\right]=\left[\begin{array}{rr}
8 & 2 \\
3 & 8 \\
1 & 11
\end{array}\right] \text {. }\)

Hence, (A + B) + C = A + (B + C).

Example 4 Find the additive inverse of the matrix A = \(\left[\begin{array}{rrr}
2 & -5 & 0 \\
4 & 3 & -1
\end{array}\right]\).

Solution

The additive inverse of the given matrix A is the matrix -A, given by

\(-A=\left[\begin{array}{ccc}
-2 & -(-5) & 0 \\
-4 & -3 & -(-1)
\end{array}\right]=\left[\begin{array}{ccc}
-2 & 5 & 0 \\
-4 & -3 & 1
\end{array}\right]\)

Subtraction Of Matrices If A and B are two comparable matrices then we define (A – B) = A + (-B).

Example 5 If A = \(\left[\begin{array}{rrr}
2 & -3 & 1 \\
0 & 7 & -9
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
1 & 2 & -3 \\
4 & 8 & -4
\end{array}\right]\), find (A – B).

Solution

We have, (-B) = \(\left[\begin{array}{lll}
-1 & -2 & 3 \\
-4 & -8 & 4
\end{array}\right]\)

∴ (A – B) = A + (-B)

= \(\left[\begin{array}{rrr}
2 & -3 & 1 \\
0 & 7 & -9
\end{array}\right]+\left[\begin{array}{lll}
-1 & -2 & 3 \\
-4 & -8 & 4
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
2+(-1) & -3+(-2) & 1+3 \\
0+(-4) & 7+(-8) & -9+4
\end{array}\right]=\left[\begin{array}{rrr}
1 & -5 & 4 \\
-4 & -1 & -5
\end{array}\right]\)

Hence, (A – B) = \(\left[\begin{array}{rrr}
1 & -5 & 4 \\
-4 & -1 & -5
\end{array}\right] \text {. }\)

Example 6 Prove that the sum of two symmetric matrices is symmetric.

Solution

Let A and B be two symmetric matrices of the same order.

Then, A’ = A and B’ = B.

∴ (A + B)’ = A’ + B’ = (A + B) [∵ A’ = A and B’ = B].

Hence, (A + B) is symmetric.

Example 7 Prove that the sum of two skew-symmetric matrices is a skew-symmetric matrix.

Solution

Let A and B be two skew-symmetric matrices. Then,

A’ = -A and B’ = -B.

∴ (A + B)’ = (A’ + B’) = (-A) + (-B) = -(A + B).

Hence, (A + B) is skew-symmetric.

Example 8 For any square matrix A with real entries, prove that (1) (A + A’) is symmetric (2) (A – A’) is skew-symmetric.

Solution

Let A and B be two skew-symmetric matrices. Then,

(1) (A + A’)’ = A’ + (A’)’ [∵ (A + B)’ = A’ + B’]

= A’ = A [∵ (A’)’ = A]

= (A + A’) [∵ A + B = B + A].

Hence, (A + A’) is symmetric.

(2) (A – A’)’ = A’ – (A’)’ [∵ (A – B)’ = A’ – B’

= A’ – A [∵ (A’)’ = A]

= (A – A’).

Thus, (A – A’)’ = -(A – A’).

Hence, (A – A’) is skew-symmetric.

Multiplication of Matrices

For two given matrices A and B, we say that the product AB exists only when the number of rows in A equals the number of columns in B.

When AB exists, we say that A is conformable to B for multiplication.

Product Of Matrices

Let A = [aij]mxn and B = [bij]nxp be two matrices such that the number of columns in A equals the number of rows in B.

Then, AB exists and it is an (mxp) matrix, given by

\(A B=\left[c_{i 2}\right]_{-\times} \text {where } c_{i k}=\left(a_{i i} b_{1 k}+a_{i k} b_{2 k}+\ldots a_{i k} b_{n k}\right)=\sum_{j=1}^n a_i b_{j k}\)

∴ (i,k)th element of AB

= sum of the products of corresponding elements of ith row of A and kth column of B.

Remarks For two given matrices A and B:

(1) AB may exist and BA may not exist;

(2) BA may exist and AB may not exist;

(3) AB and BA both may not exist;

(4) AB and BA both may exist.

Solved Examples

Example 1 If A = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
b_{31} & b_{32}
\end{array}\right]\) then show that AB and BA both exist. Find AB and BA.

Solution

Here, A is a 2 x 3 matrix and B is a 3 x 2 matrix.

∴ AB exists and it is a 2 x 2 matrix.

Let AB = \(\left[\begin{array}{ll}
c_{11} & c_{12} \\
c_{21} & c_{22}
\end{array}\right]\). Then,

c11 = (1st row of A) x (1st column of B)

= a11b11 + a12b21 + a13b31;

c12 = (1st row of A)x(2nd column of B)

= a11b12 + a12b22 + a13b32;

c21 = (2nd row of A) x (1st column of B)

= a22b11 + a22b21 + a23b31;

and c22 = (2nd row of A) x (2nd column of B)

= a21b12 + a22b22 + a23b32.

∴ AB = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right]\left[\begin{array}{ll}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
b_{31} & b_{32}
\end{array}\right]\)

= \(\left[\begin{array}{ll}
a_{11} b_{11}+a_{12} b_{21}+a_{13} b_{31} & a_{11} b_{12}+a_{12} b_{22}+a_{13} b_{32} \\
a_{21} b_{11}+a_{22} b_{21}+a_{23} b_{31} & a_{21} b_{12}+a_{22} b_{22}+a_{23} b_{32}
\end{array}\right]\)

Again, B is 3 x 2 matrix and A is a 2 x 3 matrix. So, BA exists and it is a 3 x 3 matrix.

Proceeding as above, we get

BA = \(\left[\begin{array}{ll}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
b_{31} & b_{32}
\end{array}\right]\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right]\)

= \(\left[\begin{array}{lll}
b_{11} a_{11}+b_{12} a_{21} & b_{11} a_{12}+b_{12} a_{22} & b_{11} a_{13}+b_{12} a_{23} \\
b_{21} a_{11}+b_{22} a_{21} & b_{21} a_{12}+b_{22} a_{22} & b_{21} a_{13}+b_{22} a_{23} \\
b_{31} a_{11}+b_{32} a_{21} & b_{31} a_{12}+b_{32} a_{22} & b_{31} a_{13}+b_{32} a_{23}
\end{array}\right] .\)

Example 2 If A = \(\left[\begin{array}{rr}
2 & -1 \\
3 & 4 \\
1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
-1 & 3 \\
2 & 1
\end{array}\right]\), find AB. Does BA exist?

Solution

Here A is a 3 x 2 matrix and B is a 2 x 2 matrix.

Clearly, the number of columns in A equals the number of rows in B.

∴ AB exists and it is a 3 x 2 matrix.

Now, AB = \(\left[\begin{array}{rr}
2 & -1 \\
3 & 4 \\
1 & 5
\end{array}\right]\left[\begin{array}{rr}
-1 & 3 \\
2 & 1
\end{array}\right]\)

= \(\left[\begin{array}{rl}
2 \cdot(-1)+(-1) \cdot 2 & 2 \cdot 3+(-1) \cdot 1 \\
3 \cdot(-1)+4 \cdot 2 & 3 \cdot 3+4 \cdot 1 \\
1 \cdot(-1)+5 \cdot 2 & 1 \cdot 3+5 \cdot 1
\end{array}\right]\)

= \(\left[\begin{array}{rr}
-4 & 5 \\
5 & 13 \\
9 & 8
\end{array}\right]\)

Further, B is a 2 x 2 matrix and A is a 3 x 2 matrix. So, the number of columns in B is not equal to the number of rows in A.

So, BA does not exist.

Example 3 Let A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
2 & 3 \\
4 & 5 \\
-2 & 1
\end{array}\right]\). Find AB and BA,a nd show that AB ≠ BA.

Solution

Here A is a 2 x 3 matrix and B is a 3 x 2 matrix.

So, AB exists and it is a 2 x 2 matrix.

Now, AB = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\left[\begin{array}{rr}
2 & 3 \\
4 & 5 \\
-2 & 1
\end{array}\right]\)

= \(\left[\begin{array}{ll}
1 \cdot 2+(-2) \cdot 4+3 \cdot(-2) & 1 \cdot 3+(-2) \cdot 5+3 \cdot 1 \\
(-4) \cdot 2+2 \cdot 4+5 \cdot(-2) & (-4) \cdot 3+2 \cdot 5+5 \cdot 1
\end{array}\right]\)

= \(\left[\begin{array}{rr}
-12 & -4 \\
-10 & 3
\end{array}\right]\)

Again, B is a 3 x 2 matrix and A is a 2 x 3 matrix.

So, BA exists and it is a 3 x 3 matrix.

Now, BA = \(\left[\begin{array}{rr}
2 & 3 \\
4 & 5 \\
-2 & 1
\end{array}\right]\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
2 \cdot 1+3 \cdot(-4) & 2 \cdot(-2)+3 \cdot 2 & 2 \cdot 3+3 \cdot 5 \\
4 \cdot 1+5 \cdot(-4) & 4 \cdot(-2)+5 \cdot 2 & 4 \cdot 3+5 \cdot 5 \\
(-2) \cdot 1+1 \cdot(-4) & (-2) \cdot(-2)+1 \cdot 2 & (-2) \cdot 3+1 \cdot 5
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
-10 & 2 & 21 \\
-16 & 2 & 37 \\
-6 & 6 & -1
\end{array}\right] \text {. }\)

Hence, AB ≠ BA.

Properties of Matrix Multiplication

1. Commutativity

Matrix multiplication is not commutative in general.

Proof Let A and B be two given matrices.

If AB exists then it is quite possible that BA may not exist.

For example, if A is a 3 x 2 matrix and B is a 2 x 2 matrix then clearly, AB exists but BA does not exist.

Similarly, if BA exists then AB may not exist.

For example, if A is a 2 x 3 matrix an dB is a 2 x 2 matrix then clearly, BA exists but AB does not exist.

Further, if AB and BA both exist, then may not be comparable. For example, if A is a 2 x 3 matrix and B is a 3 x 2 matrix then clearly, AB as well ad BA exists. But, AB is a 2 x 2 matrix while BA is a 3 x 3 matrix.

Again, if AB and BA both exist and they are comparable, even then they may not be equal.

For example, if A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]\) then AB and BA are both defined and each one is a 2 x 2 matrix.

But, AB = \(\left[\begin{array}{ll}
1 & 5 \\
1 & 8
\end{array}\right]\) and BA = \(\left[\begin{array}{ll}
3 & 5 \\
3 & 6
\end{array}\right]\)

This shows that AB ≠ BA.

Hence, in general, AB ≠ BA.

Remarks (1) When AB = BA, we say that A and B commute.

(2) When AB = -BA, we say that A and B anticommute.

Properties of Matrices Overview

2. Associative law

For any matrices A, B, C for which (AB)C and A(BC) both exist, we have (AB)C = A(BC).

3. Distributive laws of multiplication over addition We have:

(1) A.(B + c) = (AB + AC)

(2) (A + B).C = (AC + BC)

4. The product of two nonzero matrices can be a zero matrix.

Example Let A = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 2 \\
0 & 0
\end{array}\right]\).

Then, A ≠ O and B ≠ O. But, AB = O.

Left Zero divisor If AB = O and A ≠ O then A is called a leftzerodivisior of AB.

Right Zero divisor If AB = O and B ≠ O then B is called a right zero divisor of AB.

5. If A is a given square matrix and I is an identity matrix of the same order as A then we have A.I = I.A = A.

6. If A is a given square matrix and O is the null matrix of the same order as A then O.A = A.O = O.

Positive Integral Powers of a Square Matrix

Let A be a square matrix of order n. Then, we define:

A2 = A.A;

A3 = A.A.A = A2.A;

A4 = A.A.A.A = A3.A, and so on.

∴ An = (A. A. A……n times).

Theorem 1 If A and B are square matrices of the same order then (A+B)2 = A2 + AB + BA + B2. Also, when AB = BA then (A+B)2 = A2 + 2AB + B2.

Proof Let A and B be n-rowed square matrices.

Then, clearly, (A + B) is a square matrix of order n.

So, (A + B)2 is defined.

Now, (A + B)2 = (A + B).(A + B)

= A.(A + B) + B.(A + B) [by distributive law]

= AA + AB + BA + BB [by distributive law]

=A2 + AB + BA + B2.

Hence, (A + B)2 = (A2 + AB + BA + B2).

Particular case When AB = BA

In this case, we have

(A + B)2 = (A2 + AB + AB + B2) = (A2 + 2AB + B2) [∵ BA = AB].

Theorem 2 If A and B are square matrices of the same order then (A + B)(A – B) = A2 – AB + BA – B2.

Also, when AB = BA then (A + B)(A – B) = A2 – B2.

Proof

We have

(A + B).(A – B) = A(A – B) + B(A – B) [by distributive law]

= AA – AB + BA – BB [∵ A(B – c) = AB – AC]

= A2 – AB + BA – B2.

Hence, (A + B)(A – B) = A2 – AB + BA – B2.

Particular case When AB = BA

In this case, (A + B)(A – B) = (A2 – B2) [∵ BA = AB].

Solved Examples

Example 1 If A = \(\left[\begin{array}{ll}
5 & 4 \\
2 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 5 & 1 \\
6 & 8 & 4
\end{array}\right]\), find AB and BA whichever exists.

Solution

Here, A is a 2 x 2 matrix and B is a 2 x 3 matrix.

Clearly, the number of columns in A = number of rows in B.

∴ AB exists and it is a 2 x 3 matrix.

AB = \(\left[\begin{array}{ll}
5 & 4 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
3 & 5 & 1 \\
6 & 8 & 4
\end{array}\right]\)

= \(\left[\begin{array}{lll}
5 \cdot 3+4 \cdot 6 & 5 \cdot 5+4 \cdot 8 & 5 \cdot 1+4 \cdot 4 \\
2 \cdot 3+3 \cdot 6 & 2 \cdot 5+3 \cdot 8 & 2 \cdot 1+3 \cdot 4
\end{array}\right]\)

= \(\left[\begin{array}{ccc}
15+24 & 25+32 & 5+16 \\
6+18 & 10+24 & 2+12
\end{array}\right]=\left[\begin{array}{lll}
39 & 57 & 21 \\
24 & 34 & 14
\end{array}\right]\)

Again, B is a 2 x 3 matrix and A is a 2 x 2 matrix.

∴ number of columns in B ≠ number of rows in A.

So, BA does not exist.

Example 2 If A = \(\left[\begin{array}{rrr}
2 & -1 & 3 \\
-4 & 5 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
2 & 3 \\
4 & -2 \\
1 & 5
\end{array}\right]\) then find AB and BA. Show that AB ≠ BA.

Solution

Here, A is a 2 x 3 matrix and B is a 3 x 2 matrix.

So, number of columns in A = number of rows in B.

∴ AB exists and it is a 2 x 2 matrix.

AB = \(\left[\begin{array}{rrr}
2 & -1 & 3 \\
-4 & 5 & 1
\end{array}\right]\left[\begin{array}{rr}
2 & 3 \\
4 & -2 \\
1 & 5
\end{array}\right]\)

= \(\left[\begin{array}{ll}
2 \cdot 2+(-1) \cdot 4+3 \cdot 1 & 2 \cdot 3+(-1) \cdot(-2)+3 \cdot 5 \\
-4 \cdot 2+5 \cdot 4+1 \cdot 1 & -4 \cdot 3+5 \cdot(-2)+1 \cdot 5
\end{array}\right]\)

= \(\left[\begin{array}{rr}
4-4+3 & 6+2+15 \\
-8+20+1 & -12-10+5
\end{array}\right]=\left[\begin{array}{rr}
3 & 23 \\
13 & -17
\end{array}\right] .\)

Again, B is a 3 x 2 matrix and A is a 2 x 3 matrix.

So, number of columns in B = number of rows in A.

∴ BA exists and it is a 3 x 3 matrix.

BA = \(\left[\begin{array}{rr}
2 & 3 \\
4 & -2 \\
1 & 5
\end{array}\right]\left[\begin{array}{rrr}
2 & -1 & 3 \\
-4 & 5 & 1
\end{array}\right]\)

= \(\left[\begin{array}{lll}
2 \cdot 2+3 \cdot(-4) & 2 \cdot(-1)+3 \cdot 5 & 2 \cdot 3+3 \cdot 1 \\
4 \cdot 2+(-2) \cdot(-4) & 4 \cdot(-1)+(-2) \cdot 5 & 4 \cdot 3+(-2) \cdot 1 \\
1 \cdot 2+5 \cdot(-4) & 1 \cdot(-1)+5 \cdot 5 & 1 \cdot 3+5 \cdot 1
\end{array}\right]\)

= \(\left[\begin{array}{llr}
4-12 & -2+15 & 6+3 \\
8+8 & -4-10 & 12-2 \\
2-20 & -1+25 & 3+5
\end{array}\right]=\left[\begin{array}{rrr}
-8 & 13 & 9 \\
16 & -14 & 10 \\
-18 & 24 & 8
\end{array}\right]\)

Clearly, AB ≠ BA.

Example 3 If A = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
3 & 2 & 0 \\
-2 & 0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-2 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{lll}
2 & 1 & -3 \\
3 & 0 & -1
\end{array}\right]\) then verify that (AB)C = A(BC).

Solution

We have

AB = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
3 & 2 & 0 \\
-2 & 0 & 1
\end{array}\right]\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-2 & 5
\end{array}\right]\)

= \(\left[\begin{array}{rr}
3-0-4 & 1-2+10 \\
9+0-0 & 3+4+0 \\
-6+0-2 & -2+0+5
\end{array}\right]=\left[\begin{array}{rr}
-1 & 9 \\
9 & 7 \\
-8 & 3
\end{array}\right]\)

⇒ (AB)C = \(\left[\begin{array}{rr}
-1 & 9 \\
9 & 7 \\
-8 & 3
\end{array}\right]\left[\begin{array}{lll}
2 & 1 & -3 \\
3 & 0 & -1
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
-2+27 & -1+0 & 3-9 \\
18+21 & 9+0 & -27-7 \\
-16+9 & -8+0 & 24-3
\end{array}\right]=\left[\begin{array}{rrr}
25 & -1 & -6 \\
39 & 9 & -34 \\
-7 & -8 & 21
\end{array}\right] \text {. }\)

Also, BC = \(\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-2 & 5
\end{array}\right]\left[\begin{array}{lll}
2 & 1 & -3 \\
3 & 0 & -1
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
6+3 & 3+0 & -9-1 \\
0+6 & 0+0 & 0-2 \\
-4+15 & -2+0 & 6-5
\end{array}\right]=\left[\begin{array}{rrr}
9 & 3 & -10 \\
6 & 0 & -2 \\
11 & -2 & 1
\end{array}\right]\)

⇒ (AB)C = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
3 & 2 & 0 \\
-2 & 0 & 1
\end{array}\right]\left[\begin{array}{rrr}
9 & 3 & -10 \\
6 & 0 & -2 \\
11 & -2 & 1
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
9-6+22 & 3-0-4 & -10+2+2 \\
27+12+0 & 9+0-0 & -30-4+0 \\
-18+0+11 & -6+0-2 & 20-0+1
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
25 & -1 & -6 \\
39 & 9 & -34 \\
-7 & -8 & 21
\end{array}\right]\)

Hence, (AB)C = A(BC).

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Example 4 If A = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
1 & -2 & 5 \\
0 & 7 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{rrr}
8 & 1 & -6 \\
2 & -5 & 0
\end{array}\right]\) verify that A(B + C) = (AB + AC).

Solution

We have

A(B + C) = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right] \cdot\left\{\left[\begin{array}{rrr}
1 & -2 & 5 \\
0 & 7 & 3
\end{array}\right]+\left[\begin{array}{rrr}
8 & 1 & -6 \\
2 & -5 & 0
\end{array}\right]\right\}\)

= \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right]\left[\begin{array}{rrr}
9 & -1 & -1 \\
2 & 2 & 3
\end{array}\right]\)

= \(\left[\begin{array}{lll}
3 \cdot 9+2 \cdot 2 & 3 \cdot(-1)+2 \cdot 2 & 3 \cdot(-1)+2 \cdot 3 \\
1 \cdot 9+0 \cdot 2 & 1 \cdot(-1)+0 \cdot 2 & 1 \cdot(-1)+0 \cdot 3
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
31 & 1 & 3 \\
9 & -1 & -1
\end{array}\right] \text {. }\)

Now, AB = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right]\left[\begin{array}{rrr}
1 & -2 & 5 \\
0 & 7 & 3
\end{array}\right]\)

= \(\left[\begin{array}{lll}
3 \cdot 1+2 \cdot 0 & 3 \cdot(-2)+2 \cdot 7 & 3 \cdot 5+2 \cdot 3 \\
1 \cdot 1+0 \cdot 0 & 1 \cdot(-2)+0 \cdot 7 & 1 \cdot 5+0 \cdot 3
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
3 & 8 & 21 \\
1 & -2 & 5
\end{array}\right]\)

And, AC = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right]\left[\begin{array}{rrr}
8 & 1 & -6 \\
2 & -5 & 0
\end{array}\right]\)

= \(\left[\begin{array}{lll}
3 \cdot 8+2 \cdot 2 & 3 \cdot 1+2 \cdot(-5) & 3 \cdot(-6)+2 \cdot 0 \\
1 \cdot 8+0 \cdot 2 & 1 \cdot 1+0 \cdot(-5) & 1 \cdot(-6)+0 \cdot 0
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
28 & -7 & -18 \\
8 & 1 & -6
\end{array}\right] .\)

∴ (AB + AC) = \(\left[\begin{array}{rrr}
3 & 8 & 21 \\
1 & -2 & 5
\end{array}\right]+\left[\begin{array}{rrr}
28 & -7 & -18 \\
8 & 1 & -6
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
31 & 1 & 3 \\
9 & -1 & -1
\end{array}\right]\)

Hence, A(B + C) = (AB + AC).

Applications of Matrices in Real Life

Example 5 Give an example of two matrices A and B such that A ≠ O, B ≠ O and AB = BA = O.

Solution

Let A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right]\). Then,

AB = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 \cdot 1+1 \cdot(-1) & 1 \cdot(-1)+1 \cdot 1 \\
1 \cdot 1+1 \cdot(-1) & 1 \cdot(-1)+1 \cdot 1
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)=0 .

BA = \(\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 \cdot 1+1 \cdot(-1) & 1 \cdot 1+1 \cdot(-1) \\
(-1) \cdot 1+1 \cdot 1 & (-1) \cdot 1+1 \cdot 1
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)=O

Hence, AB = BA = O.

Example 6 Let A = \(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\) and I is the identity matrix of order 2. Show that (I + A) = \((I-A) \cdot\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)

Solution

Let \(\tan \frac{\alpha}{2}=t\)

Then, \(\cos \alpha=\frac{1-\tan ^2(\alpha / 2)}{1+\tan ^2(\alpha / 2)}=\frac{1-t^2}{1+t^2}\)

and \(\sin \alpha=\frac{2 \tan (\alpha / 2)}{1+\tan ^2(\alpha / 2)}=\frac{2 t}{1+t^2}\)

∴ \((I+A)=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]+\left[\begin{array}{rr}
0 & -t \\
t & 0
\end{array}\right]=\left[\begin{array}{rr}
1 & -t \\
t & 1
\end{array}\right]\)

And, \((I-A)=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{rr}
0 & -t \\
t & 0
\end{array}\right]=\left[\begin{array}{rr}
1 & t \\
-t & 1
\end{array}\right]\)

∴ \((I-A) \cdot\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)

= \(\left[\begin{array}{rr}
1 & t \\
-t & 1
\end{array}\right]\left[\begin{array}{ll}
\frac{1-t^2}{1+t^2} & \frac{-2 t}{1+t^2} \\
\frac{2 t}{1+t^2} & \frac{1-t^2}{1+t^2}
\end{array}\right]\)

= \(\left[\begin{array}{cc}
\frac{1-t^2}{1+t^2}+\frac{2 t^2}{1+t^2} & \frac{-2 t}{1+t^2}+\frac{t\left(1-t^2\right)}{1+t^2} \\
\frac{-t\left(1-t^2\right)}{1+t^2}+\frac{2 t}{1+t^2} & \frac{2 t^2}{1+t^2}+\frac{1-t^2}{1+t^2}
\end{array}\right]\)

= \(\left[\begin{array}{rr}
1 & -t \\
t & 1
\end{array}\right]\)=(I+A) .

Hence, (I + A) = \((I-A) \cdot\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)

Example 7 If A = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) then prove that \(A^n=\left[\begin{array}{rr}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right], n \in N .\)

Solution

We shall prove the result by using the principle of mathematical induction.

When n = 1, we have

\(A^1=\left[\begin{array}{rr}
\cos 1 \cdot \theta & \sin 1 \cdot \theta \\
-\sin 1 \cdot \theta & \cos 1 \cdot \theta
\end{array}\right]=\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right] .\)

Thus, the result is true for n = 1.

Let the result be true for n = k.

Then, \(A^k=\left[\begin{array}{rr}
\cos k \theta & \sin k \theta \\
-\sin k \theta & \cos k \theta
\end{array}\right]\)

∴ A^{k+1}

= \(A \cdot A^k=\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{rr}
\cos k \theta & \sin k \theta \\
-\sin k \theta & \cos k \theta
\end{array}\right]\)

= \(\left[\begin{array}{rr}
\cos \theta \cos k \theta-\sin \theta \sin k \theta & \cos \theta \sin k \theta+\sin \theta \cos k \theta \\
-\sin \theta \cos k \theta-\cos \theta \sin k \theta & -\sin \theta \sin k \theta+\cos \theta \cos k \theta
\end{array}\right]\)

= \(\left[\begin{array}{rr}
\cos (\theta+k \theta) & \sin (\theta+k \theta) \\
-\sin (\theta+k \theta) & \cos (\theta+k \theta)
\end{array}\right]\)

= \(\left[\begin{array}{rr}
\cos (k+1) \theta & \sin (k+1) \theta \\
-\sin (k+1) \theta & \cos (k+1) \theta
\end{array}\right]\)

Thus, the result is true for n = (k+1), whenever it is true for n = k.

Hence, \(A^n=\left[\begin{array}{rr}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\) for all values of n ∈ N.

Example 8 Let A and B be symmetric matrices of the same order. Show that AB is symmetric if and only if AB = BA.

Solution

Since A and B are symmetric, we have A’ = A and B’ = B.

Let AB be symmetric. Then,

\((A B)^t=A B \Rightarrow B^t A^t=A B\left[(A B)^t=B^t A^t\right]\)

⇒ BA = AB [∵ B’ = B and a’ = A].

Thus, AB = BA.

Conversely, let AB = BA. Then,

AB = BA ⇒ \((A B)^t=(B A)^t=A^t B^t=A B\) [∵ A’ = A and B’ = B]

⇒ AB is symmetric.

Hence, AB is symmetric ⇔ AB = BA.

Example 9 If A and B are symmetric matrices, prove that (AB – BA) is skew-symmetric.

Solution

Since A and B are symmetric, we have At = A and Bt = B.

∴ \((A B-B A)^t=(A B)^t-(B A)^t\)

= \(B^t A^t-A^t B^t\) [∵ \((A B)^t=B^t A^t \text { and }(B A)^t=A^t B^t\)]

= BA – AB [∵ \(A^t=A \text { and } B^t=B\)]

= -(AB-BA).

This shows that (AB-BA) is skew-symmetric.

Example 10 Find the matrix A such that \(\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right]-A=\left[\begin{array}{rrr}
-1 & -8 & -10 \\
1 & -2 & -5 \\
9 & 22 & 15
\end{array}\right]\)

Solution

Clearly, the product is a (3×3) matrix and the prefactor is a (3×2) matrix. So, A must be a (2×3) matrix.

Let A = \(\left[\begin{array}{lll}
x & y & z \\
u & v & w
\end{array}\right]\)

Then, the given equation becomes

\(\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right] \cdot\left[\begin{array}{lll}
x & y & z \\
u & v & w
\end{array}\right]=\left[\begin{array}{rrr}
-1 & -8 & -10 \\
1 & -2 & -5 \\
9 & 22 & 15
\end{array}\right]\)

⇒ \(\left[\begin{array}{ccc}
2 x-u & 2 y-v & 2 z-w \\
x & y & z \\
-3 x+4 u & -3 y+4 v & -3 z+4 w
\end{array}\right]=\left[\begin{array}{ccc}
-1 & -8 & -10 \\
1 & -2 & -5 \\
9 & 22 & 15
\end{array}\right]\)

⇒ 2x – u = -1, 2y – v = -8, 2z – w = -10, x = 1, y = -2, z = -5

⇒ x = 1, y = -2, x = -5, u = 3, v = 4 and w = 0.

