WBCHSE Class 12 Maths Solutions For Adjoint And Inverse Of A Matrix

Chapter 3 Adjoint And Inverse Of A Matrix

Adjoint Of A Matrix Let A = [aij] be a square matrix of order n and let Aij denote the cofactor of aij in |A|. Then , the adjoint of A, denoted by adj A, is defined as

adj A = [Aij]n x n.

Thus adj A is the transpose of the matrix of the corresponding cofactors of elements of |A|.

If A = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\), then

adj A = \(\left[\begin{array}{lll}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{array}\right]^{\prime}=\left[\begin{array}{lll}
A_{11} & A_{21} & A_{31} \\
A_{12} & A_{22} & A_{32} \\
A_{13} & A_{23} & A_{33}
\end{array}\right]\),

where Aij denotes the cofactor of aij in |A|.

Example 1 If A = \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\), find adj A.

Solution

Clearly, |A| = \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\).

The cofactors of the elements of |A| are given by

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A11 = 3, A12 = -1;

A21 = -5, A22 = 2.

∴ adj A = \(\left[\begin{array}{rr}
3 & -1 \\
-5 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
3 & -5 \\
-1 & 2
\end{array}\right]\)

Example 2 If A = \(\left[\begin{array}{rrr}
1 & -2 & 4 \\
0 & 2 & 1 \\
-4 & 5 & 3
\end{array}\right]\), find adj A.

Solution

We have, |A| = \(\left|\begin{array}{rrr}
1 & -2 & 4 \\
0 & 2 & 1 \\
-4 & 5 & 3
\end{array}\right|\).

The cofactors of the elements of |A| are given by

A11 = \(\left|\begin{array}{ll}
2 & 1 \\
5 & 3
\end{array}\right|\) = 1;

A12 = \(-\left|\begin{array}{rr}
0 & 1 \\
-4 & 3
\end{array}\right|\) = -4;

A13 = \(\left|\begin{array}{rr}
0 & 2 \\
-4 & 5
\end{array}\right|\) = 8;

A21 = \(-\left|\begin{array}{rr}
-2 & 4 \\
5 & 3
\end{array}\right|\) = 26;

A22 = \(\left|\begin{array}{rr}
1 & 4 \\
-4 & 3
\end{array}\right|\) = 19;

A23 = \(-\left|\begin{array}{rr}
1 & -2 \\
-4 & 5
\end{array}\right|\) = 3;

A31 = \(\left|\begin{array}{rr}
-2 & 4 \\
2 & 1
\end{array}\right|\) = -10;

A32 = \(-\left|\begin{array}{ll}
1 & 4 \\
0 & 1
\end{array}\right|\) = -1;

A33 = \(\left|\begin{array}{rr}
1 & -2 \\
0 & 2
\end{array}\right|\) = 3;

∴ adj A = \(\left[\begin{array}{rrr}
1 & -4 & 8 \\
26 & 19 & 3 \\
-10 & -1 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
1 & 26 & -10 \\
-4 & 19 & -1 \\
8 & 3 & 2
\end{array}\right]\).

Theorem 1 If A is a square matrix of order n then prove that A . (adj A) = (adj A) . A = |A| . I.

Proof

Let A = \(\left[\begin{array}{cccc}
a_{11} & a_{12} & \ldots & a_{1 n} \\
a_{21} & a_{22} & \ldots & a_{2 n} \\
\ldots & \ldots & \ldots & \ldots \\
\ldots & \ldots & \ldots & \ldots \\
a_{i 1} & a_{i 2} & \ldots & a_{i n} \\
\ldots & \ldots & \ldots & \ldots \\
\ldots & \ldots & \ldots & \ldots \\
a_{n 1} & a_{n 2} & \ldots & a_{n n}
\end{array}\right]\). Then,

adj A = \(\left[\begin{array}{cccc}
A_{11} & A_{12} & \ldots & A_{1 n} \\
A_{21} & A_2 & \ldots & A_{2 n} \\
\ldots & \ldots & \ldots & \ldots \\
\ldots & \ldots & \ldots & \ldots \\
A_{k 1} & A_{k 2} & \ldots & \ldots \\
\ldots & \ldots & \ldots & \ldots \\
A_{n 1} & A_{n 2} & \ldots & \ldots \\
A_{n n}
\end{array}\right]=\left[\begin{array}{cccccc}
A_{11} & A_{21} & \ldots & A_{k 1} & \ldots & A_{m 1} \\
A_{12} & A_{22} & \ldots & A_{k 2} & \ldots & A_{n 2} \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
A_{1 n} & A_{2 n} & \ldots & A_{k n} & \ldots & A_{n n}
\end{array}\right] .\)

Now, the (i,k)th element of A>(adj A) = ai1 Ak1 + ai2 Ak2 + … + ain Akn

= \(\left\{\begin{array}{cc}
|A|, & \text { when } i=k \\
0, & \text { when } i \neq k
\end{array}\right.\).

This shows that each diagonal element of A.(adj A) is |A| and each one of its nondiagonal elements is 0.

Thus, A(adj A) = |A| . I.

Similarly, (adj A)A = |A| . I.

Hence, A(adj A) = (adj A)A = |A| . I.

Summary

For every square matrix A, we have

A . (adj A) = (adj A) . A = |A| . I.

Example 3 If A = \(\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]\), verify that A . (adj A) = (adj A) . A = |A| . I.

Solution

We have

\(|A|=\left|\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right|=(15-7)=8 \neq 0\)

The cofactors of the elements of |A| are given by

A11 = 5, A12 = -7; A21 = -1, A22 = 3.

∴ \(({adj} A)=\left[\begin{array}{rr}
5 & -7 \\
-1 & 3
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
5 & -1 \\
-7 & 3
\end{array}\right] .\)

∴ \(A \cdot({adj} A)=\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]\left[\begin{array}{rr}
5 & -1 \\
-7 & 3
\end{array}\right]\)

= \(\left[\begin{array}{ll}
15-7 & -3+3 \\
35-35 & -7+15
\end{array}\right]=\left[\begin{array}{ll}
8 & 0 \\
0 & 8
\end{array}\right]\)

= \(8 \cdot\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=8 I=|A| \cdot I\) [∵ |A| = 8].

And, (adj A) . A = \(\left[\begin{array}{rr}
5 & -1 \\
-7 & 3
\end{array}\right]\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]\)

= \(\left[\begin{array}{rr}
15-7 & 5-5 \\
-21+21 & -7+15
\end{array}\right]=\left[\begin{array}{ll}
8 & 0 \\
0 & 8
\end{array}\right]\)

= \(8 \cdot\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=8 I=|A| \cdot I\) [∵ |A| = 8].

Hence, A.(adj A) . A = |A| . I.

Example 4 If A = \(\left[\begin{array}{rrr}
1 & 0 & -1 \\
3 & 4 & 5 \\
0 & -6 & -7
\end{array}\right]\), verify that A.(adj A) = (adj A).A = |A|.I.

Solution

We have

\(|A|=\left|\begin{array}{rrr}
1 & 0 & -1 \\
3 & 4 & 5 \\
0 & -6 & -7
\end{array}\right|\) = [1. (-28+30) – 1. (-18-0)] = 20.

Now, the cofactors of the elements of |A| are given by

A11 = \(\left|\begin{array}{rr}
4 & 5 \\
-6 & -7
\end{array}\right|\) = 3.

A12 = \(-\left|\begin{array}{rr}
3 & 5 \\
0 & -7
\end{array}\right|\) = 21,

A13 = \(\left|\begin{array}{rr}
3 & 4 \\
0 & -6
\end{array}\right|\) = -18;

A21 = \(-\left|\begin{array}{rr}
0 & -1 \\
-6 & -7
\end{array}\right|\) = 6,

A22 = \(\left|\begin{array}{ll}
1 & -1 \\
0 & -7
\end{array}\right|\) = -7,

A23 = \(-\left|\begin{array}{rr}
1 & 0 \\
0 & -6
\end{array}\right|\) = 6;

A31 = \(\left|\begin{array}{rr}
0 & -1 \\
4 & 5
\end{array}\right|\) = 4,

A32 = \(-\left|\begin{array}{rr}
1 & -1 \\
3 & 5
\end{array}\right|\) = -8,

A33 = \(\left|\begin{array}{ll}
1 & 0 \\
3 & 4
\end{array}\right|\) = 4.

