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.
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
\(M_{11}=\left|\begin{array}{rr}
-1 & 2 \\
5 & 2
\end{array}\right|=-12 ; M_{12}=\left|\begin{array}{ll}
4 & 2 \\
3 & 2
\end{array}\right|=2 ; M_{13}=\left|\begin{array}{rr}
4 & -1 \\
3 & 5
\end{array}\right|=23\) ;
\(M_{21}=\left|\begin{array}{rr}
-3 & 2 \\
5 & 2
\end{array}\right|=-16 ; M_{22}=\left|\begin{array}{rr}
1 & 2 \\
3 & 2
\end{array}\right|=-4 ; M_{23}=\left|\begin{array}{rr}
1 & -3 \\
3 & 5
\end{array}\right|=14\) ;
-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|.
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,
△ = \(\left|\begin{array}{rrr}
9 & 9 & 12 \\
1 & -3 & -4 \\
1 & 9 & 12
\end{array}\right|\)
= \(\left|\begin{array}{rrr}
0 & 36 & 48 \\
1 & -3 & -4 \\
0 & 12 & 16
\end{array}\right|\)
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,
△ = \(\left|\begin{array}{lll}
265 & 240 & 219 \\
240 & 225 & 198 \\
219 & 198 & 181
\end{array}\right|\)
= \(\left|\begin{array}{lll}
46 & 21 & 219 \\
42 & 27 & 198 \\
38 & 17 & 181
\end{array}\right|\)
= \(\left|\begin{array}{rrr}
4 & 21 & 9 \\
-12 & 27 & -72 \\
4 & 17 & 11
\end{array}\right|\)
= \(\left|\begin{array}{rrr}
0 & 4 & -2 \\
0 & 78 & -39 \\
4 & 17 & 11
\end{array}\right|\)
= \(\text { 2(39) }\left|\begin{array}{rrr}
0 & 2 & -1 \\
0 & 2 & -1 \\
4 & 17 & 11
\end{array}\right|\)
[taking 2 common from R1 and 39 common from R2]
= (78 x 0) = 0 [∵ R1 and R2 are identical].
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,
△ = \(\left|\begin{array}{ccc}
\sin \alpha & \cos \alpha & \cos \alpha \cos \delta-\sin \alpha \sin \delta \\
\sin \beta & \cos \beta & \cos \beta \cos \delta-\sin \beta \sin \delta \\
\sin \gamma & \cos \gamma & \cos \gamma \cos \delta-\sin \gamma \sin \delta
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
\sin \alpha & \cos \alpha & 0 \\
\sin \beta & \cos \beta & 0 \\
\sin \gamma & \cos \gamma & 0
\end{array}\right|\)
= 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,
△ = \(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
2 a & 3 a+2 b & 4 a+3 b+2 c \\
3 a & 6 a+3 b & 10 a+6 b+3 c
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
0 & a & 2 a+b \\
0 & 3 a & 7 a+3 b
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
0 & a & 2 a+b \\
0 & 0 & a
\end{array}\right|\)
= \(a \cdot\left|\begin{array}{cc}
a & 2 a+b \\
0 & a
\end{array}\right|\) [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,
△ = \(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
x+y+z & x+y+z & x+y+z \\
z & x & y \\
1 & 1 & 1
\end{array}\right|\)
= \((x+y+z) \cdot\left|\begin{array}{lll}
1 & 1 & 1 \\
z & x & y \\
1 & 1 & 1
\end{array}\right|\) [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,
△ = \(\left|\begin{array}{ccc}
x+y & x & x \\
5 x+4 y & 4 x & 2 x \\
10 x+8 y & 8 x & 3 x
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
x+y & x & x \\
3 x+2 y & 2 x & 0 \\
7 x+5 y & 5 x & 0
\end{array}\right|\)
= \(x \cdot\left|\begin{array}{ll}
3 x+2 y & 2 x \\
7 x+5 y & 5 x
\end{array}\right|\) [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,
△ = \(\left|\begin{array}{ccc}
0 & a b^2 & a c^2 \\
a^2 b & 0 & b c^2 \\
a^2 c & c b^2 & 0
\end{array}\right|\)
= \(\left(a^2 b^2 c^2\right) \cdot\left|\begin{array}{lll}
0 & a & a \\
b & 0 & b \\
c & c & 0
\end{array}\right|\)
\text { taking } a^2, b^2, c^2 \text { common from } C_1, C_2 \\
\text { and } C_3 \text { respectively }
\end{array}\right]\)
= \(\left(a^3 b^3 c^3\right) \cdot\left|\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right|\)
\text { taking } a, b, c \text { common from } R_1, R_2 \\
\text { and } R_3 \text { respectively }
\end{array}\right]\)
= \(\left(a^3 b^3 c^3\right) \cdot\left|\begin{array}{rrr}
0 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & -1
\end{array}\right|\)
= \(\left(a^3 b^3 c^3\right) \cdot(-1) \cdot\left|\begin{array}{rr}
1 & 1 \\
1 & -1
\end{array}\right|\) [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,
△ = \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|\)