Hence, A = \(\left[\begin{array}{rrr}
1 & -2 & -5 \\
3 & 4 & 0
\end{array}\right]\)

Scalar Multiplication

Let A be a given matrix and k be a nonzero real number. Then, the matrix obtained by multiplying each element of A by k is called the scalar multiple of A by k and it is denoted by kA.

Here, k is called a scalar.

If A is an (mxn) matrix then kA is also an (mxn) matrix.

If A = [aij]mxn then kA = [k.aij]mxn.

Example 1 If A = \(\left[\begin{array}{rrr}
5 & 6 & -4 \\
8 & -3 & 2
\end{array}\right]\), find (1) 3A (2) -5A (3)\frac{1}{2}A.

Solution We have:

(1) 3A = \(\left[\begin{array}{ccc}
3 \times 5 & 3 \times 6 & 3 \times(-4) \\
3 \times 8 & 3 \times(-3) & 3 \times 2
\end{array}\right]=\left[\begin{array}{rrr}
15 & 18 & -12 \\
24 & -9 & 6
\end{array}\right]\)

(2) -5A = \(\left[\begin{array}{ccc}
(-5) \times 5 & (-5) \times 6 & (-5) \times(-4) \\
(-5) \times 8 & (-5) \times(-3) & (-5) \cdot 2
\end{array}\right]=\left[\begin{array}{rrr}
-25 & -30 & 20 \\
-40 & 15 & -10
\end{array}\right] \text {. }\)

(3) \(\frac{1}{2} A=\left[\begin{array}{lll}
\frac{1}{2} \times 5 & \frac{1}{2} \times 6 & \frac{1}{2} \times(-4) \\
\frac{1}{2} \times 8 & \frac{1}{2} \times(-3) & \frac{1}{2} \times 2
\end{array}\right]=\left[\begin{array}{rrr}
\frac{5}{2} & 3 & -2 \\
4 & -\frac{3}{2} & 1
\end{array}\right] \text {. }\)

Some Properties of Scalar Multiplication

Theorem 1 If A and B are two matrices of the same order and k is a scalar then prove that k(A+B) = kA + kB.

Proof Let A = [aij]mxn and B = [bij]mxn. Then,

k(A+B) = k.([aij]mxn + [bij]mxn)

= k.[aij + bij]mxn [by definition of addition of matrices]

= [k(aij+bij)]mxn [by definition of scalar multiplication]

= [k.aij + k.bij]mxn [by distributive law]

= kA + kB.

Hence, k(A + B) = kA + kB.

Theorem 2 If A is any matrix and k1, k2 are any scalars then prove that \(\left(k_1+k_2\right) A=k_1 A+k_2 A\).

Proof

Let A = [aij]mxn. Then,

\(\left(k_1+k_2\right) A=\left(k_1+k_2\right) \cdot\left[a_{i j}\right]_{m \times n}\)

= \(\left[\left(k_1+k_2\right) \cdot a_{i j}\right]_{m \times n}\) [by definition of scalar multiplication]

= \(\left[k_1 \cdot a_{i j}+k_2 \cdot a_{i j}\right]_{m \times n}\) [by distributive law]

= \(\left[k_1 \cdot a_{i j}\right]_{m \times n}+\left[k_2 \cdot a_{i j}\right]_{m \times n}\) [by definition of addition of matrices]

= \(k_1 A+k_2 A .\)

Hence, \(\left(k_1+k_2\right) A=k_1 A+k_2 A\).

Theorem 3 If A is any matrix and k1, k2 are any scalars then prove that \(k_1\left(k_2 A\right)=\left(k_1 k_2\right) A .\)

Proof

Let A = \(\left[a_{i j}\right]_{m \times n}\). Then,

\(k_1\left(k_2 A\right)=k_1\left[k_2 \cdot a_{i j}\right]_{m \times n}=\left[k_1\left(k_2 \cdot a_{i j}\right)\right]_{m \times n}\)

= \(\left[\left(k_1 k_2\right) \cdot a_{i j}\right]_{m \times n}\) [by associativity of multiplication in numbers]

= \(\left(k_1 k_2\right) \cdot\left[a_{i j}\right]_{m \times n}=\left(k_1 k_2\right) A .\)

Hence, \(k_1\left(k_2 A\right)=\left(k_1 k_2\right) A .\)

Real-Life Applications of Matrix Theory

Theorem 4 Let A be a given matrix and let k be a scalar. Then, prove that \((k A)^t=k A^t .\)

Proof

Let A = [aij]mxn be a given matrix and let k be a scalar. Then,

A is an (mxn) matrix ⇒ kA is an (mxn) matrix

⇒ \((k A)^t\) is an (nxm) matrix.

Also, A is an (mxn) matrix ⇒ A^t is an (nxm) matrix

⇒ \(kA^t\) is an (nxm) matrix.

Thus, \((k A)^t\) and \(\left(k A^t\right)\) are comparable matrices.

Also, (j,i)th element of \((k A)^t\) = (i,j)th element of kA

= k times (i,j)th element of \(A^t\)

= (j,i)th element of \((k A)^t\).

Thus, \((k A)^t and \left(k A^t\right)\) are comparable matrices whose corresponding elements are equal.

Hence, \((k A)^t=k A^t .\)

Theorem 5 Let A and B be two matrices such that AB exists and let c be a nonzero scalar. Then, prove that c(A x B) = (cA) x B = A x (cB).

Proof

Let A = [aij]mxn and B = [bjk]nxp. Then,

A is an (mxn) matrix, B is an (nxp) matrix

⇒ (A x B) is an (m x p) matrix

⇒ c(A x B) is an (m x p) matrix.

Again, A is an (m x n) matrix, B is an (n x p) matrix

⇒ cA is an (m x n) matrix, B is an (n x p) matrix

⇒ (cA) x B is an (m x p) matrix.

∴ c(A x B) and (cA) x B are comparable matrices.

Now, (i,k)th element of c(A x B)

= c time (i,k)th element of (A x B)

= \(c \cdot \sum_{j=1}^n a_{i j} b_{j k}\)

= \(c \cdot\left(a_{i 1} b_{1 k}+a_{i 2} b_{2 k}+\ldots+a_{i n} b_{n k}\right)\)

= \(\left(c a_{i 1}\right) b_{1 k}+\left(c a_{i 2}\right) b_{2 k}+\ldots+\left(c a_{i n}\right) b_{n k}\)

= (i,j)th element of (cA) x B.

Thus, c(A x B) are (cA) x B are comparable and their corresponding elements are equal.

Hence, c(A x B) = (cA x B).

Again, A is an (m x n) matrix and B is an (n x p) matrix

⇒ A is an (m x n) matrix and cB is an (n x p) matrix

⇒ A x (cB) is an (m x p) matrix.

Thus, c(A x B) and A x (cB) are comparable.

Also, (i,k)th element of A x (cB)

= \(\sum_{j=1}^n a_{i j}\left(c b_{j k}\right)\)

= \(a_{i 1}\left(c b_{1 k}\right)+a_{i 2}\left(c b_{2 k}\right)+\ldots+a_{i n}\left(c b_{n k}\right)\)

= \(c\left(a_{i 1} b_{1 k}+a_{i 2} b_{2 k}+\ldots+a_{i n} b_{n k}\right)\)

= \(c \sum_{j=1}^n a_{i j} b_{j k}\) = (i,k)th element of c(A x B).

∴ c(A x B) = A x (cB)

Hence, c(A x B) = (cA) x B = A x (cB).

Summary of properties of Matrix Operations

1. (1) \(\text { (i) }\left(A^t\right)^t=A\)

(2) \((A+B)^t=\left(A^t+B^t\right) .\)

(3) \((A \times B)^t \neq\left(A^t \times B^t\right) \text { and }(A \times B)^t \neq\left(B^t \times A^t\right) \text {. }\)

2. (1) (A + B) = (B + A)

(2) (A + B) + C = A + (B + C)

(3) For all comparable matrices A there exists a null matrix O such that A + O = O + A = A.

3. (1) (A x B) ≠ (B x A).

(2) (A x B) x C = A x (B x C)

(3) A x (B + C) = (A x B) + (A x C)

(4) (A + B) x C = (A x C) + (B x C)

(5) For all square matrices A of the same order there exists a unit matrix I such that (A x I) = (I x A) = A.

4. (1) (C A)^t=C A^{\prime} \text {. }

(2) c(A + B) = (cA + cB)

(3) c(A x B) = (cA x B) = A x (cB).

(4) ∃ a matrix S such that St = S(called a symmetric matrix)

More Solved Examples

Example 2 If A = \(\left[\begin{array}{rr}
3 & 5 \\
7 & -9
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
6 & -4 \\
2 & 3
\end{array}\right]\), find (4A – 3B).

Solution

We have: 4A – 3B = 4A + (-3)B.

Now, \(4 A=\left[\begin{array}{ll}
4 \times 3 & 4 \times 5 \\
4 \times 7 & 4 \times(-9)
\end{array}\right]=\left[\begin{array}{rr}
12 & 20 \\
28 & -36
\end{array}\right]\)

and (-3)B = \(\left[\begin{array}{ll}
(-3) \times 6 & (-3) \times(-4) \\
(-3) \times 2 & (-3) \times 3
\end{array}\right]=\left[\begin{array}{rr}
-18 & 12 \\
-6 & -9
\end{array}\right] .\)

∴ 4A – 3B = 4A + (-3)B-B

= \(\left[\begin{array}{rr}
12 & 20 \\
28 & -36
\end{array}\right]+\left[\begin{array}{rr}
-18 & 12 \\
-6 & -9
\end{array}\right]\)

= \(\left[\begin{array}{cc}
12+(-18) & 20+12 \\
28+(-6) & (-36)+(-9)
\end{array}\right]=\left[\begin{array}{rr}
-6 & 32 \\
22 & -45
\end{array}\right] \text {. }\)

Hence, (4A – 3B) = \(\left[\begin{array}{rr}
-6 & 32 \\
22 & -45
\end{array}\right]\)

Example 3 Let A = diag[3,-5,7] and B = diag[-1, 2,4]. Find (1) (A + B), (2) (A – B), (3) -5A, (4) (2A + 3B).

Solution

We have

\(A=\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & -5 & 0 \\
0 & 0 & 7
\end{array}\right] \text { and } B=\left[\begin{array}{rrr}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 4
\end{array}\right]\)

∴ (1) \(A+B=\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & -5 & 0 \\
0 & 0 & 7
\end{array}\right]+\left[\begin{array}{rrr}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 4
\end{array}\right]=\left[\begin{array}{rrr}
2 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & 11
\end{array}\right]\)

(2) (A – B) = A + (-B)

= \(\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & -5 & 0 \\
0 & 0 & 7
\end{array}\right]+\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & -4
\end{array}\right]=\left[\begin{array}{rrr}
4 & 0 & 0 \\
0 & -7 & 0 \\
0 & 0 & 3
\end{array}\right]\)

(3) -5A = (-5).A = \((-5) \cdot\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & -5 & 0 \\
0 & 0 & 7
\end{array}\right]=\left[\begin{array}{rrr}
-15 & 0 & 0 \\
0 & 25 & 0 \\
0 & 0 & -35
\end{array}\right]\)

(4) 2A + 3B = \(\left[\begin{array}{rrr}
6 & 0 & 0 \\
0 & -10 & 0 \\
0 & 0 & 14
\end{array}\right]+\left[\begin{array}{rrr}
-3 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 12
\end{array}\right]=\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & -4 & 0 \\
0 & 0 & 26
\end{array}\right]\)

Example 4 Simplify \(\cos \theta \cdot\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta \cdot\left[\begin{array}{rr}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right] \text {. }\)

Solution

We have

\(\cos \theta \cdot\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta \cdot\left[\begin{array}{rr}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right] \text {. }\)

= \(\left[\begin{array}{cc}
\cos ^2 \theta & \sin \theta \cos \theta \\
-\sin \theta \cos \theta & \cos ^2 \theta
\end{array}\right]+\left[\begin{array}{cc}
\sin ^2 \theta & -\sin \theta \cos \theta \\
\sin \theta \cos \theta & \sin ^2 \theta
\end{array}\right]\)

= \(\left[\begin{array}{cc}
\cos ^2 \theta+\sin ^2 \theta & \sin \theta \cos \theta+(-\sin \theta \cos \theta) \\
-\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^2 \theta+\sin ^2 \theta
\end{array}\right]\)

= \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] .\)

Example 5 If \(2\left[\begin{array}{cc}
x & 5 \\
7 & y-3
\end{array}\right]+\left[\begin{array}{ll}
3 & 4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rc}
7 & 14 \\
15 & 14
\end{array}\right]\), find the values of x and y.

Solution

We have

\(2\left[\begin{array}{cc}
x & 5 \\
7 & y-3
\end{array}\right]+\left[\begin{array}{ll}
3 & 4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rr}
7 & 14 \\
15 & 14
\end{array}\right]\)

⇒ \(\left[\begin{array}{cc}
2 x & 10 \\
14 & 2 y-6
\end{array}\right]+\left[\begin{array}{ll}
3 & 4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rc}
7 & 14 \\
15 & 14
\end{array}\right]\)

⇒ \(\left[\begin{array}{cc}
2 x+3 & 14 \\
15 & 2 y-4
\end{array}\right]=\left[\begin{array}{rr}
7 & 14 \\
15 & 14
\end{array}\right]\)

⇒ 2x + 3 = 7 and 2y – 4 = 14

⇒ x = 2 and y = 9.

The values of x and y are 3 and 9

Example 6 Find matrix X such that \(X+\left[\begin{array}{rr}
4 & 6 \\
-3 & 7
\end{array}\right]=\left[\begin{array}{ll}
3 & -6 \\
5 & -8
\end{array}\right]\)

Solution

Let A = \(\left[\begin{array}{rr}
4 & 6 \\
-3 & 7
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
3 & -6 \\
5 & -8
\end{array}\right]\)

Then, the given matrix equation is X + A = B.

Now, X + A = B ⇒ X = B – A = B + (-A)

= \(\left[\begin{array}{rr}
3 & -6 \\
5 & -8
\end{array}\right]+\left[\begin{array}{rr}
-4 & -6 \\
3 & -7
\end{array}\right]\)

= \(\left[\begin{array}{ll}
3+(-4) & -6+(-6) \\
5+3 & -8+(-7)
\end{array}\right]=\left[\begin{array}{rr}
-1 & -12 \\
8 & -15
\end{array}\right] \text {. }\)

Hence, X = \(\left[\begin{array}{rr}
-1 & -12 \\
8 & -15
\end{array}\right]\)

Example 7 Find a matrix X such that 2A + B + X = O, where A = \(\left[\begin{array}{rr}
-1 & 2 \\
3 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
3 & -2 \\
1 & 5
\end{array}\right] \text {. }\)

Solution

We have

2A + B + X = O ⇒ X = -(2A + B).

Now, (2A + B) = \(2 \cdot\left[\begin{array}{rr}
-1 & 2 \\
3 & 4
\end{array}\right]+\left[\begin{array}{rr}
3 & -2 \\
1 & 5
\end{array}\right]\)

= \(\left[\begin{array}{rr}
-2 & 4 \\
6 & 8
\end{array}\right]+\left[\begin{array}{rr}
3 & -2 \\
1 & 5
\end{array}\right]\)

= \(\left[\begin{array}{rl}
-2+3 & 4+(-2) \\
6+1 & 8+5
\end{array}\right]=\left[\begin{array}{rr}
1 & 2 \\
7 & 13
\end{array}\right]\)

∴ X = -(2A + B) = \(\left[\begin{array}{cc}
-1 & -2 \\
-7 & -13
\end{array}\right]\)

Example 8 Find matrices X and Y, if X + Y = \(\left[\begin{array}{ll}
5 & 2 \\
0 & 9
\end{array}\right]\) and X – Y = \(\left[\begin{array}{rr}
3 & 6 \\
0 & -1
\end{array}\right]\).

Solution

Adding the given matrices, we get

(X + Y) + (X – Y) = \(\left[\begin{array}{ll}
5 & 2 \\
0 & 9
\end{array}\right]+\left[\begin{array}{rr}
3 & 6 \\
0 & -1
\end{array}\right]\)

⇒ \(2 X=\left[\begin{array}{ll}
5+3 & 2+6 \\
0+0 & 9+(-1)
\end{array}\right]\)

⇒ \(2 X=\left[\begin{array}{ll}
8 & 8 \\
0 & 8
\end{array}\right]\) ⇒ \(X=\frac{1}{2} \cdot\left[\begin{array}{ll}
8 & 8 \\
0 & 8
\end{array}\right]=\left[\begin{array}{ll}
4 & 4 \\
0 & 4
\end{array}\right]\)

On substracting the given matrices, we get

(X + Y) – (X – Y)=\(\left[\begin{array}{ll}
5 & 2 \\
0 & 9
\end{array}\right]-\left[\begin{array}{rr}
3 & 6 \\
0 & -1
\end{array}\right]\)

⇒ \(2 Y=\left[\begin{array}{ll}
5-3 & 2-6 \\
0-0 & 9-(-1)
\end{array}\right]=\left[\begin{array}{ll}
2 & -4 \\
0 & 10
\end{array}\right]\)

⇒ \(Y=\frac{1}{2}\left[\begin{array}{rr}
2 & -4 \\
0 & 10
\end{array}\right]=\left[\begin{array}{rr}
1 & -2 \\
0 & 5
\end{array}\right] .\)

Hence, X = \(\left[\begin{array}{ll}
4 & 4 \\
0 & 4
\end{array}\right] and Y = \left[\begin{array}{rr}
1 & -2 \\
0 & 5
\end{array}\right] \text {. }\)

Matrix Polynomial Let f(x) = \(a_0 x^m+a_1 x^{m-1}+a_2 x^{m-2}+\ldots+a_{m-1} x+a_m\) be a polynomial of degree m and let A be a square matrix of order n. Then, the corresponding matrix polynomial is:

f(A) = \(a_0 A^{m+}+a_1 A^{m-1}+a_2 A^{m-2}+\ldots+a_{m-1} A+a_m I\), where I is a unit matrix of order n.

Example 9 If f(x) = \(x^2-5 x+7\) and A = \(\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\), find f(A).

Solution

f(x) = \(x^2-5 x+7\) ⇒ f(A) = \(A^2-5 A+7 I\)

Now, \(A^2=\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\)

= \(\left[\begin{array}{rr}
9-1 & 3+2 \\
-3-2 & -1+4
\end{array}\right]=\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]\)

-5A = \(\left[\begin{array}{ll}
(-5) \cdot 3 & (-5) \cdot 1 \\
(-5) \cdot(-1) & (-5) \cdot 2
\end{array}\right]=\left[\begin{array}{rr}
-15 & -5 \\
5 & -10
\end{array}\right]\)

7I = \(7 \cdot\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right] .\)

∴ f(A) = \(A^2-5 A+7 I\)

= \(\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]+\left[\begin{array}{rr}
-15 & -5 \\
5 & -10
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right]\)

= \(\left[\begin{array}{cc}
8+(-15)+7 & 5+(-5)+0 \\
-5+5+0 & 3+(-10)+7
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

Hence, f(A) = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=0 .\)

Example 10 If A = \(\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]\), show that \(A^2-5 A-14 I=0 .\)

Solution

We have

\(A^2=\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]\)

= \(\left[\begin{array}{cc}
3 \cdot 3+(-5)(-4) & 3 \cdot(-5)+(-5) \cdot 2 \\
(-4) \cdot 3+2 \cdot(-4) & (-4) \cdot(-5)+2 \cdot 2
\end{array}\right]=\left[\begin{array}{rr}
29 & -25 \\
-20 & 24
\end{array}\right]\)

-5A = \((-5)\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{rr}
-15 & 25 \\
20 & -10
\end{array}\right] \text {; }\)

-14I = \((-14)\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
-14 & 0 \\
0 & -14
\end{array}\right] \text {. }\)

∴ \(A^2-5 A-14 I=A^2+(-5) A+(-14 I)\)

= \(\left[\begin{array}{rr}
29 & -25 \\
-20 & 24
\end{array}\right]+\left[\begin{array}{rr}
-15 & 25 \\
20 & -10
\end{array}\right]+\left[\begin{array}{rr}
-14 & 0 \\
0 & -14
\end{array}\right]\)

= \(\left[\begin{array}{cc}
29+(-15)+(-14) & -25+25+0 \\
-20+20+0 & 24+(-10)+(-14)
\end{array}\right]\)

= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=0 \text {. }\)

Hence, \(A^2-5 A-14 I=0 .\)

Examples of Matrix Operations

Example 11 If A = \(\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]\), find k so that \(A^2=8 A+k I\).

Solution

We have

\(A^2=\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]=\left[\begin{array}{rl}
1-0 & 0+0 \\
-1-7 & 0+49
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-8 & 49
\end{array}\right]\)

(8A + kI) = \(8 \cdot\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]+k \cdot\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

= \(\left[\begin{array}{rr}
8 & 0 \\
-8 & 56
\end{array}\right]+\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]=\left[\begin{array}{cc}
8+k & 0 \\
-8 & 56+k
\end{array}\right]\)

∴ \(A^2=8 A+k I\) ⇒ \(\left[\begin{array}{rr}
1 & 0 \\
-8 & 49
\end{array}\right]=\left[\begin{array}{cc}
8+k & 0 \\
-8 & 56+k
\end{array}\right]\)

⇒ 8 + k = 1 and 56 + k = 49

⇒ k = -7.

Hence, k = -7.

Example 12 If A = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\), prove that for all n ∈ N, \((a I+b A)^n=a^n I+n d^{n-1} b A\), where I is the identity matrix of order 2.

Solution

We shall prove the result by mathematical induction.

When n = 1, we have:

LHS = \((a I+b A)^1=(a I+b A)=\left(a^1 I+1 a^0 b A\right)\) = RHS.

So, the result is true for n = 1.

Let it be true for n = m, so that

\((a I+b A)^m=a^m I+m a^{m-1} b A\) …(1)

∴ \((a I+b A)^{m+1}\)

= \((a I+b A) \cdot(a I+b A)^m=(a I+b A) \cdot\left(a^m I+m a^{m-1} b A\right)\) [using(1)]

= \(a I\left(a^m I+m a^{m-1} b A\right)+b A\left(a^m I+m a^{m-1} b A\right)\)

= \(a^{m+1} I+m d^m b A+a^m b A+m a^{m-1} b^2 A^2\) [∵ II = I, IA = A = AI]

= \(a^{m+1} I+(m+1) a^m b A\) [∵ \(A^2=\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right] \times\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=O]\).

Thus, whenever the result is true for n = m, then it is also true for n = (m + 1).

Hence, by mathematical induction, it is true for all n ∈ N.

Theorem 6 If A is a symmetric matrix, then prove that kA is symmetric.

Proof Since A is symmetric, we have \(A^t=A .\)

∴ \((k A)^t=k A^t=k A\) [∵ \(A^t=A\)].

Hence, (kA) is symmetric.

Theorem 7 If A is a skew-symmetric matrix, then prove that kA is skew-symmetric.

Proof Since A is skew-symmetric, we have \(A^t=-A .\)

∴ \((k A)^t=k \cdot A^t=k \cdot(-A)=-(k A)\) [∵ \(A^t=-A\)].

Hence, (kA) is skew-symmetric.

Theorem 8 Prove that every square matrix is expressible as the sum of a symmetric matrix and a skew-symmetric matrix.

Proof Let A be any square matrix. Then, we can write

\(A=\frac{1}{2}\left(A+A^t\right)+\frac{1}{2}\left(A-A^t\right)=P+Q \text { (say). }\)

Then, it is easy to verify that P is symmetric and Q is skew-symmetric.

Hence, the theorem follows.

Example 13 Express the matrix A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\) as the sum of a symmetric matrix and a skew-symmetric matrix.

Solution

We know that \(A=\frac{1}{2}\left(A+A^{\dagger}\right)+\frac{1}{2}\left(A-A^{\dagger}\right)=P+Q\), where P is a symmetric and Q is skew-symmetric.

∴ \(P=\frac{1}{2}\left(A+A^t\right)=\frac{1}{2} \cdot\left\{\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]+\left[\begin{array}{rr}
3 & 1 \\
-4 & -1
\end{array}\right]\right\}\)

= \(\frac{1}{2} \cdot\left[\begin{array}{cc}
3+3 & -4+1 \\
1+(-4) & -1+(-1)
\end{array}\right]=\frac{1}{2} \cdot\left[\begin{array}{rr}
6 & -3 \\
-3 & -2
\end{array}\right]=\left[\begin{array}{cc}
3 & \frac{-3}{2} \\
\frac{-3}{2} & -1
\end{array}\right]\)

And, \(Q=\frac{1}{2}\left(A-A^t\right)=\frac{1}{2} \cdot\left\{\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]-\left[\begin{array}{rr}
3 & 1 \\
-4 & -1
\end{array}\right]\right\}\)

= \(\frac{1}{2} \cdot\left[\begin{array}{cc}
3-3 & -4-1 \\
1-(-4) & -1-(-1)
\end{array}\right]=\frac{1}{2} \cdot\left[\begin{array}{cc}
0 & -5 \\
1+4 & -1+1
\end{array}\right]\)

= \(\frac{1}{2} \cdot\left[\begin{array}{rr}
0 & -5 \\
5 & 0
\end{array}\right]=\left[\begin{array}{rr}
0 & -\frac{5}{2} \\
\frac{5}{2} & 0
\end{array}\right] \text {. }\)

Hence, A = P + Q, where P is symmetric and Q is skew-symmetric.

Example 14 Express the matrix A = \(\left[\begin{array}{rrr}
1 & 3 & 5 \\
-6 & 8 & 3 \\
-4 & 6 & 5
\end{array}\right]\) as the sum of a symmetric matrix and a skew-symmetric matrix.

Solution

We know that \(A=\frac{1}{2}\left(A+A^t\right)+\frac{1}{2}\left(A-A^t\right)=P+Q\), where P is symmetric and Q is skew-symmetric.

Now, \(P=\frac{1}{2}\left(A+A^t\right)\)

= \(\frac{1}{2} \cdot\left\{\left[\begin{array}{rrr}
1 & 3 & 5 \\
-6 & 8 & 3 \\
-4 & 6 & 5
\end{array}\right]+\left[\begin{array}{rrr}
1 & -6 & -4 \\
3 & 8 & 6 \\
5 & 3 & 5
\end{array}\right]\right\}\)

= \(\frac{1}{2} \cdot\left[\begin{array}{rrr}
2 & -3 & 1 \\
-3 & 16 & 9 \\
1 & 9 & 10
\end{array}\right]=\left[\begin{array}{rrr}
1 & -\frac{3}{2} & \frac{1}{2} \\
-\frac{3}{2} & 8 & \frac{9}{2} \\
\frac{1}{2} & \frac{9}{2} & 5
\end{array}\right]\)

And, \(Q=\frac{1}{2}\left(A-A^t\right)\)

= \(\frac{1}{2} \cdot\left\{\left[\begin{array}{rrr}
1 & 3 & 5 \\
-6 & 8 & 3 \\
-4 & 6 & 5
\end{array}\right]-\left[\begin{array}{rrr}
1 & -6 & -4 \\
3 & 8 & 6 \\
5 & 3 & 5
\end{array}\right]\right\}\)

= \(\frac{1}{2} \cdot\left[\begin{array}{rrr}
0 & 9 & 9 \\
-9 & 0 & -3 \\
-9 & 3 & 0
\end{array}\right]=\left[\begin{array}{rrr}
0 & \frac{9}{2} & \frac{9}{2} \\
-\frac{9}{2} & 0 & -\frac{3}{2} \\
-\frac{9}{2} & \frac{3}{2} & 0
\end{array}\right]\)

Hence, A = P + Q, where P is symmetric and Q is skew-symmetric.

Elementary Operations on Matrices

Given below are three row operations and three column operations on a matrix, which are called elementary operations or transformations.