∴ \({adj} A=\left[\begin{array}{rrr}
2 & 21 & -18 \\
6 & -7 & 6 \\
4 & -8 & 4
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
2 & 6 & 4 \\
21 & -7 & -8 \\
-18 & 6 & 4
\end{array}\right] \text {. }\)

So, A . (adj A) = \(\left[\begin{array}{rrr}
1 & 0 & -1 \\
3 & 4 & 5 \\
0 & -6 & -7
\end{array}\right]\left[\begin{array}{rrr}
2 & 6 & 4 \\
21 & -7 & -8 \\
-18 & 6 & 4
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
20 & 0 & 0 \\
0 & 20 & 0 \\
0 & 0 & 20
\end{array}\right]\) = 20 . \(\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = |A| . I.

Thus, A . (adj A) = |A| . I.

Further, (adj A).A = \(\left[\begin{array}{rrr}
2 & 6 & 4 \\
21 & -7 & -8 \\
-18 & 6 & 4
\end{array}\right] \cdot\left[\begin{array}{rrr}
1 & 0 & -1 \\
3 & 4 & 5 \\
0 & -6 & -7
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
20 & 0 & 0 \\
0 & 20 & 0 \\
0 & 0 & 20
\end{array}\right]\) = 20 . \(\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = |A| . I.

Thus, (adj A) . A = |A| . I.

Hence, A . (adj A) = (adj A) . A = |A|.I.

Singular And Nonsingular Matrices A square matrix A is said to be

(1) Singular if |A| = 0, (2) nonsingular if |A| ≠ 0.

Examples

(1) Let A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 8
\end{array}\right]\). Then,

|A| = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 8
\end{array}\right]\) = (8-8) = 0 and therefore, A is singular.

(2) Let B = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\). Then,

|B| = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) = (4-6) = -2 ≠ 0, and therefore, B is nonsingular.

Invertible Matrix A nonzero 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.

We say that the inverse of A is B and we write, A-1 = B.

Results On Invertible Matrices

Theorem 2 An invertible matrix possesses 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.

And, 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.

Theorem 3 A square matrix A is invertible if and only if A is nonsingular, i.e., A is invertible ⇔ |A| ≠ 0.

Proof

Let A be an invertible square matrix of order n.

Then, there exists a square matrix B of order n such that

AB = BA = I.

Now, AB = I ⇒ |AB| = |I|

⇒ |A| . |B| = 1

[∵ |AB| = |A| . |B| and |I| = 1]

⇒ |A| ≠ 0 [∵ ab = 1 ⇒ a ≠ 0 and b ≠ 0].

This shows that A is nonsingular.

Conversely, Let A be non singular. Then, |A| ≠ 0.

∴ A.(adj A) = (adj A) . A = |A| . I and |A| ≠ 0

⇒ \(A \cdot\left(\frac{1}{|A|} \cdot {adj} A\right)=\left(\frac{1}{|A|} \cdot {adj} A\right) \cdot A=I\)

⇒ \(A^{-1}=\frac{1}{|A|} \cdot \text { adj } A\)

This shows that A is invertible.

Hence, A is invertible ⇔ A is nonsingular.

Formula For Finding A-1

Let A be a square matrix such that |A| ≠ 0.

Then, \(A^{-1}=\frac{1}{|A|} \cdot \text { adj } A\)

Example 5 Find the inverse of the matrix, A = \(\left[\begin{array}{rr}
2 & -3 \\
-4 & 7
\end{array}\right]\)

Solution

We have

|A| = \(\left[\begin{array}{rr}
2 & -3 \\
-4 & 7
\end{array}\right]\) = (14-12) = 2 ≠ 0.

So, A-1 exists.

The cofactors of the elements of |A| are given by

A11 = 7, A12 – -(-4) = 4;

A21 = -(-3) = 3, A22 = 2.

∴ \(({adj} A)=\left[\begin{array}{ll}
7 & 4 \\
3 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]\)

Hence, \(A^{-1}=\frac{1}{|A|} \cdot \text { adj } A\)

= \(\frac{1}{2} \cdot\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]=\left[\begin{array}{ll}
\frac{7}{2} & \frac{3}{2} \\
2 & 1
\end{array}\right] \text {. }\)

Example 6 If A = \(\left[\begin{array}{rr}
3 & -4 \\
-1 & 2
\end{array}\right]\), find a matrix B such that AB = I.

Solution

We have

|A| = \(\left[\begin{array}{rr}
3 & -4 \\
-1 & 2
\end{array}\right]\) = (6-4) = 2 ≠ 0.

Thus, |A| ≠ 0 and therefore, A-1 exists.

Now, AB = I ⇒ A-1(AB) = A-1 . I

⇒ (A-1A)B = A-1.

⇒ I . B = A-1 ⇒ B = A-1.

The cofactors of the elements of |A| are

A11 = 2, A12 = -(-1), A21 = -(-4) = 4, A22 = 3.

∴ \(({adj} A)=\left[\begin{array}{ll}
2 & 1 \\
4 & 3
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
2 & 4 \\
1 & 3
\end{array}\right]\)

Hence, \(A^{-1}=\frac{1}{|A|} \cdot \text { adj } A\)

= \(\frac{1}{2} \cdot\left[\begin{array}{ll}
2 & 4 \\
1 & 3
\end{array}\right]=\left[\begin{array}{cc}
1 & 2 \\
\frac{1}{2} & \frac{3}{2}
\end{array}\right]\)

Hence, B = \(\left[\begin{array}{cc}
1 & 2 \\
\frac{1}{2} & \frac{3}{2}
\end{array}\right]\).

Example 7 Find the inverse of the matrix \(\left[\begin{array}{rrr}
3 & -10 & -1 \\
-2 & 8 & 2 \\
2 & -4 & -2
\end{array}\right] \text {. }\)

Solution

Let A = \(\left[\begin{array}{rrr}
3 & -10 & -1 \\
-2 & 8 & 2 \\
2 & -4 & -2
\end{array}\right] \text {. }\) Then,

|A| = \(\left|\begin{array}{rrr}
3 & -10 & -1 \\
-2 & 8 & 2 \\
2 & -4 & -2
\end{array}\right|=\left|\begin{array}{rrr}
0 & 0 & -1 \\
4 & -12 & 2 \\
-4 & 16 & -2
\end{array}\right|\)

[C1 → C1 + 3C3 and C2 → C2 – 10C3]

= (-1) . (64-48) = -16 ≠ 0.

Thus, |A| ≠ 0 and therefore, A-1 exists.

Now, the cofactors of the elements of |A| are given by

A11 = \(\left|\begin{array}{rr}
8 & 2 \\
-4 & -2
\end{array}\right|\) = -8,

A12 = \(-\left|\begin{array}{rr}
-2 & 2 \\
2 & -2
\end{array}\right|\) = 0,

A13 = \(\left|\begin{array}{rr}
-2 & 8 \\
2 & -4
\end{array}\right|\) = -8;

A21 = \(-\left|\begin{array}{ll}
-10 & -1 \\
-4 & -2
\end{array}\right|\) = -16,

A22 = \(\left|\begin{array}{ll}
3 & -1 \\
2 & -2
\end{array}\right|\) = -4,

A23 = \(-\left|\begin{array}{rr}
3 & -10 \\
2 & -4
\end{array}\right|\) = -8;

A31 = \(\left|\begin{array}{rr}
-10 & -1 \\
8 & 2
\end{array}\right|\) = -12,

A32 = \(-\left|\begin{array}{rr}
3 & -1 \\
-2 & 2
\end{array}\right|\) = -4,

A33 = \(\left|\begin{array}{rr}
3 & -10 \\
-2 & 8
\end{array}\right|\) = 4.