= \((a b c) \cdot\left|\begin{array}{rrr}
-a & a & a \\
b & -b & b \\
c & c & -c
\end{array}\right|\)
\text { taking out } a, b, c \text { common from } \\
C_1, C_2, C_3 \text { respectively }
\end{array}\right]\)
= \((a b c) \cdot\left|\begin{array}{ccc}
-a & 0 & 0 \\
b & 0 & 2 b \\
c & 2 c & 0
\end{array}\right|\)
= \((a b c) \cdot(-a)\left|\begin{array}{cc}
0 & 2 b \\
2 c & 0
\end{array}\right|=(a b c)(-a)(-4 b c)\)
= \(4 a^2 b^2 c^2\).
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,
△ = \(\left|\begin{array}{ccc}
a+x & y & z \\
x & a+y & z \\
x & y & a+z
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
a+x+y+z & y & z \\
a+x+y+z & a+y & z \\
a+x+y+z & y & a+z
\end{array}\right|\)
= \((a+x+y+z) \cdot\left|\begin{array}{ccc}
1 & y & z \\
1 & a+y & z \\
1 & y & a+z
\end{array}\right|\)
[taking (a + x + y + z) common from C1]
=\((a+x+y+z) \cdot\left|\begin{array}{lll}
1 & y & z \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right|\)
= \((a+x+y+z) \cdot 1 \cdot\left|\begin{array}{ll}
a & 0 \\
0 & a
\end{array}\right|\)
[expanded by C1]
= \(a^2(a+x+y+z) \text {. }\)
Hence, △ = \(a^2(a+x+y+z)\).
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,
△ = \(\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\)
= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\) [taking (a + b + c) common from R1]
= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
2 b & -(a+b+c) & 0 \\
2 c & 0 & -(a+b+c)
\end{array}\right|\)
= \((a+b+c) \cdot 1 \cdot\left|\begin{array}{cc}
-(a+b+c) & 0 \\
0 & -(a+b+c)
\end{array}\right|\)
= \((a+b+c)(a+b+c)^2=(a+b+c)^3 .\)
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,
△ = \(\left|\begin{array}{ccc}
a & b & c \\
a-b & b-c & c-a \\
b+c & c+a & a+b
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
a+b+c & b & c \\
0 & b-c & c-a \\
2(a+b+c) & c+a & a+b
\end{array}\right|\)
= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & b & c \\
0 & b-c & c-a \\
2 & c+a & a+b
\end{array}\right|\) [taking (a + b + c) common from C1]
= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & b & c \\
0 & b-c & c-a \\
0 & c+a-2 b & a+b-2 c
\end{array}\right|\)
= \((a+b+c) \cdot 1 \cdot\left|\begin{array}{cc}
b-c & c-a \\
c+a-2 b & a+b-2 c
\end{array}\right|\) [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,
△ = \(\left|\begin{array}{ccc}
3 a & -a+b & -a+c \\
a-b & 3 b & c-b \\
a-c & b-c & 3 c
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
a+b+c & -a+b & -a+c \\
a+b+c & 3 b & c-b \\
a+b+c & b-c & 3 c
\end{array}\right|\)
= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & -a+b & -a+c \\
1 & 3 b & c-b \\
1 & b-c & 3 c
\end{array}\right|\)
= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & -a+b & -a+c \\
0 & 2 b+a & a-b \\
0 & a-c & 2 c+a
\end{array}\right|\)
= \((a+b+c) \cdot 1 \cdot\left|\begin{array}{cc}
2 b+a & a-b \\
a-c & 2 c+a
\end{array}\right|\) [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,
△ = \(\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
a\left(1-x^2\right) & c\left(1-x^2\right) & p\left(1-x^2\right) \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|\)
= \(\left(1-x^2\right) \cdot\left|\begin{array}{ccc}
a & c & p \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|\)
= \(\left(1-x^2\right) \cdot\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & w
\end{array}\right|\)
Hence, △ = \(\left(1-x^2\right) \cdot\left|\begin{array}{ccc}
a & c & p \\
b & d & q \\
u & v & w
\end{array}\right| .\)
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 = \(\left|\begin{array}{ccc}
y+z & z & y \\
z & z+x & x \\
y & x & x+y
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
0 & -2 x & -2 x \\
z & z+x & x \\
y & x & x+y
\end{array}\right|\)
= \((-2 x) \cdot\left|\begin{array}{ccc}
0 & 1 & 1 \\
z & z+x & x \\
y & x & x+y
\end{array}\right|\) [taking (-2x) common from R1]
= \((-2 x) \cdot\left|\begin{array}{ccc}
0 & 0 & 1 \\
z & z & x \\
y & -y & x+y
\end{array}\right|\)
= \((-2 x) \cdot 1 \cdot\left|\begin{array}{rr}
z & z \\
y & -y
\end{array}\right|\) [expanded by R1]
= (-2x).1.(-yz – yz) = (-2x)(-2yz) = 4xyz.