Equivalent Matrices Two matrices A and B are said to be equivalent if one is obtained from the other by one or more elementary operations and we write, A ~ B.

Three Elementary Row Operations

(1) Interchange of any two rows The interchange of ith and jth rows is denoted by Ri ⟷ Rj.

Example Let A = \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
\sqrt{2} & 4 & 6 \\
5 & -3 & 7
\end{array}\right]\)

Applying R2 ⟷ R3, we get A ~ \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
5 & -3 & 7 \\
\sqrt{2} & 4 & 6
\end{array}\right] \text {. }\)

(2) Multiplication of the elements of a row by a nonzero number Suppose each element of ith row of a given matrix is multiplied by a nonzero number k.

Then, we denote it by Ri → kRi.

Example Let A = \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
\sqrt{3} & -5 & 6 \\
1 & 8 & 4
\end{array}\right]\)

Applying R2 → 4R2, we get A ~ \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
4 \sqrt{3} & -20 & 24 \\
1 & 8 & 4
\end{array}\right]\)

(3) Multiplying each element of a row by a nonzero number and then adding them to the corresponding elements of another row Suppose each element of jth row of a matrix A is multiplied by a nonzero number k and then added to the corresponding elements of ith row.

We dentoe it by Ri → Ri + kRj.

Example Let A = \(\left[\begin{array}{rrr}
2 & -1 & 5 \\
-3 & 4 & \sqrt{2} \\
7 & 6 & 3
\end{array}\right]\)

Applying R1 → R1 + 2R3, we get A ~ \(\left[\begin{array}{rrr}
16 & 11 & 11 \\
-3 & 4 & \sqrt{2} \\
7 & 6 & 3
\end{array}\right]\)

Three Elementary Column Operations

(1) Interchange of any two columns The interchange of ith and jth columns is denoted by \(c_i \leftrightarrow c_j .\)

Example Let A = \(\left[\begin{array}{rrr}
2 & 1 & -3 \\
-1 & 5 & 4 \\
6 & 3 & \frac{1}{2}
\end{array}\right]\)

Applying \(c_1 \leftrightarrow c_2\), we get A ~ \(\left[\begin{array}{rrr}
1 & 2 & -3 \\
5 & -1 & 4 \\
3 & 6 & \frac{1}{2}
\end{array}\right] \text {. }\)

(2) Multiplying each element of a column by a nonzero number Suppose each element of ith column of matrix A is multiplied by a nonzero number k.

Then, we write, \(\mathcal{C}_i \rightarrow k \mathcal{C}_i\)

Example Let A = \(\left[\begin{array}{rrr}
3 & 1 & -5 \\
\sqrt{2} & -2 & 4 \\
6 & 2 & 8
\end{array}\right]\)

Applying \(\mathcal{C}_3 \rightarrow 2 \mathcal{C}_3\), we get A ~ \(\left[\begin{array}{rrr}
3 & 1 & -10 \\
\sqrt{2} & -2 & 8 \\
6 & 2 & 16
\end{array}\right]\)

(3) Multiplying each element of a column of a given matrix A by a nonzero number and then adding to the corresponding elements of another column Suppose each element of ith column of a given matrix A is multiplied by a nonzero number k and then added to the corresponding elements of ith column.

Then we write, \(C_i \rightarrow C_i+k C_j .\)

Example Let A = \(\left[\begin{array}{rrr}
2 & 0 & 4 \\
-1 & 3 & 1 \\
5 & -2 & 6
\end{array}\right]\)

Applying \(C_3 \rightarrow C_3+2 C_1\), we get A ~ \(\left[\begin{array}{rrr}
2 & 0 & 8 \\
-1 & 3 & -1 \\
5 & -2 & 16
\end{array}\right]\)

Invertible Matrices A square matrix A of order n is said to be invertible if there exists a square matrix B of order n such that AB = BA = I.

Also, then B is called the inverse of A and we write, \(A^{-1}=B\).

Example Let A = \(\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
2 & -5 \\
-1 & 3
\end{array}\right]\). Then,

AB = \(\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]\left[\begin{array}{rr}
2 & -5 \\
-1 & 3
\end{array}\right]=\left[\begin{array}{rr}
6-5 & -15+15 \\
2-2 & -5+6
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I\),

BA = \(\left[\begin{array}{rr}
2 & -5 \\
-1 & 3
\end{array}\right]\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rr}
6-5 & 10-10 \\
-3+3 & -5+6
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I\)

∴ AB = BA = I.

Hence, \(A^{-1}=B\).

Theorem 1 (Uniqueness of Inverse) Every invertible square matrix has a unique inverse.

Proof

Let A be an invertible square matrix of order n.

If possible, let B as well as C be the inverse of A.

Then, AB = BA = I and AC = CA = I.

Now, AC = I ⇒ B(AC) = B.I = B,

BA = I ⇒ (BA)C = I.C = C.

But, B(AC) = (BA)C [by associative law of multiplication]

∴ B = C.

Hence, an invertible matrix has a unique inverse.

Inverse Of A Matrix By Elementary Row Operations

Let A be a square matrix of order n.

We can write, A = I.A …(1)

Now, let a sequence of elementary row operations reduce A on LHS of (1) to I and I on RHS of (1) to a matrix B.

Then, I = BA ⇒ \(I \cdot A^{-1}=(B A) A^{-1}=B\left(A A^{-1}\right)=B I\)

⇒ \(A^{-1}=B\).

We can summarise the above method as given below.

Method Step 1. Write A = I.A.

Step 2. By using elementary row operations on A, transform it into a unit matrix.

Step 3. In the same order we apply elementary operations on I to convert it into a matrix B.

Step 4. Then, \(A^{-1}=B\).

Remark If on applying one or more elementary row operations on A, we obtain all zeros in one or more rows, then we say that \(A^{-1}\) does not exist.

Solved Examples

Example 1 By using elementary row operations, find the inverse of the matrix A = \(\left[\begin{array}{ll}
1 & -2 \\
2 & -6
\end{array}\right]\)

Solution

We have

\(\left[\begin{array}{cc}
1 & -2 \\
2 & -6
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \cdot A\)

⇒ \(\left[\begin{array}{ll}
1 & -2 \\
0 & -2
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-2 & 1
\end{array}\right] \cdot A\)

\(\left[R_2 \rightarrow R_2-2 R_1\right]\)

⇒ \(\left[\begin{array}{rr}
1 & -2 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
1 & \frac{-1}{2}
\end{array}\right] \cdot A\)

\(\left[R_2 \rightarrow\left(\frac{-1}{2}\right) R_2\right]\)

⇒ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
3 & -1 \\
1 & -\frac{1}{2}
\end{array}\right] \cdot A\)

\(\left[R_1 \rightarrow R_1+2 R_2\right] .\)

Hence, \(A^{-1}=\left[\begin{array}{rr}
3 & -1 \\
1 & -\frac{1}{2}
\end{array}\right]\)

Example 2 By using elementary row operations, find the inverse of the matrix A = \(\left[\begin{array}{rr}
3 & -1 \\
-4 & 2
\end{array}\right]\)

Solution

We have

A = \(\left[\begin{array}{rr}
3 & -1 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \cdot A\)

⇒ \(\left[\begin{array}{cc}
-1 & 1 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \cdot A\)

\(\left[R_1 \rightarrow R_1+R_2\right]\)

⇒ \(\left[\begin{array}{rr}
1 & -1 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{rr}
-1 & -1 \\
0 & 1
\end{array}\right] \cdot A\)

\(\left[R_1 \rightarrow(-1) \cdot R_1\right]\)

⇒ \(\left[\begin{array}{ll}
1 & -1 \\
0 & -2
\end{array}\right]=\left[\begin{array}{cc}
-1 & -1 \\
-4 & -3
\end{array}\right] \cdot A\)

\(\left[R_2 \rightarrow R_2+4 R_1\right]\)

⇒ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
-1 & -1 \\
2 & \frac{3}{2}
\end{array}\right] \cdot A\)

\(\left[R_2 \rightarrow\left(\frac{-1}{2}\right) R_2\right]\)

⇒ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & \frac{1}{2} \\
2 & \frac{3}{2}
\end{array}\right] \cdot A\)

\(\left[R_1 \rightarrow R_1+R_2\right] .\)

Hence, \(A^{-1}=\left[\begin{array}{ll}
1 & \frac{1}{2} \\
2 & \frac{3}{2}
\end{array}\right] \text {. }\)

Example 3 If A = \(\left[\begin{array}{rr}
6 & -3 \\
-2 & 1
\end{array}\right]\), show that \(A^{-1}\) does not exist.

Solution

We have

\(\left[\begin{array}{rr}
6 & -3 \\
-2 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \cdot A\)

⇒ \(\left[\begin{array}{rr}
1 & -\frac{1}{2} \\
-2 & 1
\end{array}\right]=\left[\begin{array}{ll}
\frac{1}{6} & 0 \\
0 & 1
\end{array}\right] \cdot A\)

\(\left[R_1 \rightarrow \frac{1}{6} R_1\right]\)

⇒ \(\left[\begin{array}{rr}
1 & -\frac{1}{2} \\
0 & 0
\end{array}\right]=\left[\begin{array}{rr}
\frac{1}{6} & 0 \\
\frac{1}{3} & 1
\end{array}\right] \cdot A\)

\(\left[R_2 \rightarrow R_2+2 R_1\right]\)

Thus, we have all zeros in second row of the left-hand side matrix. Hence, \(A^{-1}\) does not exist.

Example 4 By using elementary row operations, find the inverse of the matrix A = \(\left[\begin{array}{rrr}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)

Solution

We have

\(\left[\begin{array}{rrr}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)

⇒ \(\left[\begin{array}{rrr}
1 & 3 & -2 \\
0 & 9 & -11 \\
0 & -1 & 4
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
3 & 1 & 0 \\
-2 & 0 & 1
\end{array}\right] \cdot A\)

\(\left[\begin{array}{l}
R_2 \rightarrow R_2+3 R_1 \\
R_3 \rightarrow R_3-2 R_1
\end{array}\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & 3 & -2 \\
0 & -1 & 4 \\
0 & 9 & -11
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
-2 & 0 & 1 \\
3 & 1 & 0
\end{array}\right] \cdot A\)

\(\left[R_2 \leftrightarrow R_3\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & 0 & 10 \\
0 & -1 & 4 \\
0 & 0 & 25
\end{array}\right]=\left[\begin{array}{rrr}
-5 & 0 & 3 \\
-2 & 0 & 1 \\
-15 & 1 & 9
\end{array}\right] \cdot A\)

\(\left[\begin{array}{l}
R_1 \rightarrow R_1+3 R_2 \\
R_3 \rightarrow R_3+9 R_2
\end{array}\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & 0 & 10 \\
0 & 1 & -4 \\
0 & 0 & 25
\end{array}\right]=\left[\begin{array}{rrr}
-5 & 0 & 3 \\
2 & 0 & -1 \\
-15 & 1 & 9
\end{array}\right] \cdot A\)

\(\left[R_2 \rightarrow(-1) \cdot R_2\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & 0 & 10 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-5 & 0 & 3 \\
2 & 0 & -1 \\
\frac{-3}{5} & \frac{1}{25} & \frac{9}{25}
\end{array}\right]\)

\(\left[R_3 \rightarrow \frac{1}{25} R_3\right]\)

⇒ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & \frac{-2}{5} & \frac{-3}{5} \\
\frac{-2}{5} & \frac{4}{25} & \frac{11}{25} \\
\frac{-3}{5} & \frac{1}{25} & \frac{9}{25}
\end{array}\right] \cdot A\)

\(\left[\begin{array}{l}
R_1 \rightarrow R_1-10 R_3 \\
R_2 \rightarrow R_2+4 R_3
\end{array}\right]\)

Hence, \(A^{-1}=\left[\begin{array}{ccc}
1 & \frac{-2}{5} & \frac{-3}{5} \\
\frac{-2}{5} & \frac{4}{25} & \frac{11}{25} \\
\frac{-3}{5} & \frac{1}{25} & \frac{9}{25}
\end{array}\right] \text {. }\)

Example 5 By using elementary row operations, find the inverse of the matrix A = \(\left[\begin{array}{rrr}
3 & -1 & -2 \\
2 & 0 & -1 \\
3 & -5 & 0
\end{array}\right] .\)

Solution

We have

\(\left[\begin{array}{rrr}
3 & -1 & -2 \\
2 & 0 & -1 \\
3 & -5 & 0
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)

⇒ \(\left[\begin{array}{rrr}
1 & -1 & -1 \\
2 & 0 & -1 \\
3 & -5 & 0
\end{array}\right]=\left[\begin{array}{rrr}
1 & -1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)

\(\left[R_1 \rightarrow R_1-R_2\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & -1 & -1 \\
0 & 2 & 1 \\
0 & -2 & 3
\end{array}\right]=\left[\begin{array}{rrr}
1 & -1 & 0 \\
-2 & 3 & 0 \\
-3 & 3 & 1
\end{array}\right] \cdot A\)

\(\left[\begin{array}{l}
R_2 \rightarrow R_2-2 R_1 \\
R_3 \rightarrow R_3-3 R_1
\end{array}\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & -1 & -1 \\
0 & 1 & \frac{1}{2} \\
0 & -2 & 3
\end{array}\right]=\left[\begin{array}{rrr}
1 & -1 & 0 \\
-1 & \frac{3}{2} & 0 \\
-3 & 3 & 1
\end{array}\right] \cdot A\)

\(\left\{R_2 \rightarrow \frac{1}{2} R_2\right\}\)

⇒ \(\left[\begin{array}{ccc}
1 & 0 & \frac{-1}{2} \\
0 & 1 & \frac{1}{2} \\
0 & 0 & 4
\end{array}\right]=\left[\begin{array}{rrr}
0 & \frac{1}{2} & 0 \\
-1 & \frac{3}{2} & 0 \\
-5 & 6 & 1
\end{array}\right] \cdot A\)

\(\left\{\begin{array}{l}
R_1 \rightarrow R_1+R_2 \\
R_3 \rightarrow R_3+2 R_2
\end{array}\right\}\)

⇒ \(\left[\begin{array}{ccc}
1 & 0 & \frac{-1}{2} \\
0 & 1 & \frac{1}{2} \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
0 & \frac{1}{2} & 0 \\
-1 & \frac{3}{2} & 0 \\
\frac{-5}{4} & \frac{3}{2} & \frac{1}{4}
\end{array}\right] \cdot A\)

\(\left\{R_3 \rightarrow \frac{1}{4} R_3\right\}\)

⇒ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
\frac{-5}{8} & \frac{5}{4} & \frac{1}{8} \\
\frac{-3}{8} & \frac{3}{4} & \frac{-1}{8} \\
\frac{-5}{4} & \frac{3}{2} & \frac{1}{4}
\end{array}\right] \cdot A\)

\(\left\{\begin{array}{l}
R_1 \rightarrow R_1+\frac{1}{2} R_3 \\
R_2 \rightarrow R_2-\frac{1}{2} R_3
\end{array}\right\} .\)

Hence, \(A^{-1}=\left[\begin{array}{ccc}
\frac{-5}{8} & \frac{5}{4} & \frac{1}{8} \\
\frac{-3}{8} & \frac{3}{4} & \frac{-1}{8} \\
\frac{-5}{4} & \frac{3}{2} & \frac{1}{4}
\end{array}\right] \text {. }\)

Example 6 By using elementary row transformations, find the invers of the matrix A = \(\left[\begin{array}{rrr}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)

Solution

We have

\(\left[\begin{array}{rrr}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)

⇒ \(\left[\begin{array}{rrr}
2 & 0 & -1 \\
1 & 1 & 2 \\
0 & 1 & 3
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
-2 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)

\(\left[R_2 \rightarrow R_2-2 R_1\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & 1 & 2 \\
2 & 0 & -1 \\
0 & 1 & 3
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)

\(\left[R_1 \leftrightarrow R_2\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & 1 & 2 \\
0 & -2 & -5 \\
0 & 1 & 3
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 1 & 0 \\
5 & -2 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)

\(\left[R_2 \rightarrow R_2-2 R_1\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & 1 & 2 \\
0 & 1 & 3 \\
0 & -2 & -5
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 1 & 0 \\
0 & 0 & 1 \\
5 & -2 & 0
\end{array}\right] \cdot A\)

\(\left[R_2 \leftrightarrow R_3\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & 0 & -1 \\
0 & 1 & 3 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 1 & -1 \\
0 & 0 & 1 \\
5 & -2 & 2
\end{array}\right] \cdot A\)

\(\left[\begin{array}{l}
R_1 \rightarrow R_1-R_2 \\
R_3 \rightarrow R_3+2 R_2
\end{array}\right]\)

⇒ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right] \cdot A\)

\(\left[\begin{array}{l}
R_1 \rightarrow R_1+R_3 \\
R_2 \rightarrow R_2-3 R_3
\end{array}\right]\)

Hence, \(A^{-1}=\left[\begin{array}{rrr}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)

Example 7 If A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & -1 \\
-1 & -2 & 2
\end{array}\right]\), show that \(A^{-1}\) does not exist.

Solution

We have

\(\left[\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & -1 \\
-1 & -2 & 2
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)

⇒ \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
0 & 3 & -3 \\
0 & -3 & 3
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
-2 & 1 & 0 \\
1 & 0 & 1
\end{array}\right], A\)

\(\left[\begin{array}{l}
R_2 \rightarrow R_2-2 R_1 \\
R_3 \rightarrow R_3+R_1
\end{array}\right]\)

⇒ \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
0 & 3 & -3 \\
0 & 0 & 0
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
-2 & 1 & 0 \\
-1 & 1 & 1
\end{array}\right] \cdot A\)

\(\left[R_3 \rightarrow R_3+R_2\right] .\)

Thus, we have all zeros in 3rd row of the left-hand side matrix.

Hence, A^{-1} does not exist.

Class 12 Maths Solutions For Determinants

Determinant of a Square Matrix

Corresponding to each square Matrix

A = \(\left[\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & \ldots & a_{1 n} \\
a_{21} & a_{22} & a_{23} & \ldots & a_{2 n} \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
a_{n 1} & a_{n 2} & a_{n 3} & \ldots & a_{n n}
\end{array}\right]\)

there is associated an expression, called the determinat of A, denoted by det A or |A|, written associated

det A = |A| = \(\left|\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & \ldots & a_{1 n} \\
a_{21} & a_{22} & a_{23} & \ldots & a_{2 n} \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
a_{n 1} & a_{n 2} & a_{n 3} & \ldots & a_{n n}
\end{array}\right| .\)

A matrix is an arrangement of numbers and so it has no fixed value, while eac determinant has a fixed value.

A determinant having n rows and n columns is known as a determinant of order n.

The determiants of nonsquare matrices are not defined.

Value Of A Determinant Of Order 1 The value of a determinant of a (1×1) matrix [a] is defined as |a| = a.

Read and Learn More  Class 12 Math Solutions

Value Of A Determinant of Order 2 We defined

\(\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right|=\left(a_{11} a_{22}-a_{21} a_{12}\right) .\)

Example 1 Evaluate:

(1) \(\left|\begin{array}{ll}
6 & -3 \\
7 & -2
\end{array}\right|\)

(2) \(\left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|\)

Solution

(1) \(\left|\begin{array}{ll}
6 & -3 \\
7 & -2
\end{array}\right|=6(-2)-7(-3)=-12+21=9 \text {. }\)

(2) \(\left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|=\left(x^2-x+1\right)(x+1)-(x+1)(x-1)\)

= \(\left(x^3+1\right)-\left(x^2-1\right)=\left(x^3-x^2+2\right)\)

Example 2 Show that \(\left|\begin{array}{rr}
\sin 10^{\circ} & -\cos 10^{\circ} \\
\sin 80^{\circ} & \cos 80^{\circ}
\end{array}\right|=1 .\)

Solution

We have

\(\left|\begin{array}{rr}
\sin 10^{\circ} & -\cos 10^{\circ} \\
\sin 80^{\circ} & \cos 80^{\circ}
\end{array}\right|=\left(\sin 10^{\circ} \cos 80^{\circ}\right)-\left(-\sin 80^{\circ} \cos 10^{\circ}\right)\)

= \(\left(\sin 10^{\circ} \cos 80^{\circ}+\sin 80^{\circ} \cos 10^{\circ}\right)\)

= \(\sin \left(10^{\circ}+80^{\circ}\right)=\sin 90^{\circ}=1 .\)

Example 3 If \(\left|\begin{array}{rr}
4 & m \\
-3 & 5
\end{array}\right|=8\), find the value of m.

Solution

\(\left|\begin{array}{rr}
4 & m \\
-3 & 5
\end{array}\right|=8\) ⇔ 20 + 3m = 8

⇔ 3m = -12.

⇔ m = -4.

The value of m = -4.

Value Of A Determinant Of Order 3 Or More To find the value of a determinant of order 3 or more, we need the following definitions.

Minor Of aij in |A| The minor of an element aij in |A| is defined as the value of the determinant obtained by deleting the ith row and jth column of |A|, and it is denoted by Mij.

Cofactor Of aij In |A| The cofactor Cij of an element aij is defined as \(C_{i j}=(-1)^{i+j} \cdot M_{i j}\)

Example 1 Find the minors and cofactors of the elements of the determinant \(\Delta=\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\)

Solution

Let Mij denote the minor of aij in Δ.

Now, a11 occurs in the 1st row and 1st column. So, in order to find the minor of a11, we delete the 1st row and 1st column of Δ. The minor M11 of a11 is given by,

\(M_{11}=\left|\begin{array}{ll}
a_{22} & a_{23} \\
a_{32} & a_{33}
\end{array}\right|=\left(a_{22} a_{33}-a_{32} a_{23}\right) \text {. }\)

Similarly, we have

\(M_{12}=\left|\begin{array}{ll}
a_{21} & a_{23} \\
a_{31} & a_{33}
\end{array}\right|=\left(a_{21} a_{33}-a_{31} a_{23}\right) ;\) \(M_{13}=\left|\begin{array}{ll}
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array}\right|=\left(a_{21} a_{32}-a_{31} a_{22}\right) ;\) \(M_{21}=\left|\begin{array}{ll}
a_{12} & a_{13} \\
a_{32} & a_{33}
\end{array}\right|=\left(a_{12} a_{33}-a_{32} a_{13}\right) .\)

Similarly, we may obtain the minor of each of the remaining elements.

Now, if we denotre the cofactor of aij by cij then

\(C_{11}=(-1)^{1+1} \cdot M_{11}=M_{11}=\left(a_{22} a_{33}-a_{32} a_{23}\right) ;\) \(C_{12}=(-1)^{1+2} \cdot M_{12}=-M_{12}=\left(a_{31} a_{23}-a_{21} a_{33}\right) ;\) \(C_{13}=(-1)^{1+3} \cdot M_{13}=M_{13}=\left(a_{21} a_{32}-a_{31} a_{22}\right) ;\) \(C_{21}=(-1)^{2+1} \cdot M_{21}=-M_{21}=\left(a_{32} a_{13}-a_{12} a_{33}\right)\)

Similarly, the cofactor of each of the remaining elements of Δ can be determined.

WBBSE Class 12 Determinants Solutions

Example 2 Find the minor and cofactor of each element of Δ = \(\left|\begin{array}{rrr}
1 & -3 & 2 \\
4 & -1 & 2 \\
3 & 5 & 2
\end{array}\right| \text {. }\)

Solution

The minors of the elements of Δ are given by

\(M_{11}=\left|\begin{array}{rr}
-1 & 2 \\
5 & 2
\end{array}\right|=-12 ; M_{12}=\left|\begin{array}{ll}
4 & 2 \\
3 & 2
\end{array}\right|=2 ; M_{13}=\left|\begin{array}{rr}
4 & -1 \\
3 & 5
\end{array}\right|=23\) ;

\(M_{21}=\left|\begin{array}{rr}
-3 & 2 \\
5 & 2
\end{array}\right|=-16 ; M_{22}=\left|\begin{array}{rr}
1 & 2 \\
3 & 2
\end{array}\right|=-4 ; M_{23}=\left|\begin{array}{rr}
1 & -3 \\
3 & 5
\end{array}\right|=14\) ;

\(M_{31}=\left|\begin{array}{ll}
-3 & 2 \\
-1 & 2
\end{array}\right|=-4 ; M_{32}=\left|\begin{array}{cc}
1 & 2 \\
4 & 2
\end{array}\right|=-6 ; M_{33}=\left|\begin{array}{cc}
1 & -3 \\
4 & -1
\end{array}\right|=11 \text {. }\)

So, the cofactors of the corresponding elements of Δ are

\(C_{11}=(-1)^{1+1} \cdot M_{11}=M_{11}=-12 ; C_{12}=(-1)^{1+2} \cdot M_{12}=-M_{12}=-2\) ;

\(C_{13}=(-1)^{1+3} \cdot M_{13}=M_{13}=23 ; C_{21}=(-1)^{2+1} \cdot M_{21}=-M_{21}=16\)

\(C_{22}=(-1)^{2+2} \cdot M_{22}=M_{22}=-4 ; C_{23}=(-1)^{2+3} \cdot M_{23}=-M_{23}=-14\) ;

\(C_{31}=(-1)^{3+1} \cdot M_{31}=M_{31}=-4 ; C_{32}=(-1)^{3+2} \cdot M_{32}=-M_{32}=6\) ;

\(C_{33}=(-1)^{3+3} \cdot M_{33}=M_{33}=11 .\)

Value of a Determinant

The value of a determinant is the sum of the products of elements of a row (or a column) with their corresponding cofactors.

We may expand a determinant by any arbitrarily chosen row or column.

Expansion of a Determinant

Expanding the given determinant by 1st row, we have

\(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|=a_{11} \cdot \text { (its cofactor) }+a_{12} \cdot \text { (its cofactor) }+a_{13} \cdot \text { (its cofactor) }\)

= \(a_{11} c_{11}+a_{12} c_{12}+a_{13} c_{13}\)

= \(a_{11} M_{11}=a_{12} M_{12}+a_{13} M_{13}\) [∵ \(C_{12}=-M_{12}\)]

= \(a_{11} \cdot\left|\begin{array}{ll}
a_{22} & a_{23} \\
a_{32} & a_{33}
\end{array}\right|-a_{12} \cdot\left|\begin{array}{ll}
a_{21} & a_{23} \\
a_{31} & a_{33}
\end{array}\right|+a_{13} \cdot\left|\begin{array}{ll}
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array}\right|\)

= \(a_{11} \cdot\left(a_{22} a_{33}-a_{32} a_{23}\right)-a_{12} \cdot\left(a_{21} a_{33}-a_{31} a_{23}\right)+a_{13} \cdot\left(a_{21} a_{32}-a_{31} a_{22}\right) .\)

We may expand it by any row or column.

Remark 1 If we expand a determinant by any row or column using minors, we keep in view the following symbols for a determinant of order theree:

\(\left|\begin{array}{lll}
+ & – & + \\
– & + & – \\
+ & – & +
\end{array}\right|\)

Remark 2 If a row or a column of a determinant consists of all zeros, the value of the determinant is zero.

Example 1 Evaluate △ = \(\left|\begin{array}{rrr}
3 & 4 & 5 \\
-6 & 2 & -3 \\
8 & 1 & 7
\end{array}\right|\)

Solution

Expanding the given determinant by 1st row, we get

△ = \(3 \cdot\left|\begin{array}{rr}
2 & -3 \\
1 & 7
\end{array}\right|-4 \cdot\left|\begin{array}{rr}
-6 & -3 \\
8 & 7
\end{array}\right|+5 \cdot\left|\begin{array}{rr}
-6 & 2 \\
8 & 1
\end{array}\right|\)

= 3(14 + 3) – 4(-42 + 24) + 5.(-6-16) = 13.

Example 2 Expand the determinant △ = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right| .\)

Solution

Expanding by 1st row, we have

△ = \(a \cdot\left|\begin{array}{ll}
b & f \\
f & c
\end{array}\right|-h \cdot\left|\begin{array}{ll}
h & f \\
g & c
\end{array}\right|+g \cdot\left|\begin{array}{ll}
h & b \\
g & f
\end{array}\right|\)

= \(a\left(b c-f^2\right)-h \cdot(c h-f g)+g \cdot(f h-b g)\)

= \(\left(a b c+2 f g h-a f^2-b g^2-c h^2\right) .\)

Properties of Determinants

The properties of a determinant serve the purpose of a useful tool for finding its value. We will mention these properties and verify them for a third-order determinant.