∴ \(({adj} A)=\left[\begin{array}{rrr}
-8 & 0 & -8 \\
-16 & -4 & -8 \\
-12 & -4 & 4
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
-8 & -16 & -12 \\
0 & -4 & -4 \\
-8 & -8 & 4
\end{array}\right]\)

Hence, \(A^{-1}=\frac{1}{|A|} \cdot \text { adj } A\)

= \(\frac{1}{-16} \cdot\left[\begin{array}{rrr}
-8 & -16 & -12 \\
0 & -4 & -4 \\
-8 & -8 & 4
\end{array}\right]=\left[\begin{array}{rrr}
\frac{1}{2} & 1 & \frac{3}{4} \\
0 & \frac{1}{4} & \frac{1}{4} \\
\frac{1}{2} & \frac{1}{2} & -\frac{1}{4}
\end{array}\right]\).

Example 8 Find the inverse of the matrix \(\left[\begin{array}{rrr}
2 & 2 & 0 \\
2 & 1 & 1 \\
-7 & 2 & -3
\end{array}\right]\).

Solution

Let A = \(\left[\begin{array}{rrr}
2 & 2 & 0 \\
2 & 1 & 1 \\
-7 & 2 & -3
\end{array}\right]\). Then,

\(|A|=\left|\begin{array}{rrr}
2 & 2 & 0 \\
2 & 1 & 1 \\
-7 & 2 & -3
\end{array}\right|=\left|\begin{array}{rrr}
2 & 0 & 0 \\
2 & -1 & 1 \\
-7 & 9 & -3
\end{array}\right|\) [C2 → C2 – C1]

= \(\text { 2. }\left|\begin{array}{rr}
-1 & 1 \\
9 & -3
\end{array}\right|\) = 2(3-9) = -12 ≠ 0.

∴ A-1 exists.

Now, the cofactors of the elements of |A| are given by

A11 = \(\left|\begin{array}{rr}
1 & 1 \\
2 & -3
\end{array}\right|\) = -5,

A12 = \(-\left|\begin{array}{rr}
2 & 1 \\
-7 & -3
\end{array}\right|\) = -1,

A13 = \(\left|\begin{array}{rr}
2 & 1 \\
-7 & 2
\end{array}\right|\) = 11;

A21 = \(-\left|\begin{array}{rr}
2 & 0 \\
2 & -3
\end{array}\right|\) = 6,

A22 = \(\left|\begin{array}{rr}
2 & 0 \\
-7 & -3
\end{array}\right|\) = -6,

A23 = \(-\left|\begin{array}{rr}
2 & 2 \\
-7 & 2
\end{array}\right|\) = -18;

A31 = \(\left|\begin{array}{ll}
2 & 0 \\
1 & 1
\end{array}\right|\) = 2,

A32 = \(-\left|\begin{array}{ll}
2 & 0 \\
2 & 1
\end{array}\right|\) = -2,

A33 = \(\left|\begin{array}{ll}
2 & 2 \\
2 & 1
\end{array}\right|\) = -2.

∴ (adj A) = \(\left[\begin{array}{rrr}
-5 & -1 & 11 \\
6 & -6 & -18 \\
2 & -2 & -2
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
-5 & 6 & 2 \\
-1 & -6 & -2 \\
11 & -18 & -2
\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|} \cdot({adj} A)\)

= \(\frac{1}{(-12)} \cdot\left[\begin{array}{rrr}
-5 & 6 & 2 \\
-1 & -6 & -2 \\
11 & -18 & -2
\end{array}\right]\)

Hence, A-1 = \(\frac{-1}{12} \cdot\left[\begin{array}{rrr}
-5 & 6 & 2 \\
-1 & -6 & -2 \\
11 & -18 & -2
\end{array}\right]\).

Some Results On Invertible Matrices

Theorem 1 (Cancellation law) Let A, B, C be three square matrices, each of order n such that AB = AC. If A is nonsingular then B = C.

Proof

Let A, B, C be square matrices, each of order n such that AB = AC and A is nonsingular.

Then, |A| ≠ 0 and therefore, A-1 exists.

∴ AB = AC ⇒ A-1(AB) = A-1(AC)

⇒ (A-1A)B = (A-1A)C

⇒ IB = IC

⇒ B = C.

Thus, AB = AC and A is nonsingular ⇒ B = C.

Remark If AB = AC and |A| = 0 then B and C are not necessarily equal.

Example Let A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 6
\end{array}\right]\), B = \(\left[\begin{array}{ll}
2 & 0 \\
0 & 0
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right]\).

Then, |A| = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 6
\end{array}\right]\) = (6-6) = 0.

Here, AB = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 6
\end{array}\right]\left[\begin{array}{ll}
2 & 0 \\
0 & 0
\end{array}\right]=\left[\begin{array}{ll}
2 & 0 \\
6 & 0
\end{array}\right]\)

and AC = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 6
\end{array}\right]\left[\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right]=\left[\begin{array}{ll}
2 & 0 \\
6 & 0
\end{array}\right]\).

Thus, AB = AC but B ≠ C.

Theorem 2 (Reversal law) If A and B are invertible square matrices of the same order then AB is also invertible and (AB)-1 = B-1A-1.

Proof

Let A and B be two invertible square matrices, each of order n.

Then, |A| ≠ 0 and |B| ≠ 0.

∴ |AB| = |A|.|B| ≠ 0.

Thus, AB is invertible.

Now, (AB)(B-1A-1) = A(BB-1)A-1 [by associativity]

= (AI)A-1 [∵ BB-1 = I]

= AA-1 = I [∵ AI = A]

And, (B-1A-1)(AB) = B-1(A-1A)B [by associativity]

= B-1(IB) [∵ A-1A = I]

= B-1B = I [∵ IB = B].

∴ (AB)(B-1A-1) = (B-1A-1)(AB) = I.

Hence, (AB)-1 = B-1A-1.

Theorem 3 If A is an invertible square matrix then prove that A’ is also invertible and (A’)-1 = (A-1)’.

Proof

Let A be an invertible square matrix of order n

Then, |A| ≠ 0.

∴ |A’| = |A| ≠ 0.

This shows that A’ is invertible.

Now, AA-1 = A-1A = I

⇒ (AA-1)’ = (A-1A)’ = I’

⇒ (A-1)’.A’ = A’ . (A-1)’ = I [∵ (AB)’ = B’A’ and I’ = I]

⇒ (A’)-1 = (A-1)’ [∵ AB = BA = I ⇒ B-1 = A].

Hence, (A’)-1 = (A-1)’.

Theorem 4 If A is an invertible symmetric matrix then prove the A-1 is also symmetric.

Proof

Let A be an invertible symmetric matrix. Then, A’ = A.

We know that (A-1)’ = (A’)-1.

∴ (A-1)’ = A-1 [∵ A’ = A]

Hence, A-1 is symmetric

Theorem 5 If A and B are nonsingular matrices of the same order then prove that (adj AB) = (adj B).(adj A).

Proof

We have

|A| ≠ 0 and |B| ≠ 0.

∴ |AB| = |A|.|B| ≠ 0.

So, (AB)-1 exists.

Now, \(A^{-1}=\frac{\text { adj } A}{|A|} ; B^{-1}=\frac{\text { adj } B}{|B|} \text { and }(A B)^{-1}=\frac{{adj}(A B)}{|A B|} \text {. }\)

∴ adj (AB) = |A B| \(\cdot \frac{{adj}(A B)}{|A B|}\)

= |A|.|B|.(AB)-1

[∵ |A B|=|A| \(\cdot|B| \text { and } \frac{{adj}(A B)}{|A B|}=(A B)^{-1}\)]

= |A|.|B|.(B-1A-1) [∵ (AB)-1 = B-1A-1]

= \(|A| \cdot|B| \cdot \frac{{adj} B}{|B|} \cdot \frac{{adj} A}{|A|}=({adj} B)({adj} A) .\)

Hence, adj(AB) = (adj B)(adj A).