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,
△ = \(\left|\begin{array}{ccc}
a+b+2 c & a & b \\
c & b+c+2 a & b \\
c & a & c+a+2 b
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
a+b+c & -(a+b+c) & 0 \\
c & b+c+2 a & b \\
0 & -(a+b+c) & (a+b+c)
\end{array}\right|\)
\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]
= \((a+b+c)^2 \cdot\left|\begin{array}{ccc}
1 & -1 & 0 \\
0 & b+2 c+2 a & b \\
0 & -1 & 1
\end{array}\right|\)
= \((a+b+c)^2 \cdot 1 \cdot\left|\begin{array}{cc}
b+2 c+2 a & b \\
-1 & 1
\end{array}\right|\) [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
= \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^3 & b^3 & c^3
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
0 & 0 & 1 \\
a-c & b-c & c \\
a^3-c^3 & b^3-c^3 & c^3
\end{array}\right|\)
= \((a-c)(b-c) \cdot\left|\begin{array}{ccc}
0 & 0 & 1 \\
1 & 1 & c \\
a^2+a c+c^2 & b^2+b c+c^2 & c^3
\end{array}\right|\)
[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
= \(\left|\begin{array}{ccc}
x & y & z \\
x^2 & y^2 & z^2 \\
y+z & z+x & x+y
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
x+y+z & x+y+z & x+y+z \\
x^2 & y^2 & z^2 \\
y+z & z+x & x+y
\end{array}\right|\)
= \((x+y+z) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
x^2 & y^2 & z^2 \\
y+z & z+x & x+y
\end{array}\right|\) [taking out (x + y + z) common from R1]
= \((x+y+z)\left|\begin{array}{ccc}
1 & 0 & 0 \\
x^2 & y^2-x^2 & z^2-x^2 \\
y+z & x-y & x-z
\end{array}\right|\)
= \((x+y+z)(x-y)(z-x) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
x^2 & -(x+y) & z+x \\
y+z & 1 & -1
\end{array}\right|\)
= \((x+y+z)(x-y)(z-x) \cdot 1 \cdot\left|\begin{array}{cc}
-(x+y) & z+x \\
1 & -1
\end{array}\right|\)
= (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
△ = \(\left|\begin{array}{ccc}
a & b & c \\
a^2 & b^2 & c^2 \\
b c & c a & a b
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
a-c & b-c & c \\
a^2-c^2 & b^2-c^2 & c^2 \\
b(c-a) & a(c-b) & a b
\end{array}\right|\)
= \((a-c)(b-c) \cdot\left|\begin{array}{ccc}
1 & 1 & c \\
a+c & b+c & c^2 \\
-b & -a & a b
\end{array}\right|\)
[taking (a-c) common from C1 and (b-c) common from C2]
= \((a-c)(b-c) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
a+c & b-a & -c a \\
-b & b-a & (a+c) b
\end{array}\right|\)
= \((a-c)(b-c) \cdot 1 \cdot\left|\begin{array}{cc}
b-a & -c a \\
b-a & (a+c) b
\end{array}\right|\) [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]
= \((a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
\frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\
\frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1
\end{array}\right|\)
= \((a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)\left|\begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & \frac{1}{b} \\
-1 & -1 & \frac{1}{c}+1
\end{array}\right|\)
= \((a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right) \cdot(1) \cdot\left|\begin{array}{rr}
0 & 1 \\
-1 & -1
\end{array}\right|\) [expanding by 1st row]
= \((a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right) \cdot 1\)
= \((a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)=(b c+c a+a b+a b c) .\)
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]
= \(\text { (abc) }\left|\begin{array}{ccc}
a & c & c+a \\
0 & -2 c & -2 c \\