Theorem 1 The value of a determinant remains unchanged if its rows and columns are interchanged.

Proof

Let △ = \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) and let △’ be the determinant obtained by interchanging the rows and columns of △.

Then, △’ = \(\left|\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|\)

= \(a_1 \cdot\left|\begin{array}{ll}
b_2 & b_3 \\
c_2 & c_3
\end{array}\right|-b_1 \cdot\left|\begin{array}{ll}
a_2 & a_3 \\
c_2 & c_3
\end{array}\right|+c_1 \cdot\left|\begin{array}{ll}
a_2 & a_3 \\
b_2 & b_3
\end{array}\right|\)

[expansion by 1st column]

= \(a_1\left(b_2 c_3-b_3 c_2\right)-b_1\left(a_2 c_3-c_2 a_3\right)+c_1\left(a_2 b_3-a_3 b_2\right)\)

= △ [expanded by 1st row].

Corollary If A is a square matrix then |A’| = |A|.

Understanding Determinants in Class 12 Maths

Theorem 2 If two rows or columns of a determinant are interchanged then the determinant retains its absolute value but its sign is changed.

Proof

Let △ = \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) and let △’ be the determinant obtained by interchanging any two rows, say 1st and 3rd rows, of △. Then,

△’ = \(\left|\begin{array}{lll}
a_3 & b_3 & c_3 \\
a_2 & b_2 & c_2 \\
a_1 & b_1 & c_1
\end{array}\right|\)

= \(a_3 \cdot\left|\begin{array}{ll}
b_2 & c_2 \\
b_1 & c_1
\end{array}\right|-a_2 \cdot\left|\begin{array}{ll}
b_3 & c_3 \\
b_1 & c_1
\end{array}\right|+a_1 \cdot\left|\begin{array}{ll}
b_3 & c_3 \\
b_2 & c_2
\end{array}\right|\) [expanded by 1st column]

= \(a_3\left(b_2 c_1-b_1 c_2\right)-a_2\left(b_3 c_1-b_1 c_3\right)+a_1\left(b_3 c_2-b_2 c_3\right)\)

= \(-a_1\left(b_2 c_3-b_3 c_2\right)+a_2\left(b_1 c_3-b_3 c_1\right)-a_3\left(b_1 c_2-b_2 c_1\right)\)

= \(-\left[a_1\left(b_2 c_3-b_3 c_2\right)-a_2\left(b_1 c_3-b_3 c_1\right)+a_3\left(b_1 c_2-b_2 c_1\right)\right]\)

= -△ [expanded by 1st column].

Theorem 3 If any two rows or columns of a determinant are identical then its value is zero.

Proof

If we interchange the identical rows of the given determinant △. But, interchanging any two rows of a determinant changes its sign.

∴ △ = -△ ⇔ 2△ = 0, i.e., △ = 0.

Theorem 4 If each element of a row or a column of a determinant is multiplied by a constant k then the value of a new determinant is k times the value of the original determinant.

Proof

Let △ = \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) and let △’ be the determinant obtained by multiplying each element of a row, say the second row of △, by k. Then,

△’ = \(\left|\begin{array}{ccc}
a_1 & b_1 & c_1 \\
k a_2 & k b_2 & k c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\)

= \(a_1 \cdot\left|\begin{array}{cc}
k b_2 & k c_2 \\
b_3 & c_3
\end{array}\right|-k a_2 \cdot\left|\begin{array}{cc}
b_1 & c_1 \\
b_3 & c_3
\end{array}\right|+a_3 \cdot\left|\begin{array}{cc}
b_1 & c_1 \\
k b_2 & k c_2
\end{array}\right|\)

= \(a_1\left(k b_2 c_3-k b_3 c_2\right)-k a_2\left(b_1 c_3-b_3 c_1\right)+a_3\left(k b_1 c_2-k b_2 c_1\right)\)

= \(k\left[a_1\left(b_2 c_3-b_3 c_2\right)-a_2\left(b_1 c_3-b_3 c_1\right)+a_3\left(b_1 c_2-b_2 c_1\right)\right]=k \Delta\)

Corollary For a square matrix A of order n, \(|k A| = k^n \cdot|A| .\)

Theorem 5 If any two rows or columns of a determinant are proportional then its value is zero.

Proof

Let △ = \(\left|\begin{array}{ccc}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
k a_1 & k b_1 & k c_1
\end{array}\right|\)

= \(k\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_1 & b_1 & c_1
\end{array}\right|\)

= k x 0 = 0 [∵ 1st and 3rd rows are identical].

Theorem 6 If each element of a two (or column) of a determinant is expressed as sum of two or more terms then the determinant can be expressed as the sum of two or more determinants.

Proof

Let △ = \(\left|\begin{array}{ccc}
a_1+\alpha_1 & b_1+\beta_1 & c_1+\gamma_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\)

Then, on expanding △ by first row, we get

△ = \(\left(a_1+\alpha_1\right)\left(b_2 c_3-b_3 c_2\right)-\left(b_1+\beta_1\right)\left(a_2 c_3-a_3 c_2\right)+\left(c_1+\gamma_1\right)\left(a_2 b_3-a_3 b_2\right)\)

= \(\left[a_1\left(b_2 c_3-b_3 c_2\right)-b_1\left(a_2 c_3-a_3 c_2\right)+c_1\left(a_2 b_3-a_3 b_2\right)\right]\)
\(+\left[\alpha_1\left(b_2 c_3-b_3 c_2\right)-\beta_1\left(a_2 c_3-a_3 c_2\right)+\gamma_1\left(a_2 b_3-a_3 b_2\right)\right]\)

= \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|+\left|\begin{array}{lll}
\alpha_1 & \beta_1 & \gamma_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right| .\)

Theorem 7 If to any row or column of a determinant, a multiple of another row or column is added, the value of the determinant remains the same.

Proof

Let △ = \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right| .\)

Let △’ = \(\left|\begin{array}{ccc}
a_1+k a_3 & b_1+k b_3 & c_1+k c_3 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right| .\)

Then, △’ = \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|+\left|\begin{array}{ccc}
k a_3 & k b_3 & k c_3 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\)

= \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) [∵ 2nd det. is 0, its 1st and 3rd rows being proportional]

Theorem 8 The sum of the products of the elements of any row (or column) of a determinant with cofactors of the corresponding elements of any other row (or column) is zero.

Proof

Let △ = \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right| .\)

Let us take the sum of the products of elements of first row with the cofactors of the corresponding elements of second row.

\(a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23}\)

= \(a_{11} \cdot(-1)^{2+1} \cdot\left|\begin{array}{ll}
a_{12} & a_{13} \\
a_{32} & a_{33}
\end{array}\right|+a_{12} \cdot(-1)^{2+2} \cdot\left|\begin{array}{ll}
a_{11} & a_{13} \\
a_{31} & a_{33}
\end{array}\right|+a_{13} \cdot(-1)^{2+3} \cdot\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{31} & a_{32}
\end{array}\right|\)

= \(-a_{11}\left(a_{12} a_{33}-a_{32} a_{13}\right)+a_{12}\left(a_{11} a_{33}-a_{31} a_{13}\right)-a_{13}\left(a_{11} a_{32}-a_{31} a_{12}\right)=0 .\)

To Find the Value of a Determinant

The main theme behind the simplification of a determinant lies in obtaining the maximum possible number of zeros in a row (or a column) by using the above properties and then to expand the determinant by that row (or column). We denote the 1st, 2nd, 3rd rows of a determinant by R1, R2, R3 respectively and the 1st, 2nd, 3rd columns by C1, C2, C3 respectively. We shall also express the

(1) interchange of the ith and jth rows by Ri ⟷ Rj;

(2) addition of k times the elements of the jth row with the corresponding elements of the ith row by Ri → Ri + k Rj.

We use similar notations for operations on columns, replacing R by C.

Solved Examples

Example 1 Evaluate \(\left|\begin{array}{rrr}
9 & 9 & 12 \\
1 & -3 & -4 \\
1 & 9 & 12
\end{array}\right| \text {. }\)

Solution

Let the given determinat be △. Then,

△ = \(\left|\begin{array}{rrr}
9 & 9 & 12 \\
1 & -3 & -4 \\
1 & 9 & 12
\end{array}\right|\)

= \(\left|\begin{array}{rrr}
0 & 36 & 48 \\
1 & -3 & -4 \\
0 & 12 & 16
\end{array}\right|\)

\(\left\{\begin{array}{l}
R_1 \rightarrow R_1-9 R_2 \\
R_3 \rightarrow R_3-R_2
\end{array}\right\}\)

= \((12 \times 4) \cdot\left|\begin{array}{rrr}
0 & 3 & 4 \\
1 & -3 & -4 \\
0 & 3 & 4
\end{array}\right|\) {taking out 12 common from R1 and 4 common from R3}

= (48 x 0) = 0 [∵ R1 and R3 are identical].

Hence, △ = 0.

Example 2 Evaluate \(\left|\begin{array}{lll}
265 & 240 & 219 \\
240 & 225 & 198 \\
219 & 198 & 181
\end{array}\right| .\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{lll}
265 & 240 & 219 \\
240 & 225 & 198 \\
219 & 198 & 181
\end{array}\right|\)

= \(\left|\begin{array}{lll}
46 & 21 & 219 \\
42 & 27 & 198 \\
38 & 17 & 181
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1-C_3\right) \text { and } C_2 \rightarrow\left(C_2-C_3\right)\right]\)

= \(\left|\begin{array}{rrr}
4 & 21 & 9 \\
-12 & 27 & -72 \\
4 & 17 & 11
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1-2 C_2\right) \text { and } C_3 \rightarrow\left(C_3-10 C_2\right)\right]\)

= \(\left|\begin{array}{rrr}
0 & 4 & -2 \\
0 & 78 & -39 \\
4 & 17 & 11
\end{array}\right|\)

\(\left[R_1 \rightarrow\left(R_1-R_3\right) \text { and } R_2 \rightarrow\left(R_2+3 R_3\right)\right]\)

= \(\text { 2(39) }\left|\begin{array}{rrr}
0 & 2 & -1 \\
0 & 2 & -1 \\
4 & 17 & 11
\end{array}\right|\)

[taking 2 common from R1 and 39 common from R2]

= (78 x 0) = 0 [∵ R1 and R2 are identical].

Step-by-Step Solutions for Determinant Problems

Example 3 Without expanding, prove that

\(\left|\begin{array}{ccc}
\sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\
\sin \beta & \cos \beta & \cos (\beta+\delta) \\
\sin \gamma & \cos \gamma & \cos (\gamma+\delta)
\end{array}\right|=0\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
\sin \alpha & \cos \alpha & \cos \alpha \cos \delta-\sin \alpha \sin \delta \\
\sin \beta & \cos \beta & \cos \beta \cos \delta-\sin \beta \sin \delta \\
\sin \gamma & \cos \gamma & \cos \gamma \cos \delta-\sin \gamma \sin \delta
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
\sin \alpha & \cos \alpha & 0 \\
\sin \beta & \cos \beta & 0 \\
\sin \gamma & \cos \gamma & 0
\end{array}\right|\)

\(\left[C_3 \rightarrow C_3+(\sin \delta) C_1-(\cos \delta) C_2\right]\)

= 0 [expanded by C3].

Hence, △ = 0.

Example 4 Prove that \(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
2 a & 3 a+2 b & 4 a+3 b+2 c \\
3 a & 6 a+3 b & 10 a+6 b+3 c
\end{array}\right|=a^3\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
2 a & 3 a+2 b & 4 a+3 b+2 c \\
3 a & 6 a+3 b & 10 a+6 b+3 c
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
0 & a & 2 a+b \\
0 & 3 a & 7 a+3 b
\end{array}\right|\)

\(\left[R_2 \rightarrow R_2-2 R_1 \text { and } R_3 \rightarrow R_3-3 R_1\right]\)

= \(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
0 & a & 2 a+b \\
0 & 0 & a
\end{array}\right|\)

\(\left[R_3 \rightarrow R_3-3 R_2\right]\)

= \(a \cdot\left|\begin{array}{cc}
a & 2 a+b \\
0 & a
\end{array}\right|\) [expanding along C1]

= \(a \cdot\left(a^2-0\right)=a^3 .\)

Hence, △ = \(a^3\).

Example 5 Without expanding, prove that \(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|=0 .\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
x+y+z & x+y+z & x+y+z \\
z & x & y \\
1 & 1 & 1
\end{array}\right|\)

\(\left[R_1 \rightarrow\left(R_1+R_2\right)\right]\)

= \((x+y+z) \cdot\left|\begin{array}{lll}
1 & 1 & 1 \\
z & x & y \\
1 & 1 & 1
\end{array}\right|\) [taking (x+y+z) common from R1]

= (x + y + z) x 0 = 0 [∵ R1 and R3 are identical].

Hence, △ = 0.

Example 6 Using properties of determinants, prove that \(\left|\begin{array}{ccc}
x+y & x & x \\
5 x+4 y & 4 x & 2 x \\
10 x+8 y & 8 x & 3 x
\end{array}\right|=x^3 \text {. }\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
x+y & x & x \\
5 x+4 y & 4 x & 2 x \\
10 x+8 y & 8 x & 3 x
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
x+y & x & x \\
3 x+2 y & 2 x & 0 \\
7 x+5 y & 5 x & 0
\end{array}\right|\)

\(\left[R_2 \rightarrow\left(R_2-2 R_1\right) \text { and } R_3 \rightarrow\left(R_3-3 R_1\right)\right]\)

= \(x \cdot\left|\begin{array}{ll}
3 x+2 y & 2 x \\
7 x+5 y & 5 x
\end{array}\right|\) [expanded by C3]

= \(x^2 \cdot\left|\begin{array}{ll}
3 x+2 y & 2 \\
7 x+5 y & 5
\end{array}\right|\) [taking x common from C2]

= \(x^2 \cdot[5(3 x+2 y)-2(7 x+5 y)]=\left(x^2 \cdot x\right)=x^3 .\)

Hence, △ = \(x^3\)

Example 7 Using properties of determinants, prove that \(\left|\begin{array}{ccc}
0 & a b^2 & a c^2 \\
a^2 b & 0 & b c^2 \\
a^2 c & c b^2 & 0
\end{array}\right|=2 a^3 b^3 c^3 \text {, }\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
0 & a b^2 & a c^2 \\
a^2 b & 0 & b c^2 \\
a^2 c & c b^2 & 0
\end{array}\right|\)

= \(\left(a^2 b^2 c^2\right) \cdot\left|\begin{array}{lll}
0 & a & a \\
b & 0 & b \\
c & c & 0
\end{array}\right|\)

\(\left[\begin{array}{l}
\text { taking } a^2, b^2, c^2 \text { common from } C_1, C_2 \\
\text { and } C_3 \text { respectively }
\end{array}\right]\)

= \(\left(a^3 b^3 c^3\right) \cdot\left|\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right|\)

\(\left[\begin{array}{l}
\text { taking } a, b, c \text { common from } R_1, R_2 \\
\text { and } R_3 \text { respectively }
\end{array}\right]\)

= \(\left(a^3 b^3 c^3\right) \cdot\left|\begin{array}{rrr}
0 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & -1
\end{array}\right|\)

\(\left[R_3 \rightarrow R_3-R_2\right]\)

= \(\left(a^3 b^3 c^3\right) \cdot(-1) \cdot\left|\begin{array}{rr}
1 & 1 \\
1 & -1
\end{array}\right|\) [expanded by C1]

= \(-\left(a^3 b^3 c^3\right)(-1-1)=2 a^3 b^3 c^3\)

Hence, △ = \(2 a^3 b^3 c^3\)

Example 8 Prove that \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|=4 a^2 b^2 c^2\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|\)

= \((a b c) \cdot\left|\begin{array}{rrr}
-a & a & a \\
b & -b & b \\
c & c & -c
\end{array}\right|\)

\(\left[\begin{array}{l}
\text { taking out } a, b, c \text { common from } \\
C_1, C_2, C_3 \text { respectively }
\end{array}\right]\)

= \((a b c) \cdot\left|\begin{array}{ccc}
-a & 0 & 0 \\
b & 0 & 2 b \\
c & 2 c & 0
\end{array}\right|\)

\(\left[C_2 \rightarrow\left(C_2+C_1\right) \text { and } C_3 \rightarrow\left(C_3+C_1\right)\right]\)

= \((a b c) \cdot(-a)\left|\begin{array}{cc}
0 & 2 b \\
2 c & 0
\end{array}\right|=(a b c)(-a)(-4 b c)\)

= \(4 a^2 b^2 c^2\).

Example 9 Using properties of determinants, show that \(\left|\begin{array}{ccc}
a+x & y & z \\
x & a+y & z \\
x & y & a+z
\end{array}\right|=a^2(a+x+y+z)\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
a+x & y & z \\
x & a+y & z \\
x & y & a+z
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
a+x+y+z & y & z \\
a+x+y+z & a+y & z \\
a+x+y+z & y & a+z
\end{array}\right|\)

\(\left[C_1 \rightarrow C_1+C_2+C_3\right]\)

= \((a+x+y+z) \cdot\left|\begin{array}{ccc}
1 & y & z \\
1 & a+y & z \\
1 & y & a+z
\end{array}\right|\)

[taking (a + x + y + z) common from C1]

=\((a+x+y+z) \cdot\left|\begin{array}{lll}
1 & y & z \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right|\)

\(\left[R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3-R_1\right]\)

= \((a+x+y+z) \cdot 1 \cdot\left|\begin{array}{ll}
a & 0 \\
0 & a
\end{array}\right|\)

[expanded by C1]

= \(a^2(a+x+y+z) \text {. }\)

Hence, △ = \(a^2(a+x+y+z)\).

Key Properties of Determinants Explained

Example 10 Using Properties of determinants, prove that \(\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|=(a+b+c)^3 \text {. }\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\)

\(\left[R_1 \rightarrow\left(R_1+R_2+R_3\right)\right]\)

= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\) [taking (a + b + c) common from R1]

= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
2 b & -(a+b+c) & 0 \\
2 c & 0 & -(a+b+c)
\end{array}\right|\)

\(\left[C_2 \rightarrow\left(C_2-C_1\right) \text { and } C_3 \rightarrow\left(C_3-C_1\right)\right]\)

= \((a+b+c) \cdot 1 \cdot\left|\begin{array}{cc}
-(a+b+c) & 0 \\
0 & -(a+b+c)
\end{array}\right|\)

= \((a+b+c)(a+b+c)^2=(a+b+c)^3 .\)

Hence, △ = \((a+b+c)^3 .\)

Example 11 Using properties of determinants, prove that \(\left|\begin{array}{ccc}
a & b & c \\
a-b & b-c & c-a \\
b+c & c+a & a+b
\end{array}\right|=\left(a^3+b^3+c^3-3 a b c\right)\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
a & b & c \\
a-b & b-c & c-a \\
b+c & c+a & a+b
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
a+b+c & b & c \\
0 & b-c & c-a \\
2(a+b+c) & c+a & a+b
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1+C_2+C_3\right)\right]\)

= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & b & c \\
0 & b-c & c-a \\
2 & c+a & a+b
\end{array}\right|\) [taking (a + b + c) common from C1]

= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & b & c \\
0 & b-c & c-a \\
0 & c+a-2 b & a+b-2 c
\end{array}\right|\)

\(\left[R_3 \rightarrow R_3-2 R_1\right]\)

= \((a+b+c) \cdot 1 \cdot\left|\begin{array}{cc}
b-c & c-a \\
c+a-2 b & a+b-2 c
\end{array}\right|\) [expanded by C1]

= \((a+b+c) \cdot\left|\begin{array}{ll}
b-c & c-a \\
a-b & b-c
\end{array}\right| \quad\left[R_2 \rightarrow R_2+R_1\right]\)

= \((a+b+c) \cdot\left[(b-c)^2-(a-b)(c-a)\right]\)

= \((a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\)

= \(\left(a^3+b^3+c^3-3 a b c\right)\)

Hence, △ = \(\left(a^3+b^3+c^3-3 a b c\right)\).

Example 12 Using properties of determinants, prove that \(\left|\begin{array}{ccc}
3 a & -a+b & -a+c \\
a-b & 3 b & c-b \\
a-c & b-c & 3 c
\end{array}\right|=3(a+b+c)(a b+b c+c a) \text {. }\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
3 a & -a+b & -a+c \\
a-b & 3 b & c-b \\
a-c & b-c & 3 c
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
a+b+c & -a+b & -a+c \\
a+b+c & 3 b & c-b \\
a+b+c & b-c & 3 c
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1+C_2+C_3\right)\right]\)

= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & -a+b & -a+c \\
1 & 3 b & c-b \\
1 & b-c & 3 c
\end{array}\right|\)

= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & -a+b & -a+c \\
0 & 2 b+a & a-b \\
0 & a-c & 2 c+a
\end{array}\right|\)

\(\left[R_2 \rightarrow\left(R_2-R_1\right) \text { and } R_3 \rightarrow\left(R_3-R_1\right)\right]\)

= \((a+b+c) \cdot 1 \cdot\left|\begin{array}{cc}
2 b+a & a-b \\
a-c & 2 c+a
\end{array}\right|\) [expanded by C1]

= (a + b + c)[(2b + a)(2c + a) – (a – c)(a – b)]

= \((a+b+c)\left[\left(4 b c+2 a b+2 a c+a^2\right)-\left(a^2-a b-a c+b c\right)\right]\)

= 3(a + b + c)(Ab + bc + ca).

Hence, △ = 3(a + b + c)(Ab + bc + ca).

Example 13 Prove that \(\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^2\right) \cdot\left|\begin{array}{ccc}
a & c & p \\
b & d & q \\
u & v & w
\end{array}\right| .\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
a\left(1-x^2\right) & c\left(1-x^2\right) & p\left(1-x^2\right) \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|\)

\(\left[R_1 \rightarrow R_1-x R_2\right]\)

= \(\left(1-x^2\right) \cdot\left|\begin{array}{ccc}
a & c & p \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|\)

= \(\left(1-x^2\right) \cdot\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & w
\end{array}\right|\)

\(\left[R_2 \rightarrow\left(R_2-x R_1\right)\right]\)

Hence, △ = \(\left(1-x^2\right) \cdot\left|\begin{array}{ccc}
a & c & p \\
b & d & q \\
u & v & w
\end{array}\right| .\)

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Example 14 Prove that \(\left|\begin{array}{ccc}
y+z & z & y \\
z & z+x & x \\
y & x & x+y
\end{array}\right|=4 x y z .\)

Solution

Given determinant = \(\left|\begin{array}{ccc}
y+z & z & y \\
z & z+x & x \\
y & x & x+y
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
0 & -2 x & -2 x \\
z & z+x & x \\
y & x & x+y
\end{array}\right|\)

\(\left[R_1 \rightarrow R_1-\left(R_2+R_3\right)\right]\)

= \((-2 x) \cdot\left|\begin{array}{ccc}
0 & 1 & 1 \\
z & z+x & x \\
y & x & x+y
\end{array}\right|\) [taking (-2x) common from R1]

= \((-2 x) \cdot\left|\begin{array}{ccc}
0 & 0 & 1 \\
z & z & x \\
y & -y & x+y
\end{array}\right|\)

\(\left[C_2 \rightarrow\left(C_2-C_3\right)\right]\)

= \((-2 x) \cdot 1 \cdot\left|\begin{array}{rr}
z & z \\
y & -y
\end{array}\right|\) [expanded by R1]

= (-2x).1.(-yz – yz) = (-2x)(-2yz) = 4xyz.

Common Questions on Determinants and Their Solutions

Example 15 Prove that \(\left|\begin{array}{ccc}
a+b+2 c & a & b \\
c & b+c+2 a & b \\
c & a & c+a+2 b
\end{array}\right|=2(a+b+c)^3\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
a+b+2 c & a & b \\
c & b+c+2 a & b \\
c & a & c+a+2 b
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
a+b+c & -(a+b+c) & 0 \\
c & b+c+2 a & b \\
0 & -(a+b+c) & (a+b+c)
\end{array}\right|\)

\(\left\{\begin{array}{l}
\text { by } R_1 \rightarrow\left(R_1-R_2\right) \\
\text { and } R_3 \rightarrow\left(R_3-R_2\right)
\end{array}\right\}\)

= \((a+b+c)^2 \cdot\left|\begin{array}{ccc}
1 & -1 & 0 \\
c & b+c+2 a & b \\
0 & -1 & 1
\end{array}\right|\)

[taking (a + b + c) common from each one of R1 and R3]

= \((a+b+c)^2 \cdot\left|\begin{array}{ccc}
1 & -1 & 0 \\
0 & b+2 c+2 a & b \\
0 & -1 & 1
\end{array}\right|\)

\(\left\{R_2 \rightarrow R_2-c R_1\right\}\)

= \((a+b+c)^2 \cdot 1 \cdot\left|\begin{array}{cc}
b+2 c+2 a & b \\
-1 & 1
\end{array}\right|\) [expanded by C1]

= \((a+b+c)^2.(b+2 c+2 a+b)=2(a+b+c)^3 .\)

Example 16 Prove that \(\left|\begin{array}{lll}
1 & 1 & 1 \\
a & b & c \\
a^3 & b^3 & c^3
\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\)

Solution

The given determinant

= \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^3 & b^3 & c^3
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
0 & 0 & 1 \\
a-c & b-c & c \\
a^3-c^3 & b^3-c^3 & c^3
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1-C_3\right) \text { and } C_2 \rightarrow\left(C_2-C_3\right)\right]\)

= \((a-c)(b-c) \cdot\left|\begin{array}{ccc}
0 & 0 & 1 \\
1 & 1 & c \\
a^2+a c+c^2 & b^2+b c+c^2 & c^3
\end{array}\right|\)

[taking out (a-c) and (b-c) common from C1 and C2]

= \((a-c)(b-c) \cdot 1 \cdot\left|\begin{array}{cc}
1 & 1 \\
a^2+a c+c^2 & b^2+b c+c^2
\end{array}\right|\) [expanded by R1]

= \((a-c)(b-c) \cdot\left[\left(b^2+b c+c^2\right)-\left(a^2+a c+c^2\right)\right]\)

= \((a-c)(b-c)\left[\left(b^2-a^2\right)+(b c-a c)\right]\)

= \((a-c)(b-c)\left[\left(b^2-a^2\right)+(b-a) c\right]\)

= (a-c)(b-c)(b-a)(b+a+c)

= (a-b)(b-c)(c-a)(a+b+c).

Example 17 Prove that \(\left|\begin{array}{ccc}
x & y & z \\
x^2 & y^2 & z^2 \\
y+z & z+x & x+y
\end{array}\right|=(x+y+z)(x-y)(y-z)(z-x)\)

Solution

The given determinant

= \(\left|\begin{array}{ccc}
x & y & z \\
x^2 & y^2 & z^2 \\
y+z & z+x & x+y
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
x+y+z & x+y+z & x+y+z \\
x^2 & y^2 & z^2 \\
y+z & z+x & x+y
\end{array}\right|\)

\(\left[R_1 \rightarrow\left(R_1+R_3\right)\right]\)

= \((x+y+z) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
x^2 & y^2 & z^2 \\
y+z & z+x & x+y
\end{array}\right|\) [taking out (x + y + z) common from R1]

= \((x+y+z)\left|\begin{array}{ccc}
1 & 0 & 0 \\
x^2 & y^2-x^2 & z^2-x^2 \\
y+z & x-y & x-z
\end{array}\right|\)

\(\left[C_2 \rightarrow\left(C_2-C_1\right) \text { and } C_3 \rightarrow\left(C_3-C_1\right)\right]\)

= \((x+y+z)(x-y)(z-x) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
x^2 & -(x+y) & z+x \\
y+z & 1 & -1
\end{array}\right|\)

= \((x+y+z)(x-y)(z-x) \cdot 1 \cdot\left|\begin{array}{cc}
-(x+y) & z+x \\
1 & -1
\end{array}\right|\)

= (x + y + z)(x-y)(z-x).[(x+y)-(z+x)]

= (x + y + z)(x-y)(y-z)(z-x).