Theorem 6 For any square matrix A, prove that (adj A)’ = adj A’.

Proof

Let A be a square matrix of order n.

Then, each one of (adj A)’ and (adj A’) is a square matrix of orer n.

Also, (i,j)th element of (adj A)’

= (j,i)th element of (adj A)

= cofactor of (i,j)th element of A

= cofactor of (j,i)th element of A’

= (i,j)th element of (adj A’).

Hence, (adj A)’ = (adj A’).

Theorem 7 If A is a nonsingular square matrix of order n then prove that |adj A| = |A|n-1.

Proof

We have

A.(adj A) = |A|.I

⇒ |A.(adj A)| = ||A|.I| [∵ A = B ⇒ |A| = |B|]

⇒ |A| |adj A| = |A|n|I| [∵ |kI| = kn-1|I|]

⇒ |A| . |adj A| = |A|n [∵ |I| = 1]

⇒ |adj A| = |A|n-1

Hence, |adj A| = |A|n-1

Theorem 8 If A is a nonsingular square matrix of order n then prove that adj(adj A) = |A|n-2 A.

Proof

For any nonsingular matrix B of order n, we have B(adj B) = |B|.I

Taking B = adj A, we get

(adj A) [adj (adj A)] = |adj A|I

⇒ (adj A)[adj (adj A)] = |A|n-1 I [∵ |adj A| = |A|n-1]

⇒ A(adj A)[adj (adj A)] = |A|n-1A [∵ AIn = A]

⇒ |A| In[adj(adj A)] = |A|n-1A [∵ A(adj A) = |A|I]

⇒ [adj (adj A)] = |A|n-2 A.

Solved Examples

Example 1 If A = \(\left[\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
6 & 7 \\
8 & 9
\end{array}\right]\), verify that (AB)-1 = B-1A-1.

Solution

We have |A| = \(\left[\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right]\) = (15-14) = 1 ≠ 0.

Cofactors of the elements of |A| are

A11 = 5, A12 = -7;

A21 = -2, A22 = 3.

∴ adj A = \(\left[\begin{array}{rr}
5 & -7 \\
-2 & 3
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
5 & -2 \\
-7 & 3
\end{array}\right]\)

Hence, \(A^{-1}=\frac{1}{|A|} \cdot {adj} A=\left[\begin{array}{rr}
5 & -2 \\
-7 & 3
\end{array}\right]\) [∵ |A| = 1].

Further, |B| = \(\left[\begin{array}{ll}
6 & 7 \\
8 & 9
\end{array}\right]\) = (54 – 56) = -2 ≠ 0.

Cofactors of the elements of |B| are

B11 = 9, B12 = -8;

B21 = -7, B22 = 6.

∴ adj B = \(\left[\begin{array}{rr}
9 & -8 \\
-7 & 6
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
9 & -7 \\
-8 & 6
\end{array}\right]\)

Hence, \(B^{-1}=\frac{1}{|B|} \cdot \text { adj } B=-\frac{1}{2}\left[\begin{array}{rr}
9 & -7 \\
-8 & 6
\end{array}\right] \text {. }\)

Now, |AB| = |A|.|B| = 1 x (-2) = -2 ≠ 0,

and adj AB = (adj B).(adj A)

= \(\left[\begin{array}{rr}
9 & -7 \\
-8 & 6
\end{array}\right] \cdot\left[\begin{array}{rr}
5 & -2 \\
-7 & 3
\end{array}\right]=\left[\begin{array}{rr}
94 & -39 \\
-82 & 34
\end{array}\right]\)

∴ \((A B)^{-1}=\frac{1}{|A B|} \cdot({adj} A B)=\frac{1}{-2} \cdot\left[\begin{array}{rr}
94 & -39 \\
-82 & 34
\end{array}\right] .\)

Also, B-1A-1 = \(-\frac{1}{2} \cdot\left[\begin{array}{rr}
9 & -7 \\
-8 & 6
\end{array}\right] \cdot\left[\begin{array}{rr}
5 & -2 \\
-7 & 3
\end{array}\right]=-\frac{1}{2} \cdot\left[\begin{array}{rr}
94 & -39 \\
-82 & 34
\end{array}\right]\)

Hence, (AB)-1 = B-1A-1.

Example 2 Show that the matrix A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) satisfies the equation A2 – 4A – 5I = 0, and hence find A-1.

Solution

We leave it to the reader to show that, A2 – 4A – 5I = 0.

Now, A2 – 4A – 5I = 0

⇒ AA – 4A = 5I

⇒ (AA).A-1 – 4A.A-1 = 5I.A-1

⇒ A(AA-1) – 4I = 5A-1

⇒ AI – 4I = 5A-1

⇒ A – 4I = 5A-1

⇒ A-1 = \(\frac{1}{5}\) (A-4I).

∴ A-1 = \(\frac{1}{5} \cdot\left\{\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-4 \cdot\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right\}\)

= \(\frac{1}{5} \cdot\left\{\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-\left[\begin{array}{lll}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{array}\right]\right\}\)

= \(\frac{1}{5} \cdot\left[\begin{array}{rrr}
-3 & 2 & 2 \\
2 & -3 & 2 \\
2 & 2 & -3
\end{array}\right]\)

Example 3 Find a matrix X such that \(X \cdot\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\).

Solution

Let A = \(\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\).

Here, |A| = \(\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]\) = (-3-2) = -5 ≠ 0.

So, A is nonsingular and therefore, invertible.

The given matrix equation is XA = B.

Now, XA = B ⇒ (XA).A-1 = BA-1

⇒ X(AA-1) = BA-1

⇒ XI = BA-1

⇒ X = BA-1.

Now, cofactors of elements of |A| are

A11 = -1, A12 = -1;

A21 = -2, A22 = 3.

∴ adj A = \(\left[\begin{array}{rr}
-1 & -1 \\
-2 & 3
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
-1 & -2 \\
-1 & 3
\end{array}\right]\)

∴ \(A^{-1}=\frac{1}{|A|} \cdot ={adj} A=-\frac{1}{5} \cdot\left[\begin{array}{rr}
-1 & -2 \\
-1 & 3
\end{array}\right]\)

Hence, X = BA-1

= \(\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right] \cdot\left(-\frac{1}{5}\right) \cdot\left[\begin{array}{rr}
-1 & -2 \\
-1 & 3
\end{array}\right]\)

= \(\left(-\frac{1}{5}\right) \cdot\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\left[\begin{array}{rr}
-1 & -2 \\
-1 & 3
\end{array}\right]\)

= \(\left(-\frac{1}{5}\right) \cdot\left[\begin{array}{rr}
-5 & -5 \\
-5 & 5
\end{array}\right]=\left[\begin{array}{rr}
1 & 1 \\
1 & -1
\end{array}\right]\).

Example 4 Find the matrix A satisfying the matrix equation \(\left[\begin{array}{ll}
2 & 1 \\
3 & 2
\end{array}\right] \cdot A \cdot\left[\begin{array}{rr}
-3 & 2 \\
5 & -3
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\).

Solution

Let B = \(\left[\begin{array}{ll}
2 & 1 \\
3 & 2
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
-3 & 2 \\
5 & -3
\end{array}\right] .\)

Clearly, |B| = \(\left[\begin{array}{ll}
2 & 1 \\
3 & 2
\end{array}\right]\) = (4-3) = 1 ≠ 0.

And, |C| = \(\left[\begin{array}{rr}
-3 & 2 \\
5 & -3
\end{array}\right]\) = (9-10) = -1 ≠ 0.

This shows that B as well as C is invertible.

The given matrix equation is BAC = I.

Now, BAC = I ⇒ B-1BACC-1 = B-1IC-1

⇒ IAI = B-1C-1

⇒ A = B-1C-1.