b & b+c & c
\end{array}\right|\)
= \((a b c) \cdot\left|\begin{array}{ccc}
a & -a & c+a \\
0 & 0 & -2 c \\
b & b & c
\end{array}\right|\)
= \((a b c)(2 c) \cdot\left|\begin{array}{rr}
a & -a \\
b & b
\end{array}\right|\) [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
= \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|\)
= \(\left|\begin{array}{lll}
a+b+c & b & c \\
a+b+c & c & a \\
a+b+c & a & b
\end{array}\right|\)
= \((a+b+c) \cdot\left|\begin{array}{lll}
1 & b & c \\
1 & c & a \\
1 & a & b
\end{array}\right|\)
= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & b & c \\
0 & c-b & a-c \\
0 & a-b & b-c
\end{array}\right|\left[R_2 \rightarrow\left(R_2-R_1\right) \text { and } R_3 \rightarrow\left(R_3-R_1\right)\right]\)
= (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
△ = \(\left|\begin{array}{lll}
m & \frac{1}{2} m(m-1) & \frac{1}{6} \cdot m(m-1)(m-2) \\
n & \frac{1}{2} n(n-1) & \frac{1}{6} \cdot n(n-1)(n-2) \\
p & \frac{1}{2} p(p-1) & \frac{1}{6} \cdot p(p-1)(p-2)
\end{array}\right|\)
= \(\left(\frac{1}{2} \times \frac{1}{6} \times m n p\right) \cdot\left|\begin{array}{lll}
1 & (m-1) & (m-1)(m-2) \\
1 & (n-1) & (n-1)(n-2) \\
1 & (p-1) & (p-1)(p-2)
\end{array}\right|\)
= \(\frac{1}{12} \cdot m n p \cdot\left|\begin{array}{lll}
1 & m-1 & (m-1)(m-2) \\
0 & n-m & (n-m)(n+m-3) \\
0 & p-m & (p-m)(p+m-3)
\end{array}\right|\)
= \(\frac{1}{12} \cdot(m n p)(n-m)(p-m) \cdot\left|\begin{array}{ccl}
1 & m-1 & (m-1)(m-2) \\
0 & 1 & (n+m-3) \\
0 & 1 & (p+m-3)
\end{array}\right|\)
[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
△ = \(\left|\begin{array}{lll}
b+c & a & b \\
c+a & c & a \\
a+b & b & c
\end{array}\right|\)
= \(\left|\begin{array}{lll}
b & a & b \\
c & c & a \\
a & b & c
\end{array}\right|+\left|\begin{array}{lll}
c & a & b \\
a & c & a \\
b & b & c
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
c & c & a \\
a & b & c
\end{array}\right|+\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
a & c & a \\
b & b & c
\end{array}\right|\)
= \((a+b+c) \cdot\left|\begin{array}{lll}
1 & 1 & 1 \\
c & c & a \\
a & b & c
\end{array}\right|+(a+b+c) \cdot\left|\begin{array}{lll}
1 & 1 & 1 \\
a & c & a \\
b & b & c
\end{array}\right|\)
[taking out (a + b + c) common from R1 in each determinant]
= \((a+b+c) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
c & 0 & a-c \\
a & b-a & c-a
\end{array}\right|+(a+b+c) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
a & c-a & 0 \\
b & 0 & c-b
\end{array}\right|\)
= (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
△ = \(\left|\begin{array}{lll}
1 & a & a^2-b c \\
1 & b & b^2-c a \\
1 & c & c^2-a b
\end{array}\right|\)
= \(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|+\left|\begin{array}{ccc}
1 & a & -b c \\
1 & b & -c a \\
1 & c & -a b
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
1 & a & a^2 \\
0 & b-a & b^2-a^2 \\
0 & c-a & c^2-a^2
\end{array}\right|+\left|\begin{array}{ccc}
1 & a & -b c \\
0 & b-a & c(b-a) \\
0 & c-a & b(c-a)
\end{array}\right|\)
= \((b-a)(c-a) \cdot\left|\begin{array}{ccc}
1 & a & a^2 \\
0 & 1 & b+a \\
0 & 1 & c+a
\end{array}\right|+(b-a)(c-a) \cdot\left|\begin{array}{ccc}
1 & a & -b c \\
0 & 1 & c \\
0 & 1 & b
\end{array}\right|\)
= (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.