Example 18 Using properties of determinants, prove that \(\left|\begin{array}{ccc}
a & b & c \\
a^2 & b^2 & c^2 \\
b c & c a & a b
\end{array}\right|=(a-b)(b-c)(c-a)(a b+b c+c a) \text {. }\)

Solution

Let the given determinant be △. Then

△ = \(\left|\begin{array}{ccc}
a & b & c \\
a^2 & b^2 & c^2 \\
b c & c a & a b
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
a-c & b-c & c \\
a^2-c^2 & b^2-c^2 & c^2 \\
b(c-a) & a(c-b) & a b
\end{array}\right|\)

\(\left\{C_1 \rightarrow\left(C_1-C_3\right), C_2 \rightarrow\left(C_2-C_3\right)\right\}\)

= \((a-c)(b-c) \cdot\left|\begin{array}{ccc}
1 & 1 & c \\
a+c & b+c & c^2 \\
-b & -a & a b
\end{array}\right|\)

[taking (a-c) common from C1 and (b-c) common from C2]

= \((a-c)(b-c) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
a+c & b-a & -c a \\
-b & b-a & (a+c) b
\end{array}\right|\)

\(\left\{C_2 \rightarrow\left(C_2-C_1\right), C_3 \rightarrow\left(C_3-c C_1\right)\right\}\)

= \((a-c)(b-c) \cdot 1 \cdot\left|\begin{array}{cc}
b-a & -c a \\
b-a & (a+c) b
\end{array}\right|\) [expanded by R1]

= \((a-c)(b-c) \cdot(b-a)\left|\begin{array}{cc}
1 & -c a \\
1 & (a+c) b
\end{array}\right|\)

= (a – b)(b – c)(c – a)(ab + bc + ca).

Example 19 Prove that \(\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|=(a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)=(b c+c a+a b+a b c)\)

Solution

The given determinant

= \(\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|\)

= \((a b c) \cdot\left|\begin{array}{ccc}
\frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\
\frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\
\frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1
\end{array}\right|\)

[taking a, b, c common from R1, R2 and R3 respectively]

= \((a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
\frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\
\frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1
\end{array}\right|\)

\([performing R_1 \rightarrow R_1+R_2+R_3 \text { and taking }\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right) common outside]\)

= \((a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)\left|\begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & \frac{1}{b} \\
-1 & -1 & \frac{1}{c}+1
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1-C_3\right) \text { and } C_2 \rightarrow\left(C_2-C_3\right)\right]\)

= \((a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right) \cdot(1) \cdot\left|\begin{array}{rr}
0 & 1 \\
-1 & -1
\end{array}\right|\) [expanding by 1st row]

= \((a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right) \cdot 1\)

= \((a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)=(b c+c a+a b+a b c) .\)

Applications of Determinants in Mathematics

Example 20 Prove that \(\left|\begin{array}{ccc}
a^2 & b c & c^2+a c \\
a^2+a b & b^2 & a c \\
a b & b^2+b c & c^2
\end{array}\right|=4 a^2 b^2 c^2 .\)

Solution

\(\left|\begin{array}{ccc}
a^2 & b c & c^2+a c \\
a^2+a b & b^2 & a c \\
a b & b^2+b c & c^2
\end{array}\right|\)

= \(\text { (abc) }\left|\begin{array}{ccc}
a & c & c+a \\
a+b & b & a \\
b & b+c & c
\end{array}\right|\)

[taking a, b, c common from C1, C2 and C3 respectively]

= \(\text { (abc) }\left|\begin{array}{ccc}
a & c & c+a \\
0 & -2 c & -2 c \\
b & b+c & c
\end{array}\right|\)

\(\left[R_2 \rightarrow R_2-\left(R_1+R_3\right)\right]\)

= \((a b c) \cdot\left|\begin{array}{ccc}
a & -a & c+a \\
0 & 0 & -2 c \\
b & b & c
\end{array}\right|\)

\(\left[C_2 \rightarrow C_2-C_3\right]\)

= \((a b c)(2 c) \cdot\left|\begin{array}{rr}
a & -a \\
b & b
\end{array}\right|\) [expansion by R2]

= \(2 a b c^2 \cdot(a b+a b)=2 a b c^2(2 a b)\)

= \(4 a^2 b^2 c^2 .\)

Example 21 If a, b, c are positive and unequal, show that the value of the determinant \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|\) is negative.

Solution

The given determinant

= \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|\)

= \(\left|\begin{array}{lll}
a+b+c & b & c \\
a+b+c & c & a \\
a+b+c & a & b
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1+C_2+C_3\right)\right]\)

= \((a+b+c) \cdot\left|\begin{array}{lll}
1 & b & c \\
1 & c & a \\
1 & a & b
\end{array}\right|\)

= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & b & c \\
0 & c-b & a-c \\
0 & a-b & b-c
\end{array}\right|\left[R_2 \rightarrow\left(R_2-R_1\right) \text { and } R_3 \rightarrow\left(R_3-R_1\right)\right]\)

= (a + b + c).[(c-b)(b-c) – (a-b)(a-c)] [expanded by C1]

= \((a+b+c) \cdot\left(-a^2-b^2-c^2+a b+b c+c a\right)\)

= \(-\frac{1}{2}(a+b+c)\left(2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a\right)\)

= \(-\frac{1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\), which is negative

[∵ \((a+b+c)>0,(a-b)^2>0,(b-c)^2>0 \text { and }(c-a)^2>0\)].

Example 22 Evaluate \(\left|\begin{array}{lll}
{ }^m C_1 & { }^m C_2 & { }^m C_3 \\
{ }^n C_1 & { }^n C_2 & { }^n C_3 \\
{ }^p C_1 & { }^p C_2 & { }^m C_3
\end{array}\right| .\)

Solution

Let the given determinant be △. Then

△ = \(\left|\begin{array}{lll}
m & \frac{1}{2} m(m-1) & \frac{1}{6} \cdot m(m-1)(m-2) \\
n & \frac{1}{2} n(n-1) & \frac{1}{6} \cdot n(n-1)(n-2) \\
p & \frac{1}{2} p(p-1) & \frac{1}{6} \cdot p(p-1)(p-2)
\end{array}\right|\)

= \(\left(\frac{1}{2} \times \frac{1}{6} \times m n p\right) \cdot\left|\begin{array}{lll}
1 & (m-1) & (m-1)(m-2) \\
1 & (n-1) & (n-1)(n-2) \\
1 & (p-1) & (p-1)(p-2)
\end{array}\right|\)

= \(\frac{1}{12} \cdot m n p \cdot\left|\begin{array}{lll}
1 & m-1 & (m-1)(m-2) \\
0 & n-m & (n-m)(n+m-3) \\
0 & p-m & (p-m)(p+m-3)
\end{array}\right|\)

\(\left[R_2 \rightarrow\left(R_2-R_1\right) \text { and } R_3 \rightarrow\left(R_3-R_1\right)\right]\)

= \(\frac{1}{12} \cdot(m n p)(n-m)(p-m) \cdot\left|\begin{array}{ccl}
1 & m-1 & (m-1)(m-2) \\
0 & 1 & (n+m-3) \\
0 & 1 & (p+m-3)
\end{array}\right|\)

[taking (n – m) common from R2 and (p – m) common from R3]

= \(\frac{1}{12} \cdot(m n p)(n-m)(p-m) \cdot 1 \cdot\left|\begin{array}{ll}
1 & (n+m-3) \\
1 & (p+m-3)
\end{array}\right|\)

= \(\frac{1}{12} \cdot(m n p)(n-m)(p-m)[(p+m-3)-(n+m-3)]\)

= \(\frac{1}{12} \cdot(m n p)(n-m)(p-m)(p-n)\)

= \(\frac{1}{12} \cdot(m n p)(m-n)(n-p)(p-m) .\)

Example 23 Prove that \(\left|\begin{array}{lll}
b+c & a & b \\
c+a & c & a \\
a+b & b & c
\end{array}\right|=(a+b+c)(a-c)^2 \text {. }\)

Solution

Let the given determinant be △. Then

△ = \(\left|\begin{array}{lll}
b+c & a & b \\
c+a & c & a \\
a+b & b & c
\end{array}\right|\)

= \(\left|\begin{array}{lll}
b & a & b \\
c & c & a \\
a & b & c
\end{array}\right|+\left|\begin{array}{lll}
c & a & b \\
a & c & a \\
b & b & c
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
c & c & a \\
a & b & c
\end{array}\right|+\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
a & c & a \\
b & b & c
\end{array}\right|\)

\(\left[R_1 \rightarrow\left(R_1+R_2+R_3\right) \text { in each determinant }\right]\)

= \((a+b+c) \cdot\left|\begin{array}{lll}
1 & 1 & 1 \\
c & c & a \\
a & b & c
\end{array}\right|+(a+b+c) \cdot\left|\begin{array}{lll}
1 & 1 & 1 \\
a & c & a \\
b & b & c
\end{array}\right|\)

[taking out (a + b + c) common from R1 in each determinant]

= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
c & 0 & a-c \\
a & b-a & c-a
\end{array}\right|+(a+b+c) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
a & c-a & 0 \\
b & 0 & c-b
\end{array}\right|\)

\(\left[C_2 \rightarrow\left(C_2-C_1\right) \text { and } C_3 \rightarrow\left(C_3-C_1\right) \text { in each }\right]\)

= (a + b + c).1.{0-(b-a)(a-c)} + (a + b + c).1.{(c-a)(c-b)-0}

= (a + b + c)(a – b)(a – c) + (a + b + c)(c – a)(c – b)

= \((a+b+c)(a-c)\{(a-b)-(c-b)\}=(a+b+c)(a-c)^2\)

Hence, △ = \((a+b+c)(a-c)^2\).

Example 24 Prove that \(\left|\begin{array}{lll}
1 & a & a^2-b c \\
1 & b & b^2-c a \\
1 & c & c^2-a b
\end{array}\right|=0\)

Solution

Let the given determinant be △. Then

△ = \(\left|\begin{array}{lll}
1 & a & a^2-b c \\
1 & b & b^2-c a \\
1 & c & c^2-a b
\end{array}\right|\)

= \(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|+\left|\begin{array}{ccc}
1 & a & -b c \\
1 & b & -c a \\
1 & c & -a b
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
1 & a & a^2 \\
0 & b-a & b^2-a^2 \\
0 & c-a & c^2-a^2
\end{array}\right|+\left|\begin{array}{ccc}
1 & a & -b c \\
0 & b-a & c(b-a) \\
0 & c-a & b(c-a)
\end{array}\right|\)

\(\left[R_2 \rightarrow\left(R_2-R_1\right) \text { and } R_3 \rightarrow\left(R_3-R_1\right) \text { in each determinant }\right]\)

= \((b-a)(c-a) \cdot\left|\begin{array}{ccc}
1 & a & a^2 \\
0 & 1 & b+a \\
0 & 1 & c+a
\end{array}\right|+(b-a)(c-a) \cdot\left|\begin{array}{ccc}
1 & a & -b c \\
0 & 1 & c \\
0 & 1 & b
\end{array}\right|\)

= (b-a)(C-a).1.{(c+a)-(b+a)}+(b-a)(c-a).1.(b-c)

= (b-a)(c-a)(c-b)+(b-a)(c-a)(b-c)

= (a-b)(b-c)(c-a)-(A-b)(b-c)(C-a) = 0.

Hence, △ = 0.

Real-Life Applications of Determinants

Example 25 If x ≠ y ≠ z and \(\left|\begin{array}{lll}
x & x^2 & 1+x^3 \\
y & y^2 & 1+y^3 \\
z & z^2 & 1+z^3
\end{array}\right|=0\) then prove that xyz = -1.

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{lll}
x & x^2 & 1+x^3 \\
y & y^2 & 1+y^3 \\
z & z^2 & 1+z^3
\end{array}\right|=\left|\begin{array}{lll}
x & x^2 & 1 \\
y & y^2 & 1 \\
z & z^2 & 1
\end{array}\right|+\left|\begin{array}{lll}
x & x^2 & x^3 \\
y & y^2 & y^3 \\
z & z^2 & z^3
\end{array}\right|\)

= \(\left|\begin{array}{lll}
x & x^2 & 1 \\
y & y^2 & 1 \\
z & z^2 & 1
\end{array}\right|+(x y z) \cdot\left|\begin{array}{lll}
1 & x & x^2 \\
1 & y & y^2 \\
1 & z & z^2
\end{array}\right|\)

= \(\left|\begin{array}{lll}
x & x^2 & 1 \\
y & y^2 & 1 \\
z & z^2 & 1
\end{array}\right|+(x y z)(-1)^2 \cdot\left|\begin{array}{lll}
x & x^2 & 1 \\
y & y^2 & 1 \\
z & z^2 & 1
\end{array}\right|\)

[interchanging the columns of 2nd det. twice]

= \((1+x y z)\left|\begin{array}{lll}
x & x^2 & 1 \\
y & y^2 & 1 \\
z & z^2 & 1
\end{array}\right|\)

= \((1+x y z)\left|\begin{array}{ccc}
x & x^2 & 1 \\
(y-x) & \left(y^2-x^2\right) & 0 \\
(z-x) & \left(z^2-x^2\right) & 0
\end{array}\right|\)

\(\left[R_2 \rightarrow\left(R_2-R_1\right), R_3 \rightarrow\left(R_3-R_1\right)\right]\)

= \((1+x y z)(y-x)(z-x)\left|\begin{array}{ccc}
x & x^2 & 1 \\
1 & y+x & 0 \\
1 & z+x & 0
\end{array}\right|\)

= \((1+x y z)(y-x)(z-x) \cdot 1 \cdot\left|\begin{array}{ll}
1 & y+x \\
1 & z+x
\end{array}\right|\) [expanding by C3].

= (1 + xyz)(y – z)(z – x)(z – y).

∴ △ = 0 ⇒ (1 + xyz)(y – z)(z – x)(z – y) = 0

⇒ (1 + xyz) = 0 [∵ (y-x) ≠ 0, (z-x) ≠0, (z-y) ≠0]

⇒ xyz = -1.

Hence, xyz = -1.

Example 26 Prove that \(\left|\begin{array}{ccc}
(b+c)^2 & a^2 & a^2 \\
b^2 & (c+a)^2 & b^2 \\
c^2 & c^2 & (a+b)^2
\end{array}\right|=2 a b c(a+b+c)^3 \text {. }\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
(b+c)^2 & a^2 & a^2 \\
b^2 & (c+a)^2 & b^2 \\
c^2 & c^2 & (a+b)^2
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
(b+c)^2-a^2 & 0 & a^2 \\
0 & (c+a)^2-b^2 & b^2 \\
c^2-(a+b)^2 & c^2-(a+b)^2 & (a+b)^2
\end{array}\right|\)

\(\left[C_1 \rightarrow C_1-C_3 \text { and } C_2 \rightarrow C_2-C_3\right]\)

= \(\left|\begin{array}{ccc}
(a+b+c)(b+c-a) & 0 & a^2 \\
0 & (a+b+c)(c+a-b) & b^2 \\
(a+b+c)(c-a-b) & (a+b+c)(c-a-b) & (a+b)^2
\end{array}\right|\)

= \((a+b+c)^2 \cdot\left|\begin{array}{ccc}
(b+c-a) & 0 & a^2 \\
0 & c+a-b & b^2 \\
c-a-b & c-a-b & (a+b)^2
\end{array}\right|\)

[taking (a+b+c) common from C1 and C2 both]

= \(\left.(a+b+c)^2\left[(b+c-a) \mid(c+a-b) \cdot 2 a b+2 a b^2\right)+a^2(0+2 b(c+a-b))\right]\)

= \((a+b+c)^2\left[(b+c-a) \cdot 2 a b|(c+a-b+b)|+2 a^2 b(c+a-b)\right]\)

= \(2 a b(a+b+c)^2\{(b+c-a)(c+a)+a(c+a-b)\}\)

= \(2 a b(a+b+c)^2 \cdot\left\{b c+a b+c^2+a c-a c-a^2+a c+a^2-a b\right\}\)

= \(2 a b(a+b+c)^2\left\{b c+c^2+a c\right\}=2 a b c(a+b+c)^3 .\)

Hence, △ = \(2 a b c(a+b+c)^3 .\)

Example 27 Prove that \(\left|\begin{array}{lll}
\frac{1}{a} & a^2 & b c \\
\frac{1}{b} & b^2 & c a \\
\frac{1}{c} & c^2 & a b
\end{array}\right|=0 .\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{lll}
\frac{1}{a} & a^2 & b c \\
\frac{1}{b} & b^2 & c a \\
\frac{1}{c} & c^2 & a b
\end{array}\right|\)

= \(\frac{1}{a b c} \cdot\left|\begin{array}{lll}
1 & a^3 & a b c \\
1 & b^3 & a b c \\
1 & c^3 & a b c
\end{array}\right|\) [multiplying R1, R2, R3 by a, b, c respectively and dividing △ by abc]

= \(\frac{1}{a b c} \cdot a b c \cdot\left|\begin{array}{lll}
1 & a^3 & 1 \\
1 & b^3 & 1 \\
1 & c^3 & 1
\end{array}\right|\)

= (1 x 0) = 0 [∵ C1 and C3 are identical].

Hence, △ = 0.

Example 28 Prove that \(\left|\begin{array}{ccc}
-b c & b^2+b c & c^2+b c \\
a^2+a c & -a c & c^2+a c \\
a^2+a b & b^2+a b & -a b
\end{array}\right|=(a b+b c+a c)^3 \text {. }\)

Solution

We have

\(\left|\begin{array}{ccc}
-b c & b^2+b c & c^2+b c \\
a^2+a c & -a c & c^2+a c \\
a^2+a b & b^2+a b & -a b
\end{array}\right|=\frac{1}{a b c}\left|\begin{array}{ccc}
-a b c & a b^2+a b c & a c^2+a b c \\
a^2 b+a b c & -a b c & c^2 b+a b c \\
a^2 c+a b c & b^2 c+a b c & -a b c
\end{array}\right|\) \(\left[R_1 \rightarrow a R_1, R_2 \rightarrow b R_2 \text { and } R_3 \rightarrow c R_3 \text { and so divide the det. by } a b c\right]\)

= \(\left(\frac{a b c}{a b c}\right)\left|\begin{array}{ccc}
-b c & a b+a c & a c+a b \\
a b+b c & -a c & b c+a b \\
a c+b c & b c+a c & -a b
\end{array}\right|\)

[taking out a, b, c common from C1, C2, C3 respectively]

= \(\left|\begin{array}{ccc}
a b+b c+a c & a b+b c+a c & a b+b c+a c \\
a b+b c & -a c & b c+a b \\
a c+b c & b c+a c & -a b
\end{array}\right|\)

\(\left[R_1 \rightarrow\left(R_1+R_2+R_3\right)\right]\)

= \((a b+b c+a c) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
a b+b c & -a c & b c+a b \\
a c+b c & b c+a c & -a b
\end{array}\right|\)

= \((a b+b c+a c) \cdot\left|\begin{array}{ccc}
0 & 0 & 1 \\
0 & -(a b+b c+a c) & b c+a b \\
(a b+b c+a c) & (a b+b c+a c) & -a b
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1-C_3\right) \text { and } C_2 \rightarrow\left(C_2-C_3\right)\right]\)

= \((a b+b c+a c)^3\) [expanding by R1].

Example 29 Show that \(\left|\begin{array}{lll}
x+1 & x+2 & x+a \\
x+2 & x+3 & x+b \\
x+3 & x+4 & x+c
\end{array}\right|=0\), where a, b, c are in AP.

Solution

Since a, b, c are in AP, we have 2b = (a + c)

Let the given determinant be △. Then,

applying \(R_1 \rightarrow R_1+R_3-2 R_2\), we get

△ = \(\left|\begin{array}{ccc}
0 & 0 & (a+c)-2 b \\
x+2 & x+3 & x+b \\
x+3 & x+4 & x+c
\end{array}\right|\)

= \(\left|\begin{array}{lcc}
0 & 0 & 0 \\
x+2 & x+3 & x+b \\
x+3 & x+4 & x+c
\end{array}\right|\)

[∵ (a+c) – 2b = 0]

= 0 [expanded by R1].

Hence, △ = 0.

Example 30 Prove that \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|=2(a+b+c)\left(a b+b c+c a-a^2-b^2-c^2\right)\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\)

\(\left[R_1 \rightarrow\left(R_1+R_2+R_3\right)\right]\)

= \(2(a+b+c) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\)

= \(2(a+b+c) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
c+a & b-c & b-a \\
a+b & c-a & c-b
\end{array}\right|\)

\(\left[\begin{array}{l}
C_2 \rightarrow\left(C_2-C_1\right) \text { and } \\
C_3 \rightarrow\left(C_3-C_1\right)
\end{array}\right]\)

= \(2(a+b+c) \cdot 1 \cdot\left|\begin{array}{ll}
b-c & b-a \\
c-a & c-b
\end{array}\right|\)

2(a+b+c).[(b-c)(c-b)-(c-a)(b-a)]

= \(2(a+b+c)\left(a b+b c+c a-a^2-b^2-c^2\right)\)

Hence, △ = \(2(a+b+c)\left(a b+b c+c a-a^2-b^2-c^2\right)\)

Example 31 Prove that \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
a b & b^2+1 & b c \\
c a & c b & c^2+1
\end{array}\right|=\left(1+a^2+b^2+c^2\right)\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
a\left(a+\frac{1}{a}\right) & a b & a c \\
a b & \left(b+\frac{1}{b}\right) b & b c \\
a c & c b & \left(c+\frac{1}{c}\right) c
\end{array}\right|\)

= \((a b c) \cdot\left|\begin{array}{ccc}
a+\frac{1}{a} & a & a \\
b & b+\frac{1}{b} & b \\
c & c & c+\frac{1}{c}
\end{array}\right|\)

[taking a, b, c common from C1, C2, C3 respectively]

= \(\frac{(a b c)}{(a b c)} \cdot\left|\begin{array}{ccc}
a^2+1 & a^2 & a^2 \\
b^2 & b^2+1 & b^2 \\
c^2 & c^2 & c^2+1
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 \\
b^2 & b^2+1 & b^2 \\
c^2 & c^2 & c^2+1
\end{array}\right|\)

\(\text { \{by } R_1 \rightarrow R_1+R_2+R_3 \text { \} }\)

= \(\left(1+a^2+b^2+c^2\right) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
b^2 & b^2+1 & b^2 \\
c^2 & c^2 & c^2+1
\end{array}\right|\)

[taking out (1 + a^2 + b^2 + c^2) common from R1]

= \(\left(1+a^2+b^2+c^2\right) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
b^2 & 1 & 0 \\
c^2 & 0 & 1
\end{array}\right|\)

\(\text { (by } C_2 \rightarrow C_2-C_1 \text { and } C_3 \rightarrow C_3-C_1 \text { ) }\)

= \(\left(1+a^2+b^2+c^2\right) \cdot 1 \cdot\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) {expanded by R1}

= \(\left(1+a^2+b^2+c^2\right) \cdot 1 \cdot(1-0)=\left(1+a^2+b^2+c^2\right) .\)

Example 32 If A + B + C = π, show that \(\left|\begin{array}{lll}
\sin ^2 A & \sin A \cos A & \cos ^2 A \\
\sin ^2 B & \sin B \cos B & \cos ^2 B \\
\sin ^2 C & \sin C \cos C & \cos ^2 C
\end{array}\right|=-\sin (A-B) \sin (B-C) \sin (C-A)\)

Solution

The given determinant

= \(\left|\begin{array}{lll}
1 & (1 / 2) \sin 2 A & (1 / 2)(1+\cos 2 A) \\
1 & (1 / 2) \sin 2 B & (1 / 2)(1+\cos 2 B) \\
1 & (1 / 2) \sin 2 C & (1 / 2)(1+\cos 2 C)
\end{array}\right|\)

\(\left[C_1 \rightarrow C_1+C_3\right]\)

= \(\frac{1}{4} \cdot\left|\begin{array}{ccc}
1 & \sin 2 A & 1+\cos 2 A \\
1 & \sin 2 B & 1+\cos 2 B \\
1 & \sin 2 C & 1+\cos 2 C
\end{array}\right|\)

= \(\frac{1}{4} \cdot\left|\begin{array}{ccc}
1 & \sin 2 A & 1+\cos 2 A \\
0 & \sin 2 B-\sin 2 A & \cos 2 B-\cos 2 A \\
0 & \sin 2 C-\sin 2 A & \cos 2 C-\cos 2 A
\end{array}\right|\)

\(\left\{\begin{array}{l}
R_2 \rightarrow R_2-R_1 \\
R_3 \rightarrow R_3-R_1
\end{array}\right\}\)

= \(\frac{1}{4} \cdot\left|\begin{array}{ccc}
1 & \sin 2 A & 1+\cos 2 A \\
0 & 2 \cos (A+B) \sin (B-A) & 2 \sin (A+B) \sin (A-B) \\
0 & 2 \cos (A+C) \sin (C-A) & 2 \sin (A+C) \sin (A-C)
\end{array}\right|\)

= \(\sin (A-B) \sin (A-C) \cdot\left|\begin{array}{ccc}
1 & \sin 2 A & 1+\cos 2 A \\
0 & -\cos (A+B) & \sin (A+B) \\
0 & -\cos (A+C) & \sin (A+C)
\end{array}\right|\)

[taking 2sin(A – B) and 2sin(A – C) common factors from R2 and R3 respectively]

= \(\sin (A-B) \sin (A-C) \cdot\left|\begin{array}{ccc}
1 & \sin 2 A & 1+\cos 2 A \\
0 & -\cos (\pi-C) & \sin (\pi-C) \\
0 & -\cos (\pi-B) & \sin (\pi-B)
\end{array}\right|\) [∵ A + B + C = π]

= \(\sin (A-B) \sin (A-C)\left|\begin{array}{ccc}
1 & \sin 2 A & 1+\cos 2 A \\
0 & \cos C & \sin C \\
0 & \cos B & \sin B
\end{array}\right|\)

= sin(A – B)sin(A – C)[sin B cos C – cos B sin C]

= sin(A – B)sin(A – C)sin(B – C)

= -sin(A – B)sin(B – C)sin(C – A).

Example 33 If A + B + C = π, prove that \(\left|\begin{array}{ccc}
\sin (A+B+C) & \sin (A+C) & \cos C \\
-\sin B & 0 & \tan A \\
\cos (A+B) & \tan (B+C) & 0
\end{array}\right|=0\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
\sin (A+B+C) & \sin (A+C) & \cos C \\
-\sin B & 0 & \tan A \\
\cos (A+B) & \tan (B+C) & 0
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
\sin \pi & \sin (\pi-B) & \cos C \\
-\sin B & 0 & \tan A \\
\cos (\pi-C) & \tan (\pi-A) & 0
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
0 & \sin B & \cos C \\
-\sin B & 0 & \tan A \\
-\cos C & -\tan A & 0
\end{array}\right|\) {∵ sin π = 0, sin(π – B) = sin B; cos(π – C) = -cos C, tan(π – A) = tan A}

= \(\sin B \cdot\left|\begin{array}{cc}
\sin B & \cos C \\
-\tan A & 0
\end{array}\right|-\cos C \cdot\left|\begin{array}{cc}
\sin B & \cos C \\
0 & \tan A
\end{array}\right|\)

= (sin B.tan A cos C – sin B tan A. cos C) = 0.

Example 34 Prove that \(\left|\begin{array}{ccc}
\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha
\end{array}\right|=1\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha
\end{array}\right|\)

= \(\frac{1}{\sin \alpha \cos \alpha} \cdot\left|\begin{array}{ccc}
\sin \alpha \cos \alpha \cos \beta & \sin \alpha \cos \alpha \sin \beta & -\sin ^2 \alpha \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \alpha \cos \beta & \sin \alpha \cos \alpha \sin \beta & \cos ^2 \alpha
\end{array}\right|\)

[multiplying R1 by sin α, R3 by cos α and dividing △ by sin α cos α]

= \(\frac{1}{\sin \alpha \cos \alpha} \cdot\left|\begin{array}{ccc}
0 & 0 & -1 \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \alpha \cos \beta & \sin \alpha \cos \alpha \sin \beta & \cos ^2 \alpha
\end{array}\right|\)

\(\left[R_1 \rightarrow\left(R_1-R_3\right)\right]\)

= \(\frac{-1}{\sin \alpha \cos \alpha} \cdot\left[-\sin \alpha \cos \alpha \sin ^2 \beta-\sin \alpha \cos \alpha \cos ^2 \beta\right]\)

= \(\frac{-1}{\sin \alpha \cos \alpha} \cdot(-\sin \alpha \cos \alpha)\left[\sin ^2 \beta+\cos ^2 \beta\right]=1\)

Example 35 Prove that \(\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|\) is independent of θ.