Now, the cofactors of the element of |B| are

B11 = 2, B12 = -3; B21 = -1, B22 = 2.

∴ adj B = \(\left[\begin{array}{rr}
2 & -3 \\
-1 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
2 & -1 \\
-3 & 2
\end{array}\right]\)

So, \(B^{-1}=\frac{1}{|B|} \cdot \text { adj } B=\left[\begin{array}{rr}
2 & -1 \\
-3 & 2
\end{array}\right]\) [∵ |B| = 1].

Again, the cofactors of the elements of |C| are

C11 = -3, C12 = -5; C21 = -2, C22 = -3.

∴ adj C = \(\left[\begin{array}{ll}
-3 & -5 \\
-2 & -3
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
-3 & -2 \\
-5 & -3
\end{array}\right]\)

⇒ \(C^{-1}=\frac{1}{|C|} \cdot({adj} C)=\frac{1}{(-1)} \cdot\left[\begin{array}{ll}
-3 & -2 \\
-5 & -3
\end{array}\right]=\left[\begin{array}{ll}
3 & 2 \\
5 & 3
\end{array}\right]\)

⇒ A = \(\left(B^{-1} C^{-1}\right)=\left[\begin{array}{rr}
2 & -1 \\
-3 & 2
\end{array}\right]\left[\begin{array}{ll}
3 & 2 \\
5 & 3
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right]\)

Example 5 If A = \(\left[\begin{array}{rr}
2 & -3 \\
4 & 6
\end{array}\right]\), verify that (adj A)-1 = (adj A-1).

Solution

We have, |A| = \(\left[\begin{array}{rr}
2 & -3 \\
4 & 6
\end{array}\right]\) = (12+12) = 24 ≠ 0.

Cofactors of the elements of |A| are

A11 = 6, A12 = -4; A21 = 3, A22 = 2.

∴ adj A = \(\left[\begin{array}{rr}
6 & -4 \\
3 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
6 & 3 \\
-4 & 2
\end{array}\right]\)

So, \(A^{-1}=\frac{1}{|A|} \cdot \text { adj } A=\frac{1}{24} \cdot\left[\begin{array}{rr}
6 & 3 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{rr}
\frac{1}{4} & \frac{1}{8} \\
-\frac{1}{6} & \frac{1}{12}
\end{array}\right]\).

Now, \(\left|A^{-1}\right|=\left|\begin{array}{cc}
\frac{1}{4} & \frac{1}{8} \\
-\frac{1}{6} & \frac{1}{12}
\end{array}\right|=\left(\frac{1}{48}+\frac{1}{48}\right)=\frac{1}{24} \text {. }\)

Cofactors of the elements of |A-1| are

C11 = \(\frac{1}{12}\), C12 = \(\frac{1}{6}\); C21 = –\(\frac{1}{8}\), C22 = \(\frac{1}{4}\).

∴ adj A-1 = \(\left[\begin{array}{cc}
\frac{1}{12} & \frac{1}{6} \\
-\frac{1}{8} & \frac{1}{4}
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
\frac{1}{12} & \frac{-1}{8} \\
\frac{1}{6} & \frac{1}{4}
\end{array}\right]\).

∴ (adj A)(adj A-1) = \(\left[\begin{array}{rr}
6 & 3 \\
-4 & 2
\end{array}\right]\left[\begin{array}{cc}
\frac{1}{12} & \frac{-1}{8} \\
\frac{1}{6} & \frac{1}{4}
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I\).

And, (adj A-1)(adj A) = \(\left[\begin{array}{cc}
\frac{1}{12} & \frac{-1}{8} \\
\frac{1}{6} & \frac{1}{4}
\end{array}\right]\left[\begin{array}{rr}
6 & 3 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I \text {. }\)

Thus, (adj A)(adj A-1) = (adj A-1)(adj A) = I.

Hence, (adj A)-1 = (adj A-1).

Example 6 If A = \(\left[\begin{array}{rrr}
1 & -2 & 1 \\
-2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right]\), verify that (adj A)-1 = (adj A-1).

Solution

We have

|A| = \(\left[\begin{array}{rrr}
1 & -2 & 1 \\
-2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right]=\left[\begin{array}{rrr}
1 & -2 & 1 \\
0 & -1 & 3 \\
0 & 3 & 4
\end{array}\right] \quad\left[\begin{array}{l}
R_2 \rightarrow R_2+2 R_1 \\
R_3 \rightarrow R_3-R_1
\end{array}\right]\)

= 1.(-4-9) = -13 ≠ 0.

So, A-1 exisits.

The cofactors of the elements of |A| are

A11 = \(\left|\begin{array}{ll}
3 & 1 \\
1 & 5
\end{array}\right|\) = 14,

A12 = \(-\left|\begin{array}{rr}
-2 & 1 \\
1 & 5
\end{array}\right|\) = 11,

A13 = \(\left|\begin{array}{rr}
-2 & 3 \\
1 & 1
\end{array}\right|\) = 5;

A21 = \(-\left|\begin{array}{rr}
-2 & 1 \\
1 & 5
\end{array}\right|\) = 11,

A22 = \(\left|\begin{array}{ll}
1 & 1 \\
1 & 5
\end{array}\right|\) = 4,

A23 = \(-\left|\begin{array}{rr}
1 & -2 \\
1 & 1
\end{array}\right|\) = -3;

A31 = \(\left|\begin{array}{rr}
-2 & 1 \\
3 & 1
\end{array}\right|\) = -5,

A32 = \(-\left|\begin{array}{rr}
1 & 1 \\
-2 & 1
\end{array}\right|\) = -3,

A33 = \(\left|\begin{array}{rr}
1 & -2 \\
-2 & 3
\end{array}\right|\) = -1.

∴ (adj A) = \(\left[\begin{array}{rrr}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|} \cdot(\text { adj } A)=\frac{1}{-13} \cdot\left[\begin{array}{rrr}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]=\left[\begin{array}{ccc}
\frac{-14}{13} & \frac{-11}{13} & \frac{5}{13} \\
\frac{-11}{13} & \frac{-4}{13} & \frac{3}{13} \\
\frac{5}{13} & \frac{3}{13} & \frac{1}{13}
\end{array}\right]\).

The cofactors of the elements of |A-1| are

C11 = \(\left|\begin{array}{cc}
\frac{-4}{13} & \frac{3}{13} \\
\frac{3}{13} & \frac{1}{13}
\end{array}\right|=\frac{-1}{13}\),

C12 = \(-\left|\begin{array}{cc}
\frac{-11}{13} & \frac{3}{13} \\
\frac{5}{13} & \frac{1}{13}
\end{array}\right|=\frac{2}{13}\),

C13 = \(\left|\begin{array}{rr}
\frac{-11}{13} & \frac{-4}{13} \\
\frac{5}{13} & \frac{3}{13}
\end{array}\right|=\frac{-1}{13}\)

C21 = \(-\left|\begin{array}{cc}
\frac{-11}{13} & \frac{5}{13} \\
\frac{3}{13} & \frac{1}{13}
\end{array}\right|=\frac{2}{13}\),

C22 = \(\left|\begin{array}{cc}
\frac{-14}{13} & \frac{5}{13} \\
\frac{5}{13} & \frac{1}{13}
\end{array}\right|=\frac{-3}{13}\),

C23 = \(-\left|\begin{array}{rr}
\frac{-14}{13} & \frac{-11}{13} \\
\frac{5}{13} & \frac{3}{13}
\end{array}\right|=\frac{-1}{13}\)

C31 = \(\left|\begin{array}{cc}
\frac{-11}{13} & \frac{5}{13} \\
\frac{-4}{13} & \frac{3}{13}
\end{array}\right|=\frac{-1}{13}\)

C32 = \(-\left|\begin{array}{cc}
\frac{-14}{13} & \frac{5}{13} \\
\frac{-11}{13} & \frac{3}{13}
\end{array}\right|=\frac{-1}{13}\)