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]
= \((1+x y z)\left|\begin{array}{lll}
x & x^2 & 1 \\
y & y^2 & 1 \\
z & z^2 & 1
\end{array}\right|\)
= \((1+x y z)\left|\begin{array}{ccc}
x & x^2 & 1 \\
(y-x) & \left(y^2-x^2\right) & 0 \\
(z-x) & \left(z^2-x^2\right) & 0
\end{array}\right|\)
= \((1+x y z)(y-x)(z-x)\left|\begin{array}{ccc}
x & x^2 & 1 \\
1 & y+x & 0 \\
1 & z+x & 0
\end{array}\right|\)
= \((1+x y z)(y-x)(z-x) \cdot 1 \cdot\left|\begin{array}{ll}
1 & y+x \\
1 & z+x
\end{array}\right|\) [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,
△ = \(\left|\begin{array}{ccc}
(b+c)^2 & a^2 & a^2 \\
b^2 & (c+a)^2 & b^2 \\
c^2 & c^2 & (a+b)^2
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
(b+c)^2-a^2 & 0 & a^2 \\
0 & (c+a)^2-b^2 & b^2 \\
c^2-(a+b)^2 & c^2-(a+b)^2 & (a+b)^2
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
(a+b+c)(b+c-a) & 0 & a^2 \\
0 & (a+b+c)(c+a-b) & b^2 \\
(a+b+c)(c-a-b) & (a+b+c)(c-a-b) & (a+b)^2
\end{array}\right|\)
= \((a+b+c)^2 \cdot\left|\begin{array}{ccc}
(b+c-a) & 0 & a^2 \\
0 & c+a-b & b^2 \\
c-a-b & c-a-b & (a+b)^2
\end{array}\right|\)
[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]
= \(\left|\begin{array}{ccc}
a b+b c+a c & a b+b c+a c & a b+b c+a c \\
a b+b c & -a c & b c+a b \\
a c+b c & b c+a c & -a b
\end{array}\right|\)
= \((a b+b c+a c) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
a b+b c & -a c & b c+a b \\
a c+b c & b c+a c & -a b
\end{array}\right|\)
= \((a b+b c+a c) \cdot\left|\begin{array}{ccc}
0 & 0 & 1 \\
0 & -(a b+b c+a c) & b c+a b \\
(a b+b c+a c) & (a b+b c+a c) & -a b
\end{array}\right|\)
= \((a b+b c+a c)^3\) [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,
△ = \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\)
= \(2(a+b+c) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\)
= \(2(a+b+c) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
c+a & b-c & b-a \\
a+b & c-a & c-b
\end{array}\right|\)
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]
= \(\frac{(a b c)}{(a b c)} \cdot\left|\begin{array}{ccc}
a^2+1 & a^2 & a^2 \\
b^2 & b^2+1 & b^2 \\
c^2 & c^2 & c^2+1
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 \\
b^2 & b^2+1 & b^2 \\
c^2 & c^2 & c^2+1
\end{array}\right|\)
= \(\left(1+a^2+b^2+c^2\right) \cdot\left|\begin{array}{ccc}
1 & 1 & 1 \\
b^2 & b^2+1 & b^2 \\
c^2 & c^2 & c^2+1
\end{array}\right|\)
[taking out (1 + a^2 + b^2 + c^2) common from R1]
= \(\left(1+a^2+b^2+c^2\right) \cdot\left|\begin{array}{ccc}
1 & 0 & 0 \\