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|\)

= \(x\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta)\)

= \(-x^3-x+x\left(\sin ^2 \theta+\cos ^2 \theta\right)=-x^3-x+x=-x^3\), which is independent of θ.

Example 36 Using properties of determinants, prove that \(\left|\begin{array}{lll}
b+c & q+r & y+z \\
c+a & r+p & z+x \\
a+b & p+q & x+y
\end{array}\right|=2\left|\begin{array}{lll}
a & p & x \\
b & q & y \\
c & r & z
\end{array}\right| .\)

Solution

We have

LHS = \(\left|\begin{array}{lll}
b+c & q+r & y+z \\
c+a & r+p & z+x \\
a+b & p+q & x+y
\end{array}\right|\)

= \(2\left|\begin{array}{ccc}
a+b+c & p+q+r & x+y+z \\
c+a & r+p & z+x \\
a+b & p+q & x+y
\end{array}\right|\)

\(\left[R_1 \rightarrow\left(R_1+R_2+R_3\right) \text { and taking } 2 \text { common }\right]\)

= \(2\left|\begin{array}{ccc}
a+b+c & p+q+r & x+y+z \\
-b & -q & -y \\
-c & -r & -z
\end{array}\right|\)

\(\left[\begin{array}{r}
R_2 \rightarrow\left(R_2-R_1\right) \\
\text { and } R_3 \rightarrow\left(R_3-R_1\right)
\end{array}\right]\)

= \(2\left|\begin{array}{ccc}
a & p & x \\
-b & -q & -y \\
-c & -r & -z
\end{array}\right|\)

\(\left[R_1 \rightarrow\left(R_1+R_2+R_3\right)\right]\)

= \(2(-1) \cdot(-1) \cdot\left|\begin{array}{lll}
a & p & x \\
b & q & y \\
c & r & z
\end{array}\right|\)

[taking (-1) common from each one of R2 and R3]

= \(2\left|\begin{array}{lll}
a & p & x \\
b & q & y \\
c & r & z
\end{array}\right|\) = RHS.

Hence, LHS = RHS.

Example 37 Without expanding the determinant, prove that \(\left|\begin{array}{lll}
a & a^2 & b c \\
b & b^2 & c a \\
c & c^2 & a b
\end{array}\right|=\left|\begin{array}{lll}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right| .\)

Solution

We have

LHS = \(\left|\begin{array}{lll}
a & a^2 & b c \\
b & b^2 & c a \\
c & c^2 & a b
\end{array}\right|\)

= \(\frac{1}{a b c} \cdot\left|\begin{array}{lll}
a^2 & a^3 & a b c \\
b^2 & b^3 & a b c \\
c^2 & c^3 & a b c
\end{array}\right|\)

\(\left[R_1 \rightarrow a R_1, R_2 \rightarrow b R_2, R_3 \rightarrow c R_3 \text { and dividing the whole det. by } a b c\right]\)

= \(\frac{1}{a b c} \cdot a b c=\left|\begin{array}{lll}
a^2 & a^3 & 1 \\
b^2 & b^3 & 1 \\
c^2 & c^3 & 1
\end{array}\right|\) [taking abc common from C3]

= \(\left|\begin{array}{lll}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right|\)

\(\left[C_2 \leftrightarrow C_3 \text { and } C_1 \leftrightarrow C_2\right]\)

= RHS.

Hence, LHS = RHS.

Comparative Analysis of Different Types of Matrices and Their Determinants

Example 38 Solve for x: \(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|=0 .\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\)

= \(\left|\begin{array}{lll}
3 a-x & a-x & a-x \\
3 a-x & a+x & a-x \\
3 a-x & a-x & a+x
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1+C_2+C_3\right)\right]\)

= \((3 a-x) \cdot\left|\begin{array}{lll}
1 & a-x & a-x \\
1 & a+x & a-x \\
1 & a-x & a+x
\end{array}\right|\) [taking (3a – x) common from C1]

= \((3 a-x) \cdot\left|\begin{array}{ccc}
1 & a-x & a-x \\
0 & 2 x & 0 \\
0 & 0 & 2 x
\end{array}\right|\)

\(\left[R_2 \rightarrow\left(R_2-R_1\right) \text { and } R_3 \rightarrow\left(R_3-R_1\right)\right]\)

= \((3 a-x) \cdot 1 \cdot\left|\begin{array}{cc}
2 x & 0 \\
0 & 2 x
\end{array}\right|\) [expanding by C1]

= 4(3 a-x) x^2

∴ △ = 0 ⇔ 4(3 a-x) x^2=0

⇔ x = 0 or x = 3a.

Example 39 Solve \(\left|\begin{array}{lll}
x-2 & 2 x-3 & 3 x-4 \\
x-4 & 2 x-9 & 3 x-16 \\
x-8 & 2 x-27 & 3 x-64
\end{array}\right|=0 .\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{lll}
x-2 & 2 x-3 & 3 x-4 \\
x-4 & 2 x-9 & 3 x-16 \\
x-8 & 2 x-27 & 3 x-64
\end{array}\right|\)

= \(\left|\begin{array}{rrr}
x-2 & 1 & 2 \\
x-4 & -1 & -4 \\
x-8 & -11 & -40
\end{array}\right|\)

\(\left[C_2 \rightarrow\left(C_2-2 C_1\right), C_3 \rightarrow\left(C_3-3 C_1\right)\right]\)

= \(\left|\begin{array}{crr}
x-2 & 1 & 2 \\
-2 & -2 & -6 \\
-6 & -12 & -42
\end{array}\right|\)

\(\left[R_2 \rightarrow\left(R_2-R_1\right), R_3 \rightarrow\left(R_3-R_1\right)\right]\)

= \((-2) \cdot(-6) \cdot\left|\begin{array}{ccc}
x-2 & 1 & 2 \\
1 & 1 & 3 \\
1 & 2 & 7
\end{array}\right|\)

= \(\text { 12. }\left|\begin{array}{ccc}
x-3 & 1 & 2 \\
0 & 1 & 3 \\
-1 & 2 & 7
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1-C_2\right)\right]\)

= 12.[(x-3)(7-6) – 1.(3-2)]

= 12.(x-4).

∴ △ = 0 ⇔ 12(x-4) = 0 ⇔ x = 4.

Hence, solution set = {4}.

Example 40 If a + b + c = 0 and \(\left|\begin{array}{ccc}
a-x & c & b \\
c & b-x & a \\
b & a & c-x
\end{array}\right|=0\) then show that x = 0 or \(x=\sqrt{(3 / 2)\left(a^2+b^2+c^2\right)} .\)

Solution

Let the given determinant be △. Then,

△ = \(\left|\begin{array}{ccc}
a+b+c-x & c & b \\
a+b+c-x & b-x & a \\
a+b+c-x & a & c-x
\end{array}\right|\)

\(\left[C_1 \rightarrow\left(C_1+C_2+C_3\right)\right]\)

= \((a+b+c-x)\left|\begin{array}{ccc}
1 & c & b \\
1 & b-x & a \\
1 & a & c-x
\end{array}\right|\)

= \((a+b+c-x)\left|\begin{array}{ccc}
1 & c & b \\
0 & b-x-c & a-b \\
0 & a-c & c-x-b
\end{array}\right|\)

\(\left[R_2 \rightarrow\left(R_2-R_1\right) \text { and } R_3 \rightarrow\left(R_3-R_1\right)\right]\)

= (a + b + c – x)[(b – x – c)(c – x – b)-(a – c)(a – b)]

= \((a+b+c-x)\left[x^2-\left(a^2+b^2+c^2-a b-b c-c a\right)\right] .\)

Now, △ = 0 ⇔ \((a+b+c-x)\left[x^2-\left(a^2+b^2+c^2-a b-b c-a c\right)\right]=0\)

⇔ \(x=a+b+c \text { or } x= \pm \sqrt{\left(a^2+b^2+c^2-a b-b c-c a\right)}\)

⇔ \(x=0 \text { or } x= \pm \sqrt{(3 / 2)\left(a^2+b^2+c^2\right)}\)

[∵ a + b + c = 0 ⇔ \(\left(a^2+b^2+c^2\right)\)=\(-2(a b+b c+c a) \Leftrightarrow(a b+b c+c a)=-(1 / 2)\left(a^2+b^2+c^2\right)\)].

Applications of Determinants

Area Of A Triangle In Determinant Form We know that the area of a triangle whose vertices are (x1, y1), (x2,y2) and (x3,y3) is given by

△ = \(\frac{1}{2}\left[x_1\left(y_2-y_3\right)-x_2\left(y_1-y_3\right)+x_3\left(y_1-y_2\right)\right]\)

= \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|\) (in determinant form).

Remark 1 Since area is a positive quantity, we always take the absolute value of the above determinant for the area.

Remark 2. If three points A, B, C are collinear then ar(△ ABC) = 0.

Condition For Collinearity Of Three Points

Let A(x1, y1), B(x2,y2) and C(x3, y3) be three given points.

Then, A, B, C are collinear

⇔ ar(△ ABC) = 0

⇔ \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=0\) ⇔ \(\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=0\)

⇔ \(x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)=0 .\)

∴ A, B, C are collinear ⇔ \(x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)=0 .\)

Solved Examples

Example 1 Find the area of a triangle whose vertices are A(-2,-3,), B(3,2) and C(-1,-8).

Solution

Here, \(\left(x_1=-2, y_1=-3\right),\left(x_2=3, y_2=2\right) \text { and }\left(x_3=-1, y_3=-8\right) \text {. }\)

∴ \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2} \cdot\left|\begin{array}{rrr}
-2 & -3 & 1 \\
3 & 2 & 1 \\
-1 & -8 & 1
\end{array}\right|\)

= \(\frac{1}{2} \cdot\left|\begin{array}{rrr}
-2 & -3 & 1 \\
5 & 5 & 0 \\
1 & -5 & 0
\end{array}\right|\)

\(\left[R_2 \rightarrow\left(R_2-R_1\right) \text { and } R_3 \rightarrow\left(R_3-R_1\right)\right]\)

= \(\frac{1}{2} \cdot 1 \cdot\left|\begin{array}{rr}
5 & 5 \\
1 & -5
\end{array}\right|=\frac{1}{2} \times(-25-5)=-15 .\)

Hence, ar(△ ABC) = |-15| = 15 square units.

Example 2 Show that the points A(a, b+c), B(b, c+a) and C(c, a+b) are collinear.

Solution

Points A, B, C are collinear ⇔ ar(△ ABC) = 0.

Now, ar(△ ABC) = \(\frac{1}{2} \cdot\left|\begin{array}{lll}
a & b+c & 1 \\
b & c+a & 1 \\
c & a+b & 1
\end{array}\right|\)

= \(\frac{1}{2} \cdot\left|\begin{array}{lll}
a & a+b+c & 1 \\
b & a+b+c & 1 \\
c & a+b+c & 1
\end{array}\right|\)

\(\left[C_2 \rightarrow\left(C_2+C_1\right)\right]\)

= \(\frac{1}{2}(a+b+c) \cdot\left|\begin{array}{lll}
a & 1 & 1 \\
b & 1 & 1 \\
c & 1 & 1
\end{array}\right|\)

= \(\frac{1}{2}(a+b+c) \times 0=0\) [∵ C2 and C3 are identical].

Example 3 If the points (a, b), (a’, b’) and (a-a’, b-b’) are collinear, show that ab’ = a’b.

Solution

The given points are collinear

⇔ \(\left|\begin{array}{ccc}
a & b & 1 \\
a^{\prime} & b^{\prime} & 1 \\
a-a^{\prime} & b-b^{\prime} & 1
\end{array}\right|=0\)

⇔ \(\left|\begin{array}{ccc}
a & b & 1 \\
a^{\prime}-a & b^{\prime}-b & 0 \\
-a^{\prime} & -b^{\prime} & 0
\end{array}\right|=0\)

\(\left[R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3-R_1\right]\)

⇔ \(\text { 1. }\left[\left(a^{\prime}-a\right)\left(-b^{\prime}\right)+a^{\prime}\left(b^{\prime}-b\right)\right]=0\)

[expanding by C3]

⇔ ab’ – a’b = 0

⇔ ab’ = a’b.

Example 4 Find the value of k in order that the points (5,5), (k,1) and (10,7) are collinear.

Solution

The given points are collinear

⇔ \(\left|\begin{array}{rrr}
5 & 5 & 1 \\
k & 1 & 1 \\
10 & 7 & 1
\end{array}\right|=0\)

⇔ \(\left|\begin{array}{crc}
5 & 5 & 1 \\
k-5 & -4 & 0 \\
5 & 2 & 0
\end{array}\right|=0\)

\(\left[R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3-R_1\right]\)

⇔ 1.[2(k-5) + 20] = 0

⇔ 2k + 10 = 0 ⇔ k = -5.

The value of k = -5.

Example 5 Let A(1,3), B(0,0) and C(k,0) be three points such that ar(△ ABC) = 3 sq units. Find the value of k.

Solution

We have

ar(△ ABC) = 3 sq units

⇔ \(\frac{1}{2} \cdot\left|\begin{array}{lll}
1 & 3 & 1 \\
0 & 0 & 1 \\
k & 0 & 1
\end{array}\right|= \pm 3\) ⇔ \(\left|\begin{array}{lll}
1 & 3 & 1 \\
0 & 0 & 1 \\
k & 0 & 1
\end{array}\right|= \pm 6\)

⇔\((-1) \cdot\left|\begin{array}{ll}
1 & 3 \\
k & 0
\end{array}\right|= \pm 6\) ⇔ 3k = ±6 ⇔ k = ±2.

Hence, k = ±2.

The value of k = ±2.

Example 6 Find the equation of the line joining the points A(1,2) and B(3,6), using determinants.

Solution

Let P(x,y) be a point on AB.

Then, the points A, P and B are collinear.

∴ ar(△ APB) = 0

⇒ \(\frac{1}{2} \cdot\left|\begin{array}{lll}
1 & 2 & 1 \\
x & y & 1 \\
3 & 6 & 1
\end{array}\right|=0\) ⇒ \(\left|\begin{array}{lll}
1 & 2 & 1 \\
x & y & 1 \\
3 & 6 & 1
\end{array}\right|=0\)

⇒ \(\left|\begin{array}{rrr}
1 & 2 & 1 \\
x & y & 1 \\
0 & 0 & -2
\end{array}\right|=0\)

\(\left[R_3 \rightarrow R_3-3 R_1\right]\)

⇒ \((-2) \cdot\left|\begin{array}{ll}
1 & 2 \\
x & y
\end{array}\right|=0\) ⇒ (y-2x) = 0 ⇒ y = 2x.

Hence, the required equation is y = 2x.

The equation of the line joining the points y = 2x.

Relations and Functions

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• Inverse Trigonometric Functions Solutions

Algebra

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• Adjoint and Inverse of a Matrix

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• Differential Equations Solutions

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• Vector Algebra Solutions

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• Straight Line Space Solutions
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Linear Programming

• Linear Programming Solutions

Chapter 4 Inverse Trigonometric Functions

Invertible Functions A one-one onto function is called an invertible function.

Inverse Of A Function Let f: X → Y be a one-one onto function. Then, for each y ∈ Y, there exists a unique element x ∈ X such that f(x) = y.

So, we define a new function, denoted by f^-1, called the inverse of f, as f^-1: Y → X: f^-1(y) = x ⇔ f(x) = y.

Clearly, domain(f^-1) = range(f) and range (f^-1) = domain (f).

Trigonometric functions are, in general, not one-on-one.

Therefore, their inverses do not exist.

However, if we restrict their domains, we can make them one-on-one, enabling us to have their inverses.

Example The sine function restricted to any of the intervals [-π/2, π/2], [3π/2, 5π/2], etc., is one-one onto, and in each case, the range is [-1,1].

Therefore, we can define the inverse of the sine function, denoted by sin^-1x, in each of these intervals. Corresponding to each such interval, we get a branch of sin^-1x.

The branch [-π/2. π/2], is called the principal-value branch and the value belonging to it is called the principal value.

We denote the inverse of the sine function as sin^-1x, called sine inverse x.

∴ sin^-1: [-1,1] → \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and sin^-1x = θ ⇔ x = sin θ.

Thus, sin^-1x is a function whose domain is [-1,1] and range is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Note that sin^-1x does not mean (sin x)^-1.

Similarly, we can define the principal-value branches of the remaining five trigonometrical functions.

The following table shows the inverse trigonometric functions and their principal-value branches.

Read and Learn More  Class 12 Math Solutions

Solved Examples

Example 1 Find the principal value of each of the following:

(1) \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)

(2) \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

(3) tan^-1(√3)

(4) cosec^-1(2)

Solution

(1) We know that the range of the principal-value branch of sin^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Let \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\) = θ. Then,

Sin θ = \(\frac{1}{\sqrt{2}}=\sin \frac{\pi}{4} \Rightarrow \theta=\frac{\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text {. }\)

Hence, the principal value of \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\) is \(\frac{\pi}{4}\).

(2) We know that the range of the principal-value branch of cos^-1 is [0, π].

Let \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) = θ. Then,

\(\cos \theta=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6} \in[0, \pi] .\)

Hence, the principal value of \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) is \(\frac{\pi}{6}\).

(3) We know that the range of the principal-value branch of tan^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Let tan^-1(√3) = θ. Then,

\(\tan \theta=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) .\)

Hence, the principal value of tan^-1(√3) is \(\frac{\pi}{3}\).

(4) We know that the range of the principal-value branch of cosec^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.

Let cosec^-1(2) = θ. Then,

cosec θ = 2 = \({cosec} \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\} .\)

Hence, the principal value of cosec^-1(2) is \(\frac{\pi}{6}\).

WBBSE Class 12 Inverse Trigonometric Functions Solutions

Example 2 Find the principal value of each of the following:

(1) \(\sin ^{-1}\left(\frac{-1}{2}\right)\)

(2) \(\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\)

(3) \(\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\)

(4) tan^-1(-)

(5) cosec^-1(-2)

(6) tan^-1(-√3)

Solution

(1) We know that the principal-value branch of sin^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Let \(\sin ^{-1}\left(\frac{-1}{2}\right)\) = θ. Then,

sin θ = \(-\frac{1}{2}=\sin \left(\frac{-\pi}{6}\right), \text { where } \frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text {. }\)

Hence, the principal value of \(\sin ^{-1}\left(\frac{-1}{2}\right)\) is \(\frac{-\pi}{6} .\)

(2) We know that the principal-value branch of sin^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Let \(\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\) = θ. Then,

\(\sin \theta=\frac{-\sqrt{3}}{2}=\sin \left(\frac{-\pi}{3}\right) \text {, where } \frac{-\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text {. }\)

Hence, the principal value of \(\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\) is \(\frac{-\pi}{3}\).

(3) We know that the principal-value branch of tan^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Let \(\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\) = θ. Then,

\(\tan \theta=\frac{-1}{\sqrt{3}}=\tan \left(\frac{-\pi}{6}\right) \text {, where } \frac{-\pi}{6} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \text {. }\)

Hence, the principal value of \(\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\) is \(\frac{-\pi}{6}\).

(4) We know that the principal-value branch of tan^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.

Let tan^-1(-1) = θ. Then,

tan θ = -1 = \(\tan \left(\frac{-\pi}{4}\right) \text {, where } \frac{-\pi}{4} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right).\)

Hence, the principal value of tan^-1(-1) is \(\frac{-\pi}{4}\).

(5) We know that the principal-value branch of cosec^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.

Let cosec^-1(-2) = θ. Then,

cosec θ = -2 = \({cosec}\left(\frac{-\pi}{6}\right), \text { where } \frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\} \text {. }\)

Hence, the principal value of cosec^-1(2) is \(\frac{-\pi}{6}\).

(6) We know that the principal-value branch of tan^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Let tan^-1(-√3) = θ. Then,

tan θ = -√3 = \(\tan \left(\frac{-\pi}{3}\right) \text {, where } \frac{-\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \text {. }\)

Hence, the principal value of tan^-1(-√3) is \(\frac{-\pi}{3}\).

Example 3 Find the principal value of each of the following:

(1) \(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\)

(2) \(\cos ^{-1}\left(\frac{-1}{2}\right)\)

(3) \(\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\)

Solution

(1) We know that the range of the principal value of cos^-1 is [0, π].

Let \(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\) = θ. Then,

\(\cos \theta=\frac{-1}{\sqrt{2}}=-\cos \frac{\pi}{4}=\cos \left(\pi-\frac{\pi}{4}\right)=\cos \frac{3 \pi}{4}\)

∴ \(\theta=\frac{3 \pi}{4} \in[0, \pi]\)

Hence, the principal value of \(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\) is \(\frac{3 \pi}{4}\).

(2) We know that the range of the principal value of cos^-1 is [0, π].

Let \(\cos ^{-1}\left(\frac{-1}{2}\right)\) = θ. Then,

\(\cos \theta=\frac{-1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3} .\)

∴ \(\theta=\frac{2 \pi}{3} \in[0, \pi] .\)

Hence, the principal value of \(\cos ^{-1}\left(\frac{-1}{2}\right)\) is \(\frac{2 \pi}{3}\).

(3) We know that the range of the principal value of cot^-1 is (0, π).

Let \(\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\) = θ. Then,

\(\cot \theta=\frac{-1}{\sqrt{3}}=-\cot \frac{\pi}{3}=\cot \left(\pi-\frac{\pi}{3}\right)=\cot \frac{2 \pi}{3} .\)

∴ \(\theta=\frac{2 \pi}{3} \in(0, \pi) .\)

Hence, the principal value of \(\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)\) is \(\frac{2 \pi}{3}\).

Example 4 Find the principal value of each of the following:

(1) cot^-1(-√3)

(2) sec^-1(-√2)

(3) cosec^-1(-1)

Solution

(1) We know that the range of the principal value of cot^-1 is (0, π).

Let cot^-1(-√3) = θ. Then,

\(\cot \theta=-\sqrt{3}=-\cot \frac{\pi}{6}=\cot \left(\pi-\frac{\pi}{6}\right)=\cot \frac{5 \pi}{6} .\)

∴ \(\theta=\frac{5 \pi}{6} \in(0, \pi) .\)

Hence, the principal value of cot^-1(-√3) is \(\frac{5 \pi}{6}\).

(2) We know that the range of principal value of sec^-1 is [0, π] \(– \left\{\frac{\pi}{2}\right\}\).

Let sec^-1(-√2) = θ. Then,

\(\sec \theta=-\sqrt{2}=-\sec \frac{\pi}{4}=\sec \left(\pi-\frac{\pi}{4}\right)=\sec \frac{3 \pi}{4} .\)

∴ \(\theta=\frac{3 \pi}{4} \in[0, \pi]-\left\{\frac{\pi}{2}\right\} .\)

Hence, the principal value of sec^-1(-√2) is \(\frac{3 \pi}{4}\).

(3) We know that the range of principal value of cosec^-1 is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.

Let cosec^-1(-1) = θ. Then, cosec θ = -1.

cosec θ = -1 = \(-{cosec} \frac{\pi}{2}={cosec}\left(\frac{-\pi}{2}\right)\)

∴ \(\theta=\frac{-\pi}{2} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\} .\)

Hence, the principal value of cosec^-1(-1) is \(\frac{-\pi}{2}\).

Step-by-Step Solutions to Inverse Trigonometric Problems

Example 5 Find the value of \(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\)

Solution:

Given:

\(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\)

We know that the ranges of principal values of cos^-1 and sin^-1 are [0, π] and \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) respectively.

Let \(\cos ^{-1}\left(\frac{1}{2}\right)\) = θ₁. Then,

\(\cos \theta_1=\frac{1}{2}=\cos \frac{\pi}{3} \Rightarrow \theta_1=\frac{\pi}{3} \in[0, \pi] .\)

Let \(\sin ^{-1}\left(\frac{1}{2}\right)\) = θ₂. Then,

\(\sin \theta_2=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow \theta_2=\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

∴ \(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}+\left(2 \times \frac{\pi}{6}\right)=\left(\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{2 \pi}{3} \text {. }\)

Example 6 Find the value of \(\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right) \text {. }\)

Solution

Given

\(\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right) \text {. }\)

We know that the ranges of principal values of tan^-1, cos^-1 and sin^-1 are \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), [0, π] and \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) respectively.

Let tan^-1(1) = θ₁. Then,

\(\tan \theta_1=1=\tan \frac{\pi}{4} \Rightarrow \theta_1=\frac{\pi}{4} \in[0, \pi]\)

Let \(\cos ^{-1}\left(\frac{-1}{2}\right)\) = θ₂. Then,

\(\cos \theta_2=\frac{-1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3} .\)

∴ \(\theta_2=\frac{2 \pi}{3} \in[0, \pi] .\)

Let \(\sin ^{-1}\left(\frac{-1}{2}\right)\) = θ₃. Then,

\(\sin \theta_3=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right) \Rightarrow \theta_3=\frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] .\)

∴ \(\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)=\left(\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}\right)=\frac{3 \pi}{4}\)

Example 7 Find the value of tan^-1√3 – sec^-1(-2).

Solution

Given

tan^-1√3 – sec^-1(-2)

We know that the ranges of principal values of tan^-1 and sec^-1 are \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and [0, π] – \(\left\{\frac{\pi}{2}\right\}\) respectively.

Let tan^-1√3 = θ₁. Then,

\(\tan \theta_1=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \theta_1=\frac{\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Let sec^-1(-2) = θ₂. Then,

\(\sec \theta_2=-2=-\sec \frac{\pi}{3}=\sec \left(\pi-\frac{\pi}{3}\right)=\sec \frac{2 \pi}{3}\)

∴ \(\theta_2=\frac{2 \pi}{3} \in[0, \pi]-\left\{\frac{\pi}{2}\right\} .\)

Hence, tan^-1√3 – sec^-1(-2) = \(\left(\frac{\pi}{3}-\frac{2 \pi}{3}\right)=\frac{-\pi}{3} .\)

Properties Of Inverse Functions

Theorem 1 Prove that:

(1) sin^-1(sin x) = x, x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

(2) cos^-1(cos x) = x, x ∈ [0, π]

(3) tan^-1(tan x) = x, x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

(4) cosec^-1(cosec x) = x, x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) – {0}

(5) sec^-1(sec x) = x, x ∈ [0, π] – \(\left\{\frac{\pi}{2}\right\}\)

(6) cot^-1(cot x) = x, x ∈ (0, π)

Proof

(1) Let sin x = y, where x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).

Then, sin^-1y = x ⇒ sin^-1(sin x) = x [∵ y = sin x]

∴ sin^-1(sin x) = x.

(2) Let cos x = y, where x x ∈ [0, π].

Then, cos^-1y = x ⇒ cos^-1(cos x) = x [∵ y = cos x].

∴ cos^-1(cos x) = x.

Similarly, the other results may be proved.

Theorem 2 Prove that:

(1) sin(sin^-1x) = x, x ∈ [-1,1]

(2) cos(cos^-1x) = x, x ∈ [-1,1]

(3) tan(tan^-1x) = x, x ∈ R

(4) cosec(cosec^-1x) = x, x ∈ R – [-1,1]

(5) sec(sec^-1x) = x, x ∈ R – [-1,1]

(6) cot(cot^-1x) = x, x ∈ R

Proof

(1) Let sin^-1x = θ, where x ∈ [-1,1].

Then, sin θ = x ⇒ sin(sin^-1x) = x [∵ θ = sin^-1 x].

∴ sin(sin^-1x) = x.

(2) Let cos^-1x = θ, where x ∈ [-1,1]

Then, cos θ = x ⇒ cos(cos^-1x) = x [∵ θ = cos^-1 x].

∴ cos(cos^-1x) = x.

Similarly, the other results may be proved.

Theorem 3 Prove that:

(1) \(\sin ^{-1} \frac{1}{x}={cosec}^{-1} x\), (x ≥ 1 or x ≤ -1)

(2) \(\cos ^{-1} \frac{1}{x}=\sec ^{-1} x\), (x ≥ 1 or x ≤ -1)

(3) \(\tan ^{-1} \frac{1}{x}=\cot ^{-1} x\), (x > 0)

Proof

(1) \(\sin ^{-1} \frac{1}{x}\) = θ ⇒ sin θ = \(\frac{1}{2}\)

⇒ cosec θ = x

⇒ θ = cosec^-1x

⇒ \(\sin ^{-1} \frac{1}{x}={cosec}^{-1} x\).