C33 = \(\left|\begin{array}{cc}
\frac{-14}{13} & \frac{-11}{13} \\
\frac{-11}{13} & \frac{-4}{13}
\end{array}\right|=\frac{-5}{13} \text {. }\)

∴ (adj A-1) = \(\left[\begin{array}{ccc}
\frac{-1}{13} & \frac{2}{13} & \frac{-1}{13} \\
\frac{2}{13} & \frac{-3}{13} & \frac{-1}{13} \\
\frac{-1}{13} & \frac{-1}{13} & \frac{-5}{13}
\end{array}\right]=\left[\begin{array}{ccc}
\frac{-1}{13} & \frac{2}{13} & \frac{-1}{13} \\
\frac{2}{13} & \frac{-3}{13} & \frac{-1}{13} \\
\frac{-1}{13} & \frac{-1}{13} & \frac{-5}{13}
\end{array}\right]\)

∴ (adj A)(adj A-1) = \(\left[\begin{array}{rrr}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]\left[\begin{array}{lll}
\frac{-1}{13} & \frac{2}{13} & \frac{-1}{13} \\
\frac{2}{13} & \frac{-3}{13} & \frac{-1}{13} \\
\frac{-1}{13} & \frac{-1}{13} & \frac{-5}{13}
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\).

Thus, (adj A)(adj A-1) = I. Similarly, (adj A-1)(adj A) = I.

Hence, (adj A)-1 = (adj A-1).

Example 7 If A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 5
\end{array}\right]\), verify that (A’)-1 + (A-1)’.

Solution

Given: A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 5
\end{array}\right]\), and therefore A’ = \(\left[\begin{array}{ll}
2 & 1 \\
3 & 5
\end{array}\right]\).

∴ |A| = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 5
\end{array}\right]\) = (10-3) = 7 ≠ 0.

So, A-1 exists.

The cofactors of the element of |A| are

A11 = 5, A12 = -1; A21 = -3, A22 = 2.

∴ (adj A) = \(\left[\begin{array}{rr}
5 & -1 \\
-3 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
5 & -3 \\
-1 & 2
\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|} \cdot({adj} A)=\frac{1}{7} \cdot\left[\begin{array}{rr}
5 & -3 \\
-1 & 2
\end{array}\right]=\left[\begin{array}{rr}
\frac{5}{7} & \frac{-3}{7} \\
\frac{-1}{7} & \frac{2}{7}
\end{array}\right]\)

⇒ (A-1)’ = \(\left[\begin{array}{cc}
\frac{5}{7} & \frac{-1}{7} \\
\frac{-3}{7} & \frac{2}{7}
\end{array}\right]\) …(1)

Also, |A|’ = |A| = 7 ≠ 0.

So, (A’)-1 exists.

The cofactors of elements of |A’| are

C11 = 5, C12 = -3; C21 = -1, C22 = 2.

∴ (adj A’) = \(\left[\begin{array}{rr}
5 & -3 \\
-1 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
5 & -1 \\
-3 & 2
\end{array}\right]\)

⇒ (A’)-1 = \(\frac{1}{\left|A^{\prime}\right|} \cdot\left({adj} A^{\prime}\right)=\frac{1}{7} \cdot\left[\begin{array}{rr}
5 & -1 \\
-3 & 2
\end{array}\right]\)

⇒ (A’)-1 = \(\left[\begin{array}{cc}
\frac{5}{7} & \frac{-1}{7} \\
\frac{-3}{7} & \frac{2}{7}
\end{array}\right]\) …(2)

Hence, from (1) and (2), we get (A’)-1 = (A-1)’.

Example 8 If A = \(\left[\begin{array}{cc}
1 & \tan x \\
-\tan x & 1
\end{array}\right]\), show that A’A-1 = \(\left[\begin{array}{rr}
\cos 2 x & -\sin 2 x \\
\sin 2 x & \cos 2 x
\end{array}\right]\).

Solution

We have

|A| = \(\left[\begin{array}{cc}
1 & \tan x \\
-\tan x & 1
\end{array}\right]\) = (1 + tan2x) = sec2x ≠ 0.

So, A is invertible.

The cofactors of the elements of |A| are

A11 = 1, A12 = tan x; A21 = -tan x, A22 = 1.

∴ (adj A) = \(\left[\begin{array}{cc}
1 & \tan x \\
-\tan x & 1
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
1 & -\tan x \\
\tan x & 1
\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{\sec ^2 x} \cdot\left[\begin{array}{cc}
1 & -\tan x \\
\tan x & 1
\end{array}\right]\)

= \(\cos ^2 x \cdot\left[\begin{array}{cc}
1 & -\tan x \\
\tan x & 1
\end{array}\right]=\left[\begin{array}{cc}
\cos ^2 x & -\tan x \cos ^2 x \\
\tan x \cos ^2 x & \cos ^2 x
\end{array}\right]\)

= \(\left[\begin{array}{cc}
\cos ^2 x & -\sin x \cos x \\
\sin x \cos x & \cos ^2 x
\end{array}\right]\)

⇒ A’A-1 = \(\left[\begin{array}{cc}
1 & -\tan x \\
\tan x & 1
\end{array}\right]\left[\begin{array}{cc}
\cos ^2 x & -\sin x \cos x \\
\sin x \cos x & \cos ^2 x
\end{array}\right]\)

= \(\left[\begin{array}{cc}
\cos ^2 x-\sin ^2 x & -2 \sin x \cos x \\
2 \sin x \cos x & -\sin ^2 x+\cos ^2 x
\end{array}\right]\)

= \(\left[\begin{array}{rr}
\cos 2 x & -\sin 2 x \\
\sin 2 x & \cos 2 x
\end{array}\right]\).

Hence, A’A-1 = \(\left[\begin{array}{rr}
\cos 2 x & -\sin 2 x \\
\sin 2 x & \cos 2 x
\end{array}\right]\)

Example 9 Let F(α) = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right] and G(β) = \left[\begin{array}{ccc}
\cos \beta & 0 & \sin \beta \\
0 & 1 & 0 \\
-\sin \beta & 0 & \cos \beta
\end{array}\right]\). Show that [F(α).G(β)]-1 = G(-β).F(-α).

Solution

We have

F(α).F(-α) = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{ccc}
\cos (-\alpha) & -\sin (-\alpha) & 0 \\
\sin (-\alpha) & \cos (-\alpha) & 0 \\
0 & 0 & 1
\end{array}\right]\)

= \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{ccc}
\cos \alpha & \sin \alpha & 0 \\
-\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\)

= \(\left[\begin{array}{ccc}
\cos ^2 \alpha+\sin ^2 \alpha & 0 & 0 \\
0 & \sin ^2 \alpha+\cos ^2 \alpha & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=I\)

Thus, F(α).F(-α) = I ⇒ {F(α)}-1 = F(-α).

Similarly, G(β).G(-β) = I ⇒ {G(β)}-1 = G(-β).

∴ [F(α).G(β)]-1 = {G(β)}-1.{F(α)}-1

G(-β).F(-α).

Hence, {F(α).G(β)}-1 = G(-β).F(-α).

Chapter 4 System Of Linear Equations

Solving a System of Linear Equations by Matrix Method

Consistent System Of Equations A given system of equations is said to be consistent if it has one or more solutions.

Inconsistent System Of Equations A given system of equations is said to be inconsistent if it has no solution.

Consider the system of equations

a1x + b1y + c1z = d1,

a2x + b2y + c2z = d2,

a3x + b3y + c3z = d3.

Let A = \(\left[\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
d_1 \\
d_2 \\
d_3
\end{array}\right]\).

Then, the given system can be written as

\(\left[\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
d_1 \\
d_2 \\
d_3
\end{array}\right] .\)

∴ AX = B.

Case 1 When |A| ≠ 0

In this case, A-1 exists.

∴ AX = B ⇒ A-1(AX) = A-1B

⇒ (A-1A)X = A-1B [by associative law]

⇒ I.X = A-1B

⇒ X = A-1B.