b^2 & 1 & 0 \\
c^2 & 0 & 1
\end{array}\right|\)
= \(\left(1+a^2+b^2+c^2\right) \cdot 1 \cdot\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) {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
= \(\left|\begin{array}{lll}
1 & (1 / 2) \sin 2 A & (1 / 2)(1+\cos 2 A) \\
1 & (1 / 2) \sin 2 B & (1 / 2)(1+\cos 2 B) \\
1 & (1 / 2) \sin 2 C & (1 / 2)(1+\cos 2 C)
\end{array}\right|\)
= \(\frac{1}{4} \cdot\left|\begin{array}{ccc}
1 & \sin 2 A & 1+\cos 2 A \\
1 & \sin 2 B & 1+\cos 2 B \\
1 & \sin 2 C & 1+\cos 2 C
\end{array}\right|\)
= \(\frac{1}{4} \cdot\left|\begin{array}{ccc}
1 & \sin 2 A & 1+\cos 2 A \\
0 & \sin 2 B-\sin 2 A & \cos 2 B-\cos 2 A \\
0 & \sin 2 C-\sin 2 A & \cos 2 C-\cos 2 A
\end{array}\right|\)
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]
= \(\frac{1}{\sin \alpha \cos \alpha} \cdot\left|\begin{array}{ccc}
0 & 0 & -1 \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \alpha \cos \beta & \sin \alpha \cos \alpha \sin \beta & \cos ^2 \alpha
\end{array}\right|\)
= \(\frac{-1}{\sin \alpha \cos \alpha} \cdot\left[-\sin \alpha \cos \alpha \sin ^2 \beta-\sin \alpha \cos \alpha \cos ^2 \beta\right]\)
= \(\frac{-1}{\sin \alpha \cos \alpha} \cdot(-\sin \alpha \cos \alpha)\left[\sin ^2 \beta+\cos ^2 \beta\right]=1\)
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 = \(\left|\begin{array}{lll}
b+c & q+r & y+z \\
c+a & r+p & z+x \\
a+b & p+q & x+y
\end{array}\right|\)
= \(2\left|\begin{array}{ccc}
a+b+c & p+q+r & x+y+z \\
c+a & r+p & z+x \\
a+b & p+q & x+y
\end{array}\right|\)
= \(2\left|\begin{array}{ccc}
a+b+c & p+q+r & x+y+z \\
-b & -q & -y \\
-c & -r & -z
\end{array}\right|\)
R_2 \rightarrow\left(R_2-R_1\right) \\
\text { and } R_3 \rightarrow\left(R_3-R_1\right)
\end{array}\right]\)
= \(2\left|\begin{array}{ccc}
a & p & x \\
-b & -q & -y \\
-c & -r & -z
\end{array}\right|\)
= \(2(-1) \cdot(-1) \cdot\left|\begin{array}{lll}
a & p & x \\
b & q & y \\
c & r & z
\end{array}\right|\)
[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 = \(\left|\begin{array}{lll}
a & a^2 & b c \\
b & b^2 & c a \\
c & c^2 & a b
\end{array}\right|\)
= \(\frac{1}{a b c} \cdot\left|\begin{array}{lll}
a^2 & a^3 & a b c \\
b^2 & b^3 & a b c \\
c^2 & c^3 & a b c
\end{array}\right|\)
= \(\frac{1}{a b c} \cdot a b c=\left|\begin{array}{lll}
a^2 & a^3 & 1 \\
b^2 & b^3 & 1 \\
c^2 & c^3 & 1
\end{array}\right|\) [taking abc common from C3]
= \(\left|\begin{array}{lll}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right|\)
= RHS.
Hence, LHS = RHS.