Hence, \(\sin ^{-1} \frac{1}{x}={cosec}^{-1} x\).

(2) \(\cos ^{-1} \frac{1}{x} = θ ⇒ cos θ = \frac{1}{x}\)

⇒ sec θ = x

⇒ θ = sec^-1x

⇒ \(\cos ^{-1} \frac{1}{x}=\sec ^{-1} x\).

Hence, \(\cos ^{-1} \frac{1}{x}=\sec ^{-1} x\).

(3) \(\tan ^{-1} \frac{1}{x}\) = θ ⇒ tan θ = \(\frac{1}{x}\)

⇒ cot θ = x

⇒ θ = cot^-1x

⇒ \(\tan ^{-1} \frac{1}{x}=\cot ^{-1} x\).

Hence, \(\tan ^{-1} \frac{1}{x}=\cot ^{-1} x\).

Theorem 4 Prove that:

(1) sin^-1(-x) = -sin^-1x, x ∈ [-1,1]

(2) tan^-1(-x) = -tan^-1x, x ∈ R

(3) cosec^-1(-x) = -cosec^-1x, |x| ≥ 1

Proof

(1) Let sin^-1(-x) = θ. Then,

sin^-1(-x) = θ ⇒ -x = sin θ

⇒ x = -sin θ = sin(-θ)

⇒ -θ = sin^-1x

⇒ θ = -sin^-1x

⇒ sin^-1(-x) = -sin^-1x.

∴ sin^-1(-x) = -sin^-1x.

Similarly, the other results may be proved.

Theorem 5 Prove that:

(1) cos^-1(-x) = π – cos^-1x, x ∈ [-1,1]

(2) sec^-1(-x) = π – sec^-1x, |x| ≥ 1

(3) cot^-1(-x) = π – cot^-1x, x ∈ R

Proof

(1) Let cos^-1(-x) = θ. Then,

cos^-1(-x) = θ ⇒ -x = cos θ

⇒ x = -cos θ = cos(π – θ)

⇒ cos^-1x = (π – θ) = π – cos^-1(-x)

⇒ cos^-1(-x) = π – cos^-1x.

∴ cos^-1(-x) = π – cos^-1x.

Similarly, the other results may be proved.

Properties of Inverse Trigonometric Functions Explained

Theorem 6 Prove that:

(1) \(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}, x \in[-1,1]\)

(2) \(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}, x \in R\)

(3) \({cosec}^{-1} x+\sec ^{-1} x=\frac{\pi}{2}, \quad|x| \geq 1\)

Proof

(1) Let sin^-1x = θ. Then,

sin^-1x = θ ⇒ x = sin θ = \(\cos \left(\frac{\pi}{2}-\theta\right)\)

⇒ \(\cos ^{-1} x=\frac{\pi}{2}-\theta\)

⇒ \(\cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x\)

⇒ \(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\).

∴ \(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\).

Similarly, the other results may be proved.

Theorem 7 Prove that:

(1) \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \text {, if } x y<1\)

(2) \(\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right) \text {, if } x y>-1\)

(3) \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right), \text { if }|x|<1\)

Proof

(1) Let xy < 1, tan^-1x = θ and tan^-1y = Φ. Then,

tan θ = x and tan Φ = y

⇒ \(\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\frac{x+y}{1-x y}\)

⇒ \(\theta+\phi=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)

⇒ \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)

∴ \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\).

(2) Let xy > -1.

On replacing y by -y in (1), we get

\(\tan ^{-1} x+\tan ^{-1}(-y)=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\)

⇒ \(\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\).

(3) Let |x| < 1.

Replacing y by x in (1), we get

\(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)

Theorem 8 Prove that:

(1) \(2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right),|x|<1\)

(2) \(2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right),|x| \geq 0\)

(3) \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right),|x|<1\)

Proof

Let tan^-1x = θ. Then x = tan θ.

(1) \(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) [∵ x = tan θ]

= sin^-1(sin 2θ)

= 2θ = 2 tan^-1x.

∴ \(2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)

(2) \(\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\) [∵ x = tan θ]

= cos^-1(cos 2θ)

= 2θ = 2tan^-1x.

∴ \(2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)

(3) \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)\) [∵ x = tan θ]

= tan^-1(tan 2θ)

= 2θ = 2 tan^-1x.

∴ \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\).

Theorem 9 Prove that \(2 \sin ^{-1} x=\sin ^{-1}\left[2 x \sqrt{1-x^2}\right], \frac{-1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} .\)

Proof

Given

\(2 \sin ^{-1} x=\sin ^{-1}\left[2 x \sqrt{1-x^2}\right], \frac{-1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} .\)

Let sin^-1x = θ. Then, x = sin θ.

∴ sin 2θ = 2 sinθ cosθ

= \(2 \sin \theta \cdot \sqrt{\left(1-\sin ^2 \theta\right)}\)

= \(2 x \sqrt{1-x^2}\)

⇒ \(2 \theta=\sin ^{-1}\left\{2 x \sqrt{1-x^2}\right\}\)

⇒ \(2 \sin ^{-1} x=\sin ^{-1}\left\{2 x \sqrt{1-x^2}\right\} \text {. }\)

Hence, \(2 \sin ^{-1} x=\sin ^{-1}\left\{2 x \sqrt{1-x^2}\right\} \text {. }\)

Theorem 10 Prove that \(2 \cos ^{-1} x=\cos ^{-1}\left(2 x^2-1\right), \frac{1}{\sqrt{2}} \leq x \leq 1 .\)

Proof

Let cos^-1x = θ. Then, x = cos θ.

∴ cos 2θ = (2cos²θ – 1) = (2x² – 1)

⇒ 2θ = cos^-1(2x² – 1)

⇒ 2cos^-1x = cos^-1(2x² – 1) [∵ θ = cos^-1x].

Hence, 2cos^-1x = cos^-1(2x² – 1).

Theorem 11 Prove that 3sin^-1x = sin^-1(3x – 4x³), x ∈ \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)

Proof

Given

3sin^-1x = sin^-1(3x – 4x³)

Put sin^-1x = θ. Then, x = sinθ.

Now, sin 3θ = (3 sin θ – 4 sin³θ) = (3x – 4x³)

⇒ 3θ = sin^-1(3x – 4x³)

⇒ 3sin^-1x = sin^-1(3x – 4x³) [∵ θ = sin^-1x].

Hence, 3sin^-1x = sin^-1(3x – 4x³).

Theorem 12 Prove that 3cos^-1x = cos^-1(4x³ – 3x), x ∈ [\(\frac{1}{2}\), 1].

Proof

3cos^-1x = cos^-1(4x³ – 3x)

Put cos^-1x = θ. Then, x = cosθ.

Now, cos 3θ = (4cos³θ – 3 cosθ) = (4x³ – 3x)

⇒ 3θ = cos^-1(4x³ – 3x)

⇒ 3cos^-1x = cos^-1(4x³ – 3x) [∵ θ = cos^-1x]

Hence, 3 cos^-1x = cos^-1(4x³ – 3x).

Theorem 13 Prove that:

(1) \(\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\)

(2) \(\sin ^{-1} x-\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^2}-y \sqrt{1-x^2}\right\}\)

(3) \(\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left\{x y-\sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right\}\)

(4) \(\cos ^{-1} x-\cos ^{-1} y=\cos ^{-1}\left\{x y+\sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right\}\)

Proof

(1) Let sin^-1x = θ₁, and sin^-1y = θ₂. Then,

sinθ₁ = x and sin θ₂ = y

⇒ sin(θ₁ + θ₂) = sinθ₁cosθ₂ + cosθ₁sinθ₂

= \(\left\{\sin \theta_1 \cdot \sqrt{1-\sin ^2 \theta_2}\right\}+\left\{\sqrt{1-\sin ^2 \theta_1} \cdot \sin \theta_2\right\}\)

= \(\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\)

⇒ \(\left(\theta_1+\theta_2\right)=\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\)

⇒ \(\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\).

Hence, \(\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\).

The other results can be proved similarly.

Theorem 14 Prove that:

(1) \(\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^2}=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)={cosec}^{-1}\left(\frac{1}{x}\right)\)

= \(\sec ^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)=\cot ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\)

(2) \(\cos ^{-1} x=\sin ^{-1} \sqrt{1-x^2}=\tan ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\)

= \({cosec}^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)=\sec ^{-1}\left(\frac{1}{x}\right)=\cot ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\)

(3) \(\tan ^{-1} x=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)=\cos ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\)

= \({cosec}^{-1}\left(\frac{\sqrt{1+x^2}}{x}\right)=\sec ^{-1}\left(\sqrt{1+x^2}\right)=\cot ^{-1}\left(\frac{1}{x}\right)\)

Proof

(1) Let sin^-1x = θ. Then, sin θ = x.

∴ \(\cos \theta=\sqrt{1-x^2}, \tan \theta=\frac{x}{\sqrt{1-x^2}} \cdot {cosec} \theta=\frac{1}{x}\)

\(\sec \theta=\frac{1}{\sqrt{1-x^2}} \text { and } \cot \theta=\frac{\sqrt{1-x^2}}{x} \text {. }\)

∴ \(\theta=\cos ^{-1} \sqrt{1-x^2}, \theta=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right), \theta={cosec}^{-1} \frac{1}{x}\)

\(\theta=\sec ^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right) \text { and } \theta=\cot ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \text {. }\)

Hence, \(\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^2}=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\)

= \({cosec}^{-1} \frac{1}{x}=\sec ^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)\)

= \(\cot ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \text {. }\)

Similarly, the other results can be proved.

Solved Examples

Example 1 Find the value of:

(1) \(\sin ^{-1}\left(\sin \frac{\pi}{3}\right)\)

(2) \(\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)\)

(3) \(\tan ^{-1}\left(\tan \frac{\pi}{4}\right)\)

Solution

We have

(1) \(\sin ^{-1}\left(\sin \frac{\pi}{3}\right)=\frac{\pi}{3}, \text { since } \frac{\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text {. }\)

(2) \(\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)=\frac{2 \pi}{3}, \text { since } \frac{2 \pi}{3} \in[0, \pi] \text {. }\)

(3) \(\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4}, \text { since } \frac{\pi}{4} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Example 2 Find the value of:

(1) \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\)

(2) \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)\)

(3) \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\)

Solution

(1) We know that the principal-value branch of sin-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).

∴ \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) \neq \frac{2 \pi}{3}\)

Now, \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\sin ^{-1}\left\{\sin \left(\pi-\frac{\pi}{3}\right)\right\}\)

= \(\sin ^{-1}\left(\sin \frac{\pi}{3}\right)\) [∵ sin(π – θ) = sinθ]

= \(\frac{\pi}{3}\) {∵ \(\frac{\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)}

Hence, \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\) = \(\frac{\pi}{3}\).

(2) We know that the principal-value branch of cos-1 is [0, π].

∴ \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right) \neq \frac{7 \pi}{6}\)

Now, \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left\{\cos \left(2 \pi-\frac{5 \pi}{6}\right)\right\}\)

= \(\cos ^{-1}\left\{\cos \frac{5 \pi}{6}\right\}\) {∵ cos(2π – θ) = cosθ}

= \(\frac{5 \pi}{6}\).

Hence, \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)\) = \(\frac{5 \pi}{6}\).

(3) We know that the principal-value branch of tan-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).

∴ \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right) \neq \frac{3 \pi}{4}\)

Now, \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)=\tan ^{-1}\left\{\tan \left(\pi-\frac{\pi}{4}\right)\right\}\)

= \(\tan ^{-1}\left\{-\tan \frac{\pi}{4}\right\}\) [∵ \(\tan \left(\pi-\frac{\pi}{4}\right)=-\tan \frac{\pi}{4}\)]

= \(\tan _{-\pi}^{-1}\left\{\tan \left(\frac{-\pi}{4}\right)\right\}\) [∵ -tanθ = tan(-θ)]

= \(\frac{-\pi}{4}\).

Hence, \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\) = \(\frac{-\pi}{4}\).

Common Questions on Inverse Trigonometric Functions and Their Solutions

Example 3 Find the value of:

(1) \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)\)

(2) \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)\)

(3) \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)\)

Solution

(1) We know that the principal-value branch of sin^-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).

∴ \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right) \neq \frac{3 \pi}{5}\)

Now, \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1}\left\{\sin \left(\pi-\frac{2 \pi}{5}\right)\right\}\).

= \(\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right)\) [∵ sin(π – θ) = sinθ]

= \(\frac{2 \pi}{5}\) {∵ \(\frac{2 \pi}{5} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)}.

Hence, \(\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\frac{2 \pi}{5}\)

(2) We know that the principal-value branch of cos^-1 is [0, π].

∴ \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) \neq \frac{13 \pi}{6} .\)

Now, \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left\{\cos \left(2 \pi+\frac{\pi}{6}\right)\right\}\)

= \(\cos ^{-1}\left\{\cos \frac{\pi}{6}\right\}\) [∵ cos(2π + θ) = cos θ]

= \(\frac{\pi}{6}\).

Hence, \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\frac{\pi}{6}\)

(3) We know that the principal-value branch of tan^-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).

∴ \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right) \neq \frac{7 \pi}{6}\)

Now, \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\tan ^{-1}\left\{\tan \left(\pi+\frac{\pi}{6}\right)\right\}\)

= \(\tan ^{-1}\left(\tan \frac{\pi}{6}\right)\) [∵ tan(π + θ) = tan θ]

= \(\frac{\pi}{6}\).

Hence, \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\frac{\pi}{6}\)

Example 4 (1) Show that \(\sin ^{-1}\left\{\sin \frac{3 \pi}{4}\right\} \neq \frac{3 \pi}{4}\) and find its value.

(2) Show that \(\cos ^{-1}\left\{\cos \left(\frac{-\pi}{3}\right)\right\}=\frac{-\pi}{3}\) and find its value.

(3) Show that \(\tan ^{-1}\left\{\tan \frac{5 \pi}{6}\right\} \neq \frac{5 \pi}{6}\) and find its value.

Solution

(1) We know that the principal-value branch of sin^-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).

∴ \(\sin ^{-1}\left\{\sin \frac{3 \pi}{4}\right\} \neq \frac{3 \pi}{4}\)

Now, \(\sin ^{-1}\left\{\sin \frac{3 \pi}{4}\right\}=\sin ^{-1}\left\{\sin \left(\pi-\frac{\pi}{4}\right)\right\}\)

= \(\sin ^{-1}\left\{\sin \frac{\pi}{4}\right\} [∵ \sin \left(\pi-\frac{\pi}{4}\right)=\sin \frac{\pi}{4}]\)

= \(\frac{\pi}{4}\) {∵ \(\frac{\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)}.

∴ \(\sin ^{-1}\left\{\sin \frac{3 \pi}{4}\right\}=\frac{\pi}{4}\).

(2) We know that the principal-value branch of cos^-1 is [0, π].

∴ \(\cos ^{-1}\left\{\cos \left(\frac{-\pi}{3}\right)\right\}=\frac{-\pi}{3}\)

Now, \(\cos ^{-1}\left\{\cos \left(\frac{-\pi}{3}\right)\right\}=\cos ^{-1}\left\{\cos \frac{\pi}{3}\right\}\). [∵ cos(-θ) = cos θ]

= \(\frac{\pi}{3}\) [∵ \(\frac{\pi}{3} \in[0, \pi]\)].

∴ \(\cos ^{-1}\left\{\cos \left(\frac{-\pi}{3}\right)\right\}=\frac{\pi}{3}\)

(3) We know that the principal-value branch of tan^-1 is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).

∴ \(\tan ^{-1}\left\{\tan \frac{5 \pi}{6}\right\} \neq \frac{5 \pi}{6}\).

Now, \(\tan ^{-1}\left\{\tan \frac{5 \pi}{6}\right\}=\tan ^{-1}\left\{\tan \left(\pi-\frac{\pi}{6}\right)\right\}\)

= \(\tan ^{-1}\left\{\tan \left(\frac{-\pi}{6}\right)\right\}\) [∵ tan(π – θ) = tan(-θ)]

= \(\frac{-\pi}{6}\) [∵ \(\frac{-\pi}{6} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)].

∴ \(\tan ^{-1}\left\{\tan \frac{5 \pi}{6}\right\}=\frac{-\pi}{6}\).

Example 5 Evaluate:

(1) \(\sin \left(\cos ^{-1} \frac{3}{5}\right)\)

(2) \(\cos \left(\tan ^{-1} \frac{3}{4}\right)\)

Solution

(1) Let cos^-1\(\frac{3}{5}\) = θ, where θ ∈ [0, π].

Then, cos θ = \(\frac{3}{5}\).

Since θ ∈ [0, π], we have sin θ > 0.

∴ \(\sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} \text {. }\)

Hence, \(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \theta=\frac{4}{5} .\)

(2) Let tan^-1\(\frac{3}{4}\) = θ, where θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).

Then, tan θ = \(\frac{3}{4}\).

Since θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\), so cosθ > 0.

∴ \(\cos \theta=\frac{1}{\sec \theta}=\frac{1}{\sqrt{1+\tan ^2 \theta}}=\frac{1}{\sqrt{1+9 / 16}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\).

Hence, \(\cos \left(\tan ^{-1} \frac{3}{4}\right)=\cos \theta=\frac{4}{5} \text {. }\)

Example 6 Evaluate:

(1) \(\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}\)

(2) \(\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)\)

(3) sin(cot^-1x)

(4) \(\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)\)

Solution

(1) We know that sin^-1(-θ) = -sin^-1θ.

∴ \(\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}=\sin \left\{\frac{\pi}{3}+\sin ^{-1} \frac{1}{2}\right\}\)

= \(\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)\) [∵ \(\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}\)]

= sin\(\frac{\pi}{2}\) = 1.

(2) Let cos^-1 \(\frac{4}{5}\) = θ, where θ ∈ [0, π].

Then, cosθ = \(\frac{4}{5}\).

Since θ ∈ [0, π] ⇒ \(\frac{1}{2}\)θ ∈ \(\left[0, \frac{\pi}{2}\right]\) ⇒ sin\(\frac{1}{2}\)θ > 0.

∴ \(\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=\sin \frac{1}{2} \theta=\sqrt{\frac{1-\cos \theta}{2}}=\sqrt{\frac{1-(4 / 5)}{2}}=\frac{1}{\sqrt{10}}\)

(3) Let cot^-1x = θ. Then, θ ∈ [0, π].

∴ sin(cot^-1x) = sinθ > 0.

Now, \(\sin \theta=\frac{1}{{cosec} \theta}=\frac{1}{\sqrt{1+\cot ^2 \theta}}=\frac{1}{\sqrt{1+x^2}}\).

∴ \(\sin \left(\cot ^{-1} x\right)=\sin \theta=\frac{1}{\sqrt{1+x^2}} .\)

(4) Let cos^-1\(\frac{\sqrt{5}}{3}\) = θ. Then, cos θ = \(\frac{\sqrt{5}}{3}\), where θ ∈ [0, π].

∴ \(\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)=\tan \frac{1}{2} \theta=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\sqrt{\frac{(1-\sqrt{5} / 3)}{(1+\sqrt{5} / 3)}}\)

= \(\sqrt{\frac{(3-\sqrt{5})}{(3+\sqrt{5})} \times \frac{(3-\sqrt{5})}{(3-\sqrt{5})}}=\frac{(3-\sqrt{5})}{2}\).

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Example 7 Evaluate \(\sin \left[2 \cos ^{-1}\left(\frac{-3}{5}\right)\right]\)

Solution

Given:

\(\sin \left[2 \cos ^{-1}\left(\frac{-3}{5}\right)\right]\)

Let cos^-1(\(\frac{3}{5}\)) = θ, where θ ∈ [0, π].

Then, cos θ = –\(\frac{3}{5}\).

since θ ∈ [0, π], we have sin θ > 0.

∴ \(\sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} .\)

∴ \(\sin \left[2 \cos ^{-1}\left(\frac{-3}{5}\right)\right]=\sin 2 \theta\)

= \(2 \sin \theta \cos \theta=\left\{2 \times \frac{4}{5} \times\left(\frac{-3}{5}\right)\right\}=\frac{-24}{25}\).

\(\sin \left[2 \cos ^{-1}\left(\frac{-3}{5}\right)\right]\) = \(2 \sin \theta \cos \theta=\left\{2 \times \frac{4}{5} \times\left(\frac{-3}{5}\right)\right\}=\frac{-24}{25}\).

Example 8 Evaluate \(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)\)

Solution

Given

\(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)\)

Let sin^-1\(\frac{3}{5}\) = A and sin^-1\(\frac{5}{13}\) = B. Then,

A, B ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) ⇒ cos A > 0 and cos B > 0.

∴ sin A = \(\frac{3}{5}\) and sin B = \(\frac{5}{13}\)

⇒ \(\cos A=\sqrt{1-\sin ^2 A}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)

and \(\cos B=\sqrt{1-\sin ^2 B}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13} \text {. }\)

∴ \(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)=\cos (A+B)\)

= cos A cos B – sin A sin B

= \(\left(\frac{4}{5} \times \frac{12}{13}\right)-\left(\frac{3}{5} \times \frac{5}{13}\right)\)

= \(\left(\frac{48}{65}-\frac{15}{65}\right)=\frac{33}{65}\).

\(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)\) = \(\left(\frac{48}{65}-\frac{15}{65}\right)=\frac{33}{65}\).

 

Example 9 Find the value of \(\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}\).

Solution

We have

\(\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}=\tan ^{-1}\left\{2 \cos \left(2 \times \frac{\pi}{6}\right)\right\}\) [∵ sin-1\(\frac{1}{2}\) = \(\frac{\pi}{6}\)]

= \(\tan ^{-1}\left\{2 \cos \frac{\pi}{3}\right\}\)

= \(\tan ^{-1}\left(2 \times \frac{1}{2}\right)=\tan ^{-1} 1=\frac{\pi}{4}\).

The value of \(\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}\) = \(\tan ^{-1}\left(2 \times \frac{1}{2}\right)=\tan ^{-1} 1=\frac{\pi}{4}\).

Example 10 If tan^-1\(\frac{4}{3}\) = θ, find the value of cos θ.

Solution

tan^-1\(\frac{4}{3}\) = θ, where θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).

∴ tan θ = \(\frac{4}{3}\).

We know that cos θ > 0 where θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).

∴ \(\cos \theta=\frac{1}{\sec \theta}=\frac{1}{\sqrt{1+\tan ^2 \theta}}=\frac{1}{\sqrt{1+\frac{16}{9}}}=\frac{3}{5} .\)

The value of  \(\cos \theta=\frac{1}{\sec \theta}=\frac{1}{\sqrt{1+\tan ^2 \theta}}=\frac{1}{\sqrt{1+\frac{16}{9}}}=\frac{3}{5} .\)

Applications of Inverse Trigonometric Functions in Mathematics

Example 11 If \(\cot ^{-1}\left(-\frac{1}{5}\right)=\theta\), find the value of sin θ.

Solution

Given: \(\cot ^{-1}\left(-\frac{1}{5}\right)=\theta\), where θ ∈ (0, π).

∴ cot θ = –\(\frac{1}{5}\).

Sin θ > 0 in (0, π).

\(\sin \theta=\frac{1}{{cosec} \theta}=\frac{1}{\sqrt{1+\cot ^2 \theta}}=\frac{1}{\sqrt{1+\frac{1}{25}}}=\frac{5}{\sqrt{26}} .\)

The value of \(\sin \theta=\frac{1}{{cosec} \theta}=\frac{1}{\sqrt{1+\cot ^2 \theta}}=\frac{1}{\sqrt{1+\frac{1}{25}}}=\frac{5}{\sqrt{26}} .\)

Example 12 Prove that \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}=\tan ^{-1} \frac{2}{9} \text {. }\)

Solution

We know that tan^-1 x + tan^-1 y = \(\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\), xy < 1.

Let x = \(\frac{1}{7}\) and y = \(\frac{1}{13}\). Then, xy = \(\frac{1}{91}\) < 1.

∴ \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}=\tan ^{-1}\left\{\frac{\left(\frac{1}{7}+\frac{1}{13}\right)}{1-\frac{1}{7} \times \frac{1}{13}}\right\}=\tan ^{-1}\left(\frac{20}{90}\right)=\tan ^{-1} \frac{2}{9}\)

Example 13 Prove that tan^-1\(\frac{3}{4}\) + tan^-1\(\frac{3}{5}\) – tan^-1\(\frac{8}{19}\) = \(\frac{\pi}{4}\).

Solution

We have

LHS = \(\left\{\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}\right\}-\tan ^{-1} \frac{8}{19}\)

= \(\tan ^{-1}\left\{\frac{\left(\frac{3}{4}+\frac{3}{5}\right)}{\left(1-\frac{3}{4} \times \frac{3}{5}\right)}\right\}-\tan ^{-1} \frac{8}{19}\)

= \(\tan ^{-1}\left(\frac{27}{11}\right)-\tan ^{-1} \frac{8}{19}\)

= \(\tan ^{-1} \frac{\left(\frac{27}{11}-\frac{8}{19}\right)}{\left(1+\frac{27}{11} \times \frac{8}{19}\right)}=\tan ^{-1}\left(\frac{425}{425}\right)=\tan ^{-1} 1=\frac{\pi}{4}\) = RHS.

∴ LHS = RHS.

Example 14 Prove that tan^-1\(\frac{1}{3}\) + tan^-1\(\frac{1}{5}\) + tan^-1\(\frac{1}{7}\) + tan^-1\(\frac{1}{8}\) = \(\frac{\pi}{4}\).

Solution

We have

LHS = \(\left(\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}\right)+\left(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}\right)\)

= \(\tan ^{-1} \frac{\left(\frac{1}{3}+\frac{1}{5}\right)}{\left(1-\frac{1}{3} \times \frac{1}{5}\right)}+\tan ^{-1} \frac{\left(\frac{1}{7}+\frac{1}{8}\right)}{\left(1-\frac{1}{7} \times \frac{1}{8}\right)}=\tan ^{-1} \frac{(9 / 5)}{(14 / 5)}+\tan ^{-1} \frac{(15 / 56)}{(55 / 56)}\)

= \(\tan ^{-1} \frac{8}{14}+\tan ^{-1} \frac{15}{55}=\tan ^{-1} \frac{4}{7}+\tan ^{-1} \frac{3}{11}\)

= \(\tan ^{-1} \frac{\left(\frac{4}{7}+\frac{3}{11}\right)}{\left(1-\frac{4}{7} \times \frac{3}{11}\right)}=\tan ^{-1} \frac{(65 / 77)}{(65 / 77)}=\tan ^{-1} 1=\frac{\pi}{4}\) = RHS.

∴ LHS = RHS.

Example 15 Prove that 2tan^-1\(\frac{1}{2}\) + tan^-1\(\frac{1}{7}\) = tan^-1\(\frac{31}{17}\).

Solution

We have

LHS = 2tan^-1\(\frac{1}{2}\) + tan^-1\(\frac{1}{7}\)

= \(\tan ^{-1} \frac{\left(2 \times \frac{1}{2}\right)}{\left\{1-\left(\frac{1}{2}\right)^2\right\}}+\tan ^{-1} \frac{1}{7}\) [∵ \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)]

= \(\tan ^{-1} \frac{1}{(3 / 4)}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7}\)

= \(\tan ^{-1} \frac{\left(\frac{4}{3}+\frac{1}{7}\right)}{\left(1-\frac{4}{3} \times \frac{1}{7}\right)}=\tan ^{-1} \frac{(31 / 21)}{(17 / 21)}=\tan ^{-1} \frac{31}{17}\) = RHS.

∴ LHS = RHS.

Example 16 Prove that \(2\left(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}\right)=\tan ^{-1} \frac{4}{3} \text {. }\)

Solution

We have

LHS = 2tan^-1\(\frac{1}{4}\) + 2tan^-1\(\frac{2}{9}\)

= \(\tan ^{-1} \frac{\left(2 \times \frac{1}{4}\right)}{\left\{1-\left(\frac{1}{4}\right)^2\right\}}+\tan ^{-1} \frac{\left(2 \times \frac{2}{9}\right)}{\left\{1-\left(\frac{2}{9}\right)^2\right\}}\)

= \(\tan ^{-1} \frac{(1 / 2)}{(15 / 16)}+\tan ^{-1} \frac{(4 / 9)}{(7 / 81)}=\tan ^{-1}\left(\frac{1}{2} \times \frac{16}{15}\right)+\tan ^{-1}\left(\frac{4}{9} \times \frac{81}{77}\right)\)

= \(\tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{36}{77}\)

= \(\tan ^{-1} \frac{\left(\frac{8}{15}+\frac{36}{77}\right)}{\left(1-\frac{8}{15} \times \frac{36}{77}\right)}=\tan ^{-1} \frac{(616+540)}{(1155-288)}\)

= \(\tan ^{-1}\left(\frac{1156}{867}\right)=\tan ^{-1} \frac{4}{3}\) = RHS.