Since A-1 is unique, the given system has a unique solution.

Thus, when |A| ≠ 0, then the given system is consistent and it has a unique solution.

Case 2 When |A| = 0 and (adj A) ≠ 0

In this case, the given system has no solution and hence it is inconsistent.

Case 3 When |A| = 0 and (adj A)B = 0

In this case, the given system has infinitely many solutions.

Summary

Let AX = B be the given system of equations.

(1) If |A| ≠ 0, the system has a unique solution.

(2) If |A| = 0 and (adj A)B ≠ 0 then the given system has no solution.

(3) If |A| = 0 and (adj A)B = 0 then the system has infinitely many solutions.

Solved Examples

Example 1 Use matrix method to show that the system of equations 2x + 5y = 7, 6x + 15y = 13. is inconsistent.

Solution

The given equations are

2x + 5y = 7 …(1)

6x + 15y = 13 …(2)

Let A = \(\left[\begin{array}{rr}
2 & 5 \\
6 & 15
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) and B = \(\left[\begin{array}{r}
7 \\
13
\end{array}\right]\).

Then, the given system in matrix from is AX = B.

Now, |A| = \(\left[\begin{array}{rr}
2 & 5 \\
6 & 15
\end{array}\right]\) = 0.

The system will be inconsistent if (adj A)B ≠ 0.

The minors of the elements of |A| are

M11 = 15, M12 = 6; M21 = 5, M22 = 2.

So, the cofactors of the elements of |A| are

A11 = 15, A12 = -6; A21 = -5, A22 = 2.

∴ adj A = \(\left[\begin{array}{rr}
15 & -6 \\
-5 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
15 & -5 \\
-6 & 2
\end{array}\right]\)

⇒ (adj A)B = \(\left[\begin{array}{rr}
15 & -5 \\
-6 & 2
\end{array}\right]\left[\begin{array}{r}
7 \\
13
\end{array}\right]=\left[\begin{array}{r}
105-65 \\
-42+26
\end{array}\right]=\left[\begin{array}{r}
40 \\
-16
\end{array}\right] \neq O .\)

Thus, |A| = 0 and (adj A)B ≠ 0.

Hence, the given system of equations is inconsistent.

Example 2 Use matrix method to show that the following system of equation is inconsistent: 3x – y + 2z = 3, 2x + y + 3z = 5, x – 2y – z = 1.

Solution

Let us take

A = \(\left[\begin{array}{rrr}
3 & -1 & 2 \\
2 & 1 & 3 \\
1 & -2 & -1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
3 \\
5 \\
1
\end{array}\right]\).

The given system in matrix form in AX = B.

Now, |A| = \(\left|\begin{array}{rrr}
3 & -1 & 2 \\
2 & 1 & 3 \\
1 & -2 & -1
\end{array}\right|\)

= 3(-1+6) + 1.(-2-3) + 2.(-4-1)

= (15-5-10) = 0.

So, the system will be inconsistent if (adj A).B ≠ 0.

The minors of the elements of |A| are

M11 = 5, M12 = -5, M13 = -5;

M21 = 5, M22 = -5, M23 = -5;

M31 = -5, M32 = 5, M33 = 5.

So, the cofactors of the elements of |A| are

A11 = 5, A12 = 5, A13 = -5;

A21 = -5, A22 = -5, A23 = 5;

A31 = -5, A32 = -5, A33 = 5.

∴ (adj A) = \(\left[\begin{array}{rrr}
5 & 5 & -5 \\
-5 & -5 & 5 \\
-5 & -5 & 5
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
5 & -5 & -5 \\
5 & -5 & -5 \\
-5 & 5 & 5
\end{array}\right]\)

⇒ (adj A)B = \(\left[\begin{array}{rrr}
5 & -5 & -5 \\
5 & -5 & -5 \\
-5 & 5 & 5
\end{array}\right]\left[\begin{array}{l}
3 \\
5 \\
1
\end{array}\right]\)

= \(\left[\begin{array}{l}
5 \cdot 3+(-5) \cdot 5+(-5) \cdot 1 \\
5 \cdot 3+(-5) \cdot 5+(-5) \cdot 1 \\
(-5) \cdot 3+5 \cdot 5+5 \cdot 1
\end{array}\right]=\left[\begin{array}{r}
15-25-5 \\
15-25-5 \\
-15+25+5
\end{array}\right]\)

= \(\left[\begin{array}{r}
-15 \\
-15 \\
15
\end{array}\right] \neq 0\)

Thus, |A| = 0 and (adj A)B ≠ 0.

Hence, the given system of equation is inconsistent.

Example 3 Show that the following system of equations is consistent and solve it: 2x + 5y = 1, 3x + 2y = 7.

Solution

The given system of equations is

2x + 5y = 1 …(1)

3x + 2y = 7 …(2)

Let A = \(\left[\begin{array}{ll}
2 & 5 \\
3 & 2
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) and B = \(\left[\begin{array}{l}
1 \\
7
\end{array}\right]\).

Then, the given system is AX = B.

Now, |A| = \(\left[\begin{array}{ll}
2 & 5 \\
3 & 2
\end{array}\right]\) = (4-15) = -11 ≠ 0.

Hence, the given system has a unique solution.

The minors of the elements of |A| are

M11 = 2, M12 = 3; M21 = 5, M22 = 2.

So, the cofactors of the elements of |A| are

A11 = 2, A12 = -3; A21 = -5, A22 =2.

∴ (adj A) = \(\left[\begin{array}{rr}
2 & -3 \\
-5 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
2 & -5 \\
-3 & 2
\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{-1}{11} \cdot\left[\begin{array}{rr}
2 & -5 \\
-3 & 2
\end{array}\right]=\left[\begin{array}{ll}
\frac{-2}{11} & \frac{5}{11} \\
\frac{3}{11} & \frac{-2}{11}
\end{array}\right]\)

⇒ X = A-1B

⇒ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{ll}
\frac{-2}{11} & \frac{5}{11} \\
\frac{3}{11} & \frac{-2}{11}
\end{array}\right]\left[\begin{array}{l}
1 \\
7
\end{array}\right]=\left[\begin{array}{l}
\frac{-2}{11}+\frac{35}{11} \\
\frac{3}{11}-\frac{14}{11}
\end{array}\right]=\left[\begin{array}{r}
3 \\
-1
\end{array}\right]\)

⇒ x = 3 and y = -1.

Example 4 If A = \(\left[\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right]\), find A-1 and hence solve the system of linear equations: x + 2y – 3z = -4; 2x + 3y + 2z = 2; 3x – 3y – 4z = 11.

Solution

The given equations are x + 2y – 3z = -4 …(1)

2x + 3y + 2z = 2 …(2)

3x – 3y – 4z = 11 …(3)

Let A = \(\left[\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{r}
-4 \\
2 \\
11
\end{array}\right] \text {. }\)

So, the given system in matrix form is AX = B.

Now, |A| = \(\left|\begin{array}{rrr}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right|=\left|\begin{array}{rrr}
1 & 2 & -3 \\
0 & -1 & 8 \\
0 & -9 & 5
\end{array}\right|\)

[R2 → R2 – 2R1 and R3 → R3 – 3R1]

= 1.(-5+72) = 67 ≠ 0.

Thus, A is invertible.

So, the system has a unique solution, X = A-1B.

Now, the cofactors of the elements of |A| are

A11 = -6, A12 = 14, A13 = -15; A21 = 17, A22 = 5. A23 = 9; A31 = 13, A32 = -8, A33 = -1.