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,
△ = \(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\)
= \(\left|\begin{array}{lll}
3 a-x & a-x & a-x \\
3 a-x & a+x & a-x \\
3 a-x & a-x & a+x
\end{array}\right|\)
= \((3 a-x) \cdot\left|\begin{array}{lll}
1 & a-x & a-x \\
1 & a+x & a-x \\
1 & a-x & a+x
\end{array}\right|\) [taking (3a – x) common from C1]
= \((3 a-x) \cdot\left|\begin{array}{ccc}
1 & a-x & a-x \\
0 & 2 x & 0 \\
0 & 0 & 2 x
\end{array}\right|\)
= \((3 a-x) \cdot 1 \cdot\left|\begin{array}{cc}
2 x & 0 \\
0 & 2 x
\end{array}\right|\) [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,
△ = \(\left|\begin{array}{lll}
x-2 & 2 x-3 & 3 x-4 \\
x-4 & 2 x-9 & 3 x-16 \\
x-8 & 2 x-27 & 3 x-64
\end{array}\right|\)
= \(\left|\begin{array}{rrr}
x-2 & 1 & 2 \\
x-4 & -1 & -4 \\
x-8 & -11 & -40
\end{array}\right|\)
= \(\left|\begin{array}{crr}
x-2 & 1 & 2 \\
-2 & -2 & -6 \\
-6 & -12 & -42
\end{array}\right|\)
= \((-2) \cdot(-6) \cdot\left|\begin{array}{ccc}
x-2 & 1 & 2 \\
1 & 1 & 3 \\
1 & 2 & 7
\end{array}\right|\)
= \(\text { 12. }\left|\begin{array}{ccc}
x-3 & 1 & 2 \\
0 & 1 & 3 \\
-1 & 2 & 7
\end{array}\right|\)
= 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,
△ = \(\left|\begin{array}{ccc}
a+b+c-x & c & b \\
a+b+c-x & b-x & a \\
a+b+c-x & a & c-x
\end{array}\right|\)
= \((a+b+c-x)\left|\begin{array}{ccc}
1 & c & b \\
1 & b-x & a \\
1 & a & c-x
\end{array}\right|\)
= \((a+b+c-x)\left|\begin{array}{ccc}
1 & c & b \\
0 & b-x-c & a-b \\
0 & a-c & c-x-b
\end{array}\right|\)
= (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, \(\left(x_1=-2, y_1=-3\right),\left(x_2=3, y_2=2\right) \text { and }\left(x_3=-1, y_3=-8\right) \text {. }\)
∴ \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2} \cdot\left|\begin{array}{rrr}
-2 & -3 & 1 \\
3 & 2 & 1 \\
-1 & -8 & 1
\end{array}\right|\)
= \(\frac{1}{2} \cdot\left|\begin{array}{rrr}
-2 & -3 & 1 \\
5 & 5 & 0 \\
1 & -5 & 0
\end{array}\right|\)
= \(\frac{1}{2} \cdot 1 \cdot\left|\begin{array}{rr}
5 & 5 \\
1 & -5
\end{array}\right|=\frac{1}{2} \times(-25-5)=-15 .\)
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) = \(\frac{1}{2} \cdot\left|\begin{array}{lll}
a & b+c & 1 \\
b & c+a & 1 \\
c & a+b & 1
\end{array}\right|\)
= \(\frac{1}{2} \cdot\left|\begin{array}{lll}
a & a+b+c & 1 \\
b & a+b+c & 1 \\
c & a+b+c & 1
\end{array}\right|\)
= \(\frac{1}{2}(a+b+c) \cdot\left|\begin{array}{lll}
a & 1 & 1 \\
b & 1 & 1 \\
c & 1 & 1
\end{array}\right|\)
= \(\frac{1}{2}(a+b+c) \times 0=0\) [∵ 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
⇔ \(\left|\begin{array}{ccc}
a & b & 1 \\
a^{\prime} & b^{\prime} & 1 \\
a-a^{\prime} & b-b^{\prime} & 1
\end{array}\right|=0\)
⇔ \(\left|\begin{array}{ccc}
a & b & 1 \\
a^{\prime}-a & b^{\prime}-b & 0 \\
-a^{\prime} & -b^{\prime} & 0
\end{array}\right|=0\)
⇔ \(\text { 1. }\left[\left(a^{\prime}-a\right)\left(-b^{\prime}\right)+a^{\prime}\left(b^{\prime}-b\right)\right]=0\)
[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
⇔ \(\left|\begin{array}{rrr}
5 & 5 & 1 \\
k & 1 & 1 \\
10 & 7 & 1
\end{array}\right|=0\)
⇔ \(\left|\begin{array}{crc}
5 & 5 & 1 \\
k-5 & -4 & 0 \\
5 & 2 & 0
\end{array}\right|=0\)
⇔ 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
⇒ \(\frac{1}{2} \cdot\left|\begin{array}{lll}
1 & 2 & 1 \\
x & y & 1 \\
3 & 6 & 1
\end{array}\right|=0\) ⇒ \(\left|\begin{array}{lll}
1 & 2 & 1 \\
x & y & 1 \\
3 & 6 & 1
\end{array}\right|=0\)
⇒ \(\left|\begin{array}{rrr}
1 & 2 & 1 \\
x & y & 1 \\
0 & 0 & -2
\end{array}\right|=0\)
⇒ \((-2) \cdot\left|\begin{array}{ll}
1 & 2 \\
x & y
\end{array}\right|=0\) ⇒ (y-2x) = 0 ⇒ y = 2x.
Hence, the required equation is y = 2x.
The equation of the line joining the points y = 2x.