∴ LHS = RHS.

Example 17 Prove that cot^-17 + cot^-18 + cot^-118 = cot^-13.

Solution

We have

LHS = cot^-17 + cot^-18 + cot^-118

= \(\left(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}\right)+\tan ^{-1} \frac{1}{18}\)

= \(\tan ^{-1} \frac{\left(\frac{1}{7}+\frac{1}{8}\right)}{\left(1-\frac{1}{7} \times \frac{1}{8}\right)}+\tan ^{-1} \frac{1}{18}=\tan ^{-1} \frac{(15 / 56)}{(55 / 56)}+\tan ^{-1} \frac{1}{18}\)

= \(\left(\tan ^{-1} \frac{3}{11}+\tan ^{-1} \frac{1}{18}\right)=\tan ^{-1} \frac{\left(\frac{3}{11}+\frac{1}{18}\right)}{\left(1-\frac{3}{11} \times \frac{1}{18}\right)}\)

= \(\tan ^{-1} \frac{(65 / 198)}{(195 / 198)}=\tan ^{-1} \frac{1}{3}=\cot ^{-1} 3\) = RHS.

∴ LHS = RHS.

Example 18 Prove that sin^-1\(\frac{3}{5}\) – sin^-1\(\frac{8}{17}\) = cos^-1\(\frac{84}{85}\).

Solution

Let sin^-1\(\frac{3}{5}\) = x and sin^-1\(\frac{8}{17}\) = y. Then,

sin x = \(\frac{3}{5}\) and sin y = \(\frac{8}{17}\).

∴ \(\cos x=\sqrt{1-\sin ^2 x}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)

and \(\cos y=\sqrt{1-\sin ^2 y}=\sqrt{1-\frac{64}{289}}=\sqrt{\frac{225}{289}}=\frac{15}{17} \text {. }\)

∴ cos(x-y) = cos x cos y + sin x sin y

= \(\left(\frac{4}{5} \times \frac{15}{17}\right)+\left(\frac{3}{5} \times \frac{8}{17}\right)=\left(\frac{12}{17}+\frac{24}{85}\right)=\frac{84}{85}\)

⇒ \(x-y=\cos ^{-1}\left(\frac{84}{85}\right) ⇒ \sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{84}{85} \text {. }\)

Example 19 Prove that \(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}\)

Solution

LHS = \(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}, where xy = \left(\frac{1}{4} \times \frac{2}{9}\right)=\frac{1}{18}<1\)

= \(\tan ^{-1}\left\{\frac{\left(\frac{1}{4}+\frac{2}{9}\right)}{\left(1-\frac{1}{4} \times \frac{2}{9}\right)}\right\}=\tan ^{-1}\left(\frac{17}{34}\right)=\tan ^{-1} \frac{1}{2}\)

Now, let \(\frac{1}{2}cos-1\frac{3}{5}\) = θ. Then, cos 2θ = \(\frac{3}{5}\).

∴ \(\tan \theta=\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}=\sqrt{\frac{\left(1-\frac{3}{5}\right)}{\left(1+\frac{3}{5}\right)}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)

⇒ \(\tan ^{-1} \frac{1}{2}=\theta\)

⇒ \(\tan ^{-1} \frac{1}{2}=\frac{1}{2} \cos ^{-1} \frac{3}{5}\).

Hence, \(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}\).

Example 20 Prove that \(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{3}{5}=\tan ^{-1} \frac{27}{11} \text {. }\)

Solution

Let cos^-1\(\frac{4}{5}\) = θ. Then cos θ = \(\frac{4}{5}\).

∴ \(\tan \theta=\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta}=\frac{\sqrt{1-\frac{16}{25}}}{(4 / 5)}=\left(\frac{3}{5} \times \frac{5}{4}\right)=\frac{3}{4}\)

⇒ θ = tan^-1\(\frac{3}{4}\).

∴ \(\cos ^{-1} \frac{4}{5}=\tan ^{-1} \frac{3}{4} \text {. }\)

∴ \(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}\)

= \(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{3}{5}}{1-\frac{3}{4} \times \frac{3}{5}}\right)=\tan ^{-1} \frac{27}{11} \text {. }\)

Example 21 Solve tan^-12x + tan^-13x = \(\frac{\pi}{4}\).

Solution

tan^-12x + tan^-13x = \(\frac{\pi}{4}\)

⇒ \(\tan ^{-1}\left(\frac{2 x+3 x}{1-6 x^2}\right)=\frac{\pi}{4} ⇒ \tan ^{-1}\left(\frac{5 x}{1-6 x^2}\right)=\frac{\pi}{4}\)

⇒ \(\frac{5 x}{\left(1-6 x^2\right)}=\tan \frac{\pi}{4}=1\) ⇒ 1 – 6x² = 5x

⇒ 6x² + 5x – 1 = 0 ⇒ 6x² + 6x – x – 1 = 0

⇒ 6x(x+1) – (x+1) = 0 ⇒ (x+1)(6x-1) = 0

⇒ x = -1 or x = \(\frac{1}{6}\).

⇒ x = \(\frac{1}{6}\) [∵ x = -1 makes LHS negative].

Example 22 Solve \(\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)

Solution

We have

\(\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)

⇒ \(\tan ^{-1} \frac{\left\{\frac{(x-1)}{(x-2)}+\frac{(x+1)}{(x+2)}\right\}}{\left\{1-\frac{(x-1)}{(x-2)} \cdot \frac{(x+1)}{(x+2)}\right\}}=\frac{\pi}{4}\)

⇒ \(\tan ^{-1}\left\{\frac{(x-1)(x+2)+(x+1)(x-2)}{\left(x^2-4\right)-\left(x^2-1\right)}\right\}=\frac{\pi}{4}\)

⇒ \(\tan ^{-1}\left\{\frac{\left(x^2+x-2\right)+\left(x^2-x-2\right)}{-3}\right\}=\frac{\pi}{4}\)

⇒ \(\frac{2 x^2-4}{-3}=\tan \frac{\pi}{4}=1\) ⇒ 2x² – 4 = -3

⇒ 2x² = 1 ⇒ x² = \(\frac{1}{2}\) ⇒ x = \(\pm \frac{1}{\sqrt{2}}\)

Example 23 Solve tan^-1(x+1) + tan^-1(x-1) = tan^-1\(\frac{8}{31}\).

Solution

We have

tan^-1(x+1) + tan^-1(x-1) = tan^-1\(\frac{8}{31}\)

⇒ \(\tan ^{-1}\left\{\frac{(x+1)+(x-1)}{1-(x+1)(x-1)}\right\}=\tan ^{-1} \frac{8}{31}\)

⇒ \(\tan ^{-1}\left(\frac{2 x}{2-x^2}\right)=\tan ^{-1} \frac{8}{31}\)

⇒ \(\tan \left\{\tan ^{-1}\left(\frac{2 x}{2-x^2}\right)\right\}=\frac{8}{31}\) ⇒ \(\frac{2 x}{2-x^2}=\frac{8}{31}\)

⇒ 8x² + 62x – 16 = 0 ⇒ 4x² + 31x – 8 = 0

⇒ (4x – 1)(x + 8) = 0 ⇒ x = \(\frac{1}{4}\) or x = -8.

But, x = -8 gives LHS = tan^-1(-7) + tan^-1(-9), which is negative, while RHS is positive. So, x = -8 is not possible.

Hence, x = \(\frac{1}{4}\).

Examples of Principal Values of Inverse Trigonometric Functions

Example 24 Solve \(\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x,(x>0)\)

Solution

We have

\(\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x,(x>0)\)

⇒ \(\tan ^{-1} 1-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x\)

⇒ \(\frac{3}{2} \tan ^{-1} x=\tan ^{-1} 1=\frac{\pi}{4}\)

⇒ \(\tan ^{-1} x=\left(\frac{\pi}{4} \times \frac{2}{3}\right)=\frac{\pi}{6}\) ⇒ \(x=\tan \left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}.

Hence, x = [latex]\frac{1}{\sqrt{3}}\).

Example 25 Solve 2 tan^-1(cos x) = tan^-1(2 cosec x).

Solution

We have

2 tan^-1(cos x) = tan^-1(2 cosec x)

⇒ \(\tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}(2 {cosec} x)\)

⇒ \(\tan \left[\tan ^{-1}\left(\frac{2 \cos ^x x}{\sin ^2 x}\right)\right]=2 {cosec} x\)

⇒ \(\frac{2 \cos x}{\sin ^2 x}=2 {cosec} x\) ⇒ cos x = sin x

⇒ tan x = 1 ⇒ x = \(\frac{\pi}{4}\).

Problems Based On Trigonometric Formulae

Example 26 Prove that \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\frac{x}{2}, x<\pi\)

Solution

We have

LHS = \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\tan ^{-1} \sqrt{\frac{2 \sin ^2(x / 2)}{2 \cos ^2(x / 2)}}\)

= \(\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}=\text { RHS. }\)

∴ \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\frac{x}{2} \text {. }\)

Example 27 Prove that \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\left(\frac{\pi}{4}-x\right), x<\pi \text {. }\)

Solution

We have

LHS = \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\)

= \(\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)\) [dividing num. and denom. by cos x]

= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-x\right)\right\}=\left(\frac{\pi}{4}-x\right)=\text { RHS }\).

∴ \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\left(\frac{\pi}{4}-x\right)\)

Example 28 Prove that \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\left(\frac{\pi}{4}-\frac{x}{2}\right) \text {. }\)

Solution

We have

LHS = \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\)

= \(\tan ^{-1}\left\{\frac{\sin \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}\right\}\)

= \(\tan ^{-1}\left\{\frac{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right\}\)

= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}=\left(\frac{\pi}{4}-\frac{x}{2}\right)=\text { RHS }\)

∴ \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\left(\frac{\pi}{4}-\frac{x}{2}\right) \text {. }\)

Example 29 Prove that \(\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)=\left(\frac{\pi}{4}+\frac{x}{2}\right), x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Solution

We have

LHS = \(\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)\)

= \(\tan ^{-1}\left\{\frac{\sin \left(\frac{\pi}{2}-x\right)}{1-\cos \left(\frac{\pi}{2}-x\right)}\right\}\)

= \(\tan ^{-1}\left\{\frac{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right\}\)

= \(\tan ^{-1}\left\{\cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}=\tan ^{-1}\left[\tan \left\{\frac{\pi}{2}-\left(\frac{\pi}{4}-\frac{x}{2}\right)\right]\right\}\)

= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}\)

= \(\left(\frac{\pi}{4}+\frac{x}{2}\right)\) = RHS.

Hence, \(\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right)=\left(\frac{\pi}{4}+\frac{x}{2}\right)\).

Example 30 Prove that \(\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)=3 \tan ^{-1} \frac{x}{a}\).

Solution

Putting x = atan θ, we get

\(\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)=\tan ^{-1}\left(\frac{3 a^3 \tan \theta-a^3 \tan ^3 \theta}{a^3-3 a^3 \tan ^2 \theta}\right)\)

= \(\tan ^{-1}\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)=\tan ^{-1}(\tan 3 \theta)\)

= \(3 \theta=3 \tan ^{-1} \frac{x}{a}\)

Hence, \(\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)=3 \tan ^{-1} \frac{x}{a}\)

Example 31 Prove that \(\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x-a \sin x}\right)=\tan ^{-1}\left(\frac{a}{b}\right)-x\).

Solution

We have

LHS = \(\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x-a \sin x}\right)=\tan ^{-1}\left\{\frac{\frac{a \cos x-b \sin x}{b \cos x}}{\frac{b \cos x-a \sin x}{b \cos x}}\right\}\)

[on dividing num. and denom. by cos x]

= \(\tan ^{-1}\left\{\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b} \tan x}\right\}=\tan ^{-1}\left(\frac{p-q}{1+p q}\right)\), where \(\frac{a}{b}\) = p and tan x = q

= tan^-1p – tan^-1q = tan-1\(\frac{a}{b}\) – tan^-1(tan x)

= \(\left(\tan ^{-1} \frac{a}{b}-x\right)\) = RHS.

Hence, \(\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x-a \sin x}\right)=\left(\tan ^{-1} \frac{a}{b}-x\right) \text {. }\)

Graphing Inverse Trigonometric Functions: Techniques and Examples

Example 32 Prove that \(\cot ^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)\)

Solution

We have

LHS = \(\cot ^{-1}\left\{\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})} \times \frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}\right\}\)

= \(\cot ^{-1}\left\{\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{1-\sin ^2 x}}{(1+\sin x)-(1-\sin x)}\right\}\)

= \(\cot ^{-1}\left\{\frac{2(1+\cos x)}{2 \sin x}\right\}=\cot ^{-1}\left(\frac{1+\cos x}{\sin x}\right)\)

= \(\cot ^{-1}\left\{\frac{2 \cos ^2(x / 2)}{2 \sin (x / 2) \cos (x / 2)}\right\}=\cot ^{-1}\left(\cot \frac{x}{2}\right)=\frac{x}{2}\) = RHS.

Hence, LHS = RHS.

Example 33 Prove that \(\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\)

Solution

\(\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\)

Putting x = cos θ, we get

LHS = \(\tan ^{-1}\left\{\frac{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}\right\}\)

= \(\tan ^{-1}\left\{\frac{\sqrt{2 \cos ^2(\theta / 2)}-\sqrt{2 \sin ^2(\theta / 2)}}{\sqrt{2 \cos ^2(\theta / 2)}+\sqrt{2 \sin ^2(\theta / 2)}}\right\}=\tan ^{-1}\left\{\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}\right\}\)

= \(\tan ^{-1}\left\{\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}\right\}\) [dividing num. and denom. by cos \frac{θ}{2}]

= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right\}=\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)

= \(\left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\right)\) = RHS.

Hence, LHS = RHS.

Example 34 Prove that \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{1}{2} \tan ^{-1} x\).

Solution

\(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{1}{2} \tan ^{-1} x\)

Putting x = tan θ, we get

\(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)\)

= \(\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)\)

= \(\tan ^{-1}\left\{\frac{2 \sin ^2\left(\frac{\theta}{2}\right)}{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}\right\}=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)\)

= \(\frac{θ}{2}\) = \(\frac{1}{2}tan-1x\).

∴ \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{1}{2} \tan ^{-1} x\)

Example 35 Prove that \(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)=\sin ^{-1} \frac{x}{a}\)

Solution

\(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)=\sin ^{-1} \frac{x}{a}\)

Putting x = asin θ, we get

LHS = \(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)\)

= \(\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}}\right)\)

= \(\tan ^{-1}\left(\frac{{asin} \theta}{a \cos \theta}\right)=\tan ^{-1}(\tan \theta)\)

= θ = sin^-1 \(\frac{x}{a}\) = RHS.

∴ \(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)=\sin ^{-1} \frac{x}{a}\)

Example 36 Prove that \(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right) .\)

Solution

\(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right) .\)

Putting x = tan²θ, we get

RHS = \(\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)\)

= \(\frac{1}{2} \cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\)

= \(\frac{1}{2} \cos ^{-1}(\cos 2 \theta)\)

= \(\left(\frac{1}{2} \times 2 \theta\right)=\theta=\tan ^{-1} \sqrt{x}\) = LHS

[∵ x = tan²θ ⇒ tan θ = √x ⇒ θ = tan^-1√x].

Hence, \(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right).\)

Example 37 Prove that \(\tan ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\left(\frac{\pi}{2}-\sec ^{-1} x\right)\)

Solution

\(\tan ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\left(\frac{\pi}{2}-\sec ^{-1} x\right)\)

Putting x = sec θ, we get

\(\tan ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\tan ^{-1}\left(\frac{1}{\sqrt{\sec ^2 \theta-1}}\right)\)

= \(\tan ^{-1}\left(\frac{1}{\tan \theta}\right)=\tan ^{-1}(\cot \theta)\)

= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\theta\right)\right\}=\left(\frac{\pi}{2}-\theta\right)=\left(\frac{\pi}{2}-\sec ^{-1} x\right) .\)

∴ \(\tan ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\left(\frac{\pi}{2}-\sec ^{-1} x\right)\)

Example 38 Prove that \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2\)

Solution

\(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2\)

Putting x² = cos 2θ, we get

\(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right)\)

= \(\tan ^{-1}\left(\frac{\sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta}}{\sqrt{2 \cos ^2 \theta}-\sqrt{2 \sin ^2 \theta}}\right)\)

= \(\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)\)

[dividing num. and denom. by cos θ]

= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+\theta\right)\right\}=\left(\frac{\pi}{4}+\theta\right)\)

= \(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2\)

∴ \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2\)

Example 39 Prove that \(2 \tan ^{-1} \frac{1}{x}=\sin ^{-1}\left(\frac{2 x}{x^2+1}\right) \text {. }\)

Solution

\(2 \tan ^{-1} \frac{1}{x}=\sin ^{-1}\left(\frac{2 x}{x^2+1}\right) \text {. }\)

Let tan^-1\(\frac{1}{x}\) = θ. Then, \(\frac{1}{x}\) = tan θ ⇒ x = cot θ.

∴ LHS = \(2 \tan ^{-1} \frac{1}{x}\) = 2θ.

RHS = \(\sin ^{-1}\left(\frac{2 \cot \theta}{\cot ^2 \theta+1}\right)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)

= sin^-1(sin 2θ) = 2θ.

∴ LHS = RHS.

Hence, \(2 \tan ^{-1} \frac{1}{x}=\sin ^{-1}\left(\frac{2 x}{x^2+1}\right) \text {. }\)

Example 40 Prove that \(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1-x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}=\left(\frac{x+y}{1-x y}\right)\), where |x| < 1, y > 0 and xy < 1.

Solution

\(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1-x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}=\left(\frac{x+y}{1-x y}\right)\)

Putting x = tan θ and y = tan Φ, we get

LHS = \(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1-x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}\)

= \(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-\tan ^2 \phi}{1+\tan ^2 \phi}\right)\right\}\)

= \(\tan \left\{\frac{1}{2} \sin ^{-1}(\sin 2 \theta)+\frac{1}{2} \cos ^{-1}(\cos 2 \phi)\right\}\)

= \(\tan \left\{\left(\frac{1}{2} \times 2 \theta\right)+\left(\frac{1}{2} \times 2 \phi\right)\right\}\)

= \(\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\frac{(x+y)}{(1-x y)}\) = RHS.

∴ \(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1-x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}=\left(\frac{x+y}{1-x y}\right)\)

Real-Life Applications of Inverse Trigonometric Functions

Example 41 Prove that \(\cot ^{-1}\left(\frac{a b+1}{a-b}\right)+\cot ^{-1}\left(\frac{b c+1}{b-c}\right)+\cot ^{-1}\left(\frac{c a+1}{c-a}\right)=0 \text {. }\)

Solution

We have

LHS = \(\tan ^{-1}\left(\frac{a-b}{1+a b}\right)+\tan ^{-1}\left(\frac{b-c}{1+b c}\right)+\tan ^{-1}\left(\frac{c-a}{1+c a}\right)\)

= (tan^-1a – tan^-1b) + (tan^-1b – tan^-1c) + (tan^-1c – tan^-1a)

= 0 = RHS.

∴ LHS = RHS.

Graphs Of Inverse Trigonometric Functions

1. Graph of sin-1x

Let f:[-1,1] → \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) : f(x) = sin^-1x.

Here \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) is called the principal-value branch of sin^-1x.

The other branches of sin^-1x are

\(\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right],\left[\frac{3 \pi}{2}, \frac{5 \pi}{2}\right], \ldots \text { and }\left[\frac{-3 \pi}{2}, \frac{-\pi}{2}\right],\left[\frac{-5 \pi}{2}, \frac{-3 \pi}{2}\right]\), etc

Table for sin-1x

Also, sin^-1(-x) = -sin^-1x.

∴ \(\left(x=\frac{-1}{2} \Rightarrow \sin ^{-1} x=\frac{-\pi}{6}\right)\),

\(\left(x=-0.7 \Rightarrow \sin ^{-1} x=\frac{-\pi}{4}\right)\),

\(\left(x=-0.86 \Rightarrow \sin ^{-1} x=\frac{-\pi}{3}\right)\),

\(\left(x=-1 \Rightarrow \sin ^{-1} x=\frac{-\pi}{2}\right) .\)

O(0,0), A(\(\frac{1}{2}\), \(\frac{\pi}{6}\)), B(0.7, \(\frac{\pi}{4}\)), c(0.86, \(\frac{\pi}{3}\)), D(1, \(\frac{\pi}{2}\)), E(\(-frac{1}{2}\), \(\frac{-\pi}{6}\)), F(-0.7, \(\frac{-\pi}{4}\)). G(-0.86, \(\frac{-\pi}{3}\)) and H(-1, \(\frac{-\pi}{2}\)).

Join the points OA, AB, BC, CD, and OE, EF, FG, GH successively with a freehand to get the required graph, as shown in the given figure.

Moreover, we have

\(\sin \frac{2 \pi}{3}=\sin \left(\pi-\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=0.86\) \(\sin \frac{5 \pi}{6}=\sin \left(\pi-\frac{\pi}{6}\right)=\sin \frac{\pi}{6}=0.5, \sin \pi=0 \text {. }\) \(\sin \left(\frac{-2 \pi}{3}\right)=-\sin \frac{2 \pi}{3}=-0.86 .\) \(\sin \left(\frac{-5 \pi}{6}\right)=-\sin \frac{5 \pi}{6}=-0.5\)

Now, we may extend the graph as shown in the figure.

2. Graph of cos^-1x

Let f : [-1,1] → [0, π] : f(x) = cos^-1x.

Here, [0, π] is called the principal value branch of cos^-1x.

The other branches of cos^-1x are [π, 2π], [2π, 3π], …, [-π,0], [-2π, -π], etc.

Table for cos^-1x

On a graph paper, we plot the points

A(1,0), B(0.87, \(\frac{\pi}{6}\)), C(0.7, \(\frac{\pi}{4}\)), D(0.5, \(\frac{\pi}{3}\)), E(0, \(\frac{\pi}{2}\)), F(-0.5, \(\frac{2 \pi}{3}\)), G(-0.7, \(\frac{3 \pi}{4}\)), H(-0.86, \(\frac{5 \pi}{6}\)) and K(-1, π).

Join AB, BC, CD, DE, EF, FG, GH and HK successively with a freehand to obtain the graph of cos^-1x, as shown in the given figure.

3. Graph of tan^-1x

Let f : R → \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) : f(x) = tan^-1x.

Here \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) is the principal value branch of tan^-1x.

The other branches of tan^-1x are (\(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)), (\(\frac{3 \pi}{2}\), \(\frac{5 \pi}{2}\)), … , (\(-\frac{3 \pi}{2}\), \(-\frac{\pi}{2}\)), etc.

We take the positive values and then use tan^-1(-x) = -tan^-1x.

Table for tan^-1x

Also, tan^-1(-x) = -tan^-1x.

∴ tan^-1(-0.58) = \(-\frac{\pi}{6}\), tan^-1(-1) = \(-\frac{\pi}{4}\), tan^-1(-1.73) = \(-\frac{\pi}{3}\).

On a graph paper, we plot the points O(0,0), A(0.58, \(\frac{\pi}{6}\)), B(1, \(\frac{\pi}{4}\)), C(1.73, \(\frac{\pi}{3}\)), and D = (0.58, \(-\frac{\pi}{6}\)), E(-1, \(-\frac{\pi}{4}\)), F(-1.73, \(-\frac{\pi}{3}\)).

Join OA, AB, BC, and OD, DE, EF successively to get the graph.

Now, when x → ∞, then tan^-1x → \(\frac{\pi}{2}\).

Also, when x → ∞, then tan^-1x → \(-\frac{\pi}{2}\).

4. Graph of cot^-1x

Let f : R → (0, π) : f(x) = cot^-1x.

Here, (0,π) is the principal value branch of cot^-1x.

The other branches of cot-1x are (π, 2π), (2π, 3π), …, (-π,0), etc.

Table for cot^-1x

Using cot(π – x) = -cot x, we have

\(\cot \frac{2 \pi}{3}=\cot \left(\pi-\frac{\pi}{3}\right)=-\cot \frac{\pi}{3}=-0.58, \cot \frac{3 \pi}{4}=\cot \left(\pi-\frac{\pi}{4}\right)=-\cot \frac{\pi}{4}=-1\) \(\cot \frac{5 \pi}{6}=\cot \left(\pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-1.73 \text {. }\)

On a graph paper, we plot the points \(A\left(1.73, \frac{\pi}{6}\right), B\left(1, \frac{\pi}{4}\right), C\left(0.58, \frac{\pi}{3}\right)\),

\(D\left(0, \frac{\pi}{2}\right), E\left(-0.58, \frac{2 \pi}{3}\right), F\left(-1, \frac{3 \pi}{4}\right), \text { and } G\left(-1.73, \frac{5 \pi}{6}\right) \text {. }\)

We join the points AB, BC, CD, DE, EF, FG successively to get the graph.

As x → ∞, then cot^-1x → 0.

And, as x → ∞, then cot^-1x → π.

5. Graph of sec^-1x

Let f : R – (-1,1) → [0, π] – \(\frac{\pi}{2}\) : f(x) = sec^-1x.

The other branches of sec^-1x are [π, 2π] – \(\frac{3 \pi}{2}\), …, [-π, 0] – \(\frac{\pi}{2}\), etc.

Table for sec^-1x

\(\left\{\sec \frac{2 \pi}{3}=-2 \Rightarrow \sec ^{-1}(-2)=\frac{2 \pi}{3}\right\},\left\{\sec \frac{3 \pi}{4}=-\sqrt{2}=-1.41 \Rightarrow \sec ^{-1}(-1.41)=\frac{3 \pi}{4}\right\}\) \(\left\{\sec \frac{5 \pi}{6}=\frac{-2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-2 \sqrt{3}}{3}=\frac{-2 \times 1.73}{3}=\frac{-3.46}{3}=-1.15 \Rightarrow \sec ^{-1}(-1.15)=\frac{5 \pi}{6}\right\}\)

sec π = -1 ⇒ sec^-1(-1) = π.

On a graph paper, we plot the points

A(1,0), B(1.15, \(\frac{\pi}{6}\)), C(1.41, \(\frac{\pi}{4}\)), D(2, \(\frac{\pi}{3}\)) and E(-2, \(\frac{2 \pi}{3}\)), F(-1.41, \(\frac{3 \pi}{4}\)), G(-1.15, \(\frac{5 \pi}{6}\)), H(-1, π).

We join the points AB, BC, CD, DE, and HG, GF, FE.

6. Graph of cosec^-1x

Let f : R -(-1,1) → \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) – {0} : f(x) = cosec^-1x.

The other branches of cosec^-1x are \(\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]-\{\pi\}\), …, etc.

Table for cosec^-1x

Since cosec^-1(-x) = -cosec^-1x, we have

\({cosec}^{-1}\left(\frac{-\pi}{6}\right)=-2, {cosec}^{-1}\left(\frac{-\pi}{4}\right)=-1.41, {cosec}^{-1}\left(\frac{-\pi}{3}\right)=-1.15 and {cosec}^{-1}\left(\frac{-\pi}{2}\right)=-1\)

On a graph paper, we plot the points \(A\left(2, \frac{\pi}{6}\right), B\left(1.41, \frac{\pi}{4}\right), C\left(1.15, \frac{\pi}{3}\right)\)

\(D\left(1, \frac{\pi}{2}\right) \text { and } E\left(-2, \frac{-\pi}{6}\right), F\left(-1.41, \frac{-\pi}{4}\right), G\left(-1.15, \frac{-\pi}{3}\right), H\left(-1, \frac{-\pi}{2}\right) \text {. }\)

Join the points as shown in the given figure, to get the graph.

Also as x → 0 from +ve values, then cosec^-1x → ∞.

As x → 0 from -ve values, then cosec^-1x → ∞.