∴ adj A = \(\left[\begin{array}{rrr}
-6 & 14 & -15 \\
17 & 5 & 9 \\
13 & -8 & -1
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
-6 & 17 & 13 \\
14 & 5 & -8 \\
-15 & 9 & -1
\end{array}\right]\)

So, \(A^{-1}=\frac{1}{|A|} \cdot {adj} A=\frac{1}{67} \cdot\left[\begin{array}{rrr}
-6 & 17 & 13 \\
14 & 5 & -8 \\
-15 & 9 & -1
\end{array}\right]\)

∴ X = A-1B

or \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{67} \cdot\left[\begin{array}{rrr}
-6 & 17 & 13 \\
14 & 5 & -8 \\
-15 & 9 & -1
\end{array}\right] \cdot\left[\begin{array}{r}
-4 \\
2 \\
11
\end{array}\right]\)

= \(\frac{1}{67},\left[\begin{array}{r}
201 \\
-134 \\
67
\end{array}\right]=\left[\begin{array}{r}
3 \\
-2 \\
1
\end{array}\right] .\)

∴ x = 3, y = -2 and z = 1.

Example 5 Using matrices, solve the following system of linear equations: 3x + 4y + 2z = 8, 2y – 3z 3, x – 2y + 6z = -2.

Solution

The given equations are

3x + 4y + 2z = 8 …(1)

2y – 3z = 3 …(2)

x – 2y + 6z = -2 …(3)

Let A = \(\left[\begin{array}{rrr}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{r}
8 \\
3 \\
-2
\end{array}\right]\).

So, the given system in matrix form is AX = B.

Now, |A| = \(\left|\begin{array}{rrr}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right|=\left|\begin{array}{rrr}
0 & 10 & -16 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right|\)

[R1 → R1 – 3R3]

= 1.(-30+32) = 2 ≠ 0.

So, A is invertible.

Therefore, the given system has a unique solution, X = A-1B.

Now, the minors of the elements of |A| are

M11 = 6, M12 = 3, M13 = -2; M21 = 28, M22 = 16, M23 = -10; M31 = -16, M32 = -9, M33 = 6.

The cofactors of the elements of |A| are

A11 = 6, A12 = -3, A13 = -2; A21 = -28, A22 = 16, A23 = 10; A31 = -16, A32 = 9, A33 = 6.

∴ (adj A) = \(\left[\begin{array}{rrr}
6 & -3 & -2 \\
-28 & 16 & 10 \\
-16 & 9 & 6
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
6 & -28 & -16 \\
-3 & 16 & 9 \\
-2 & 10 & 6
\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|} \cdot({adj} A)\)

= \(\frac{1}{2} \cdot\left[\begin{array}{rrr}
6 & -28 & -16 \\
-3 & 16 & 9 \\
-2 & 10 & 6
\end{array}\right]=\left[\begin{array}{rrr}
3 & -14 & -8 \\
\frac{-3}{2} & 8 & \frac{9}{2} \\
-1 & 5 & 3
\end{array}\right]\)

∴ X = A-1B ⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{rrr}
3 & -14 & -8 \\
\frac{-3}{2} & 8 & \frac{9}{2} \\
-1 & 5 & 3
\end{array}\right]\left[\begin{array}{r}
8 \\
3 \\
-2
\end{array}\right]\)

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
24-42+16 \\
-12+24-9 \\
-8+15-6
\end{array}\right]=\left[\begin{array}{r}
-2 \\
3 \\
1
\end{array}\right]\)

⇒ x = -2, y = 3 and z = 1.

Hence, x = -2, y = 3 and z = 1.

Example 6 Using matrices, solve the following system of equations:

\(\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4\);

\(\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1\);

\(\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2 .(x, y, z \neq 0)\)

Solution

Putting \(\frac{1}{x}\) = u, and \(\frac{1}{y}\) = v and \(\frac{1}{z}\) = w, the given equations become:

2u + 3v + 10w = 4 …(1)

4u – 6v + 5w = 1 …(2)

6u + 9v – 20w = 2 …(3)

Let A = \(\left[\begin{array}{rrr}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right], Y = \left[\begin{array}{l}
u \\
v \\
w
\end{array}\right] and B = \left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right]\).

Then, the given system in matrix form is AY = B.

Now, |A| = \(\left[\begin{array}{rrr}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right]\)

= \(\left|\begin{array}{rrr}
2 & 3 & 10 \\
0 & -12 & -15 \\
0 & 0 & -50
\end{array}\right|\) [R2 → R2 – 2R1; R3 → R3 – 3R1]

= (-50).(-24-0) = 1200 ≠ 0.

Thus, A is invertible.

So, the given system has a unique solution, Y = A-1B.

The minors of the elements of |A| are

M11 = 75, M12 = -110, M13 = 72; M21 = -150, M22 = -100, M23 = 0; M31 = 75, M32 = -30, M33 = -24.

So, the cofactors of the elements of |A| are

A11 = 75, A12 = 110, A13 – 72; A21 = 150, A22 = -100, A23 = 0; A31 = 75, A32 = 30, A33= -24.

∴ (adj A) = \(\left[\begin{array}{rrr}
75 & 110 & 72 \\
150 & -100 & 0 \\
75 & 30 & -24
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
75 & 150 & 75 \\
110 & -100 & 30 \\
72 & 0 & -24
\end{array}\right]\)

∴ \(A^{-1}=\frac{1}{|A|} \cdot({adj} A)=\frac{1}{1200} \cdot\left[\begin{array}{rrr}
75 & 150 & 75 \\
110 & -100 & 30 \\
72 & 0 & -24
\end{array}\right]\)

∴ Y = A-1B

⇒ \(\left[\begin{array}{c}
u \\
v \\
w
\end{array}\right]=\frac{1}{1200} \cdot\left[\begin{array}{rcr}
75 & 150 & 75 \\
110 & -100 & 30 \\
72 & 0 & -24
\end{array}\right]\left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right]\)

= \(\frac{1}{1200} \cdot\left[\begin{array}{l}
300+150+150 \\
440-100+60 \\
288+0-48
\end{array}\right]=\frac{1}{1200} \cdot\left[\begin{array}{l}
600 \\
400 \\
240
\end{array}\right]\)

= \(\left[\begin{array}{l}
\frac{600}{1200} \\
\frac{400}{1200} \\
\frac{240}{1200}
\end{array}\right]=\left[\begin{array}{l}
\frac{1}{2} \\
\frac{1}{3} \\
\frac{1}{5}
\end{array}\right]\)

⇒ u = \(\frac{1}{2}\), v = \(\frac{1}{3}\), w = \(\frac{1}{5}\)

⇒ \(\frac{1}{x}\) = \(\frac{1}{2}\), \(\frac{1}{y}\) = \(\frac{1}{3}\) and \(\frac{1}{z}\) = \(\frac{1}{5}\)

⇒ x = 2, y = 3 and z = 5.

Hence x = 2, y = 3 and z = 5.

Example 7 Use the product \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\left[\begin{array}{rrr}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]\) to solve the following system of equations: x – y + 2z = 1; 2y – 3z = 1; 3x – 2y + 4z = 2.

Solution

The given equations are

x – y + 2z + 1 …(1)

2y – 3z = 1 …(2)

3x – 2y + 4z = 2 …(3)

Let A = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right]\).

Then, the given system in matrix form is AX = B.

Now, \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\left[\begin{array}{rrr}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
-2-9+12 & 0-2+2 & 1+3-4 \\
0+18-18 & 0+4-3 & 0-6+6 \\
-6-18+24 & 0-4+4 & 3+6-8
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)

⇒ \(A \cdot\left[\begin{array}{rrr}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]=I\)

⇒ \(A^{-1}=\left[\begin{array}{rrr}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]\)

Now, AX = B

⇒ x = A-1B

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right]\)

= \(\left[\begin{array}{r}
-2+0+2 \\
9+2-6 \\
6+1-4
\end{array}\right]=\left[\begin{array}{l}
0 \\
5 \\
3
\end{array}\right]\)

⇒ x = 0, y = 5 and z = 3.

Hence, x = 0, y = 5 and z = 3.

Example 8 Given A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
1 & -2 & -2 \\
2 & 1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
-4 & 4 & 4 \\
-7 & 1 & 3 \\
5 & -3 & -1
\end{array}\right]\), find AB and use this result in solving the following system of equations: x – y + z = 4; x – 2y – 2z = 9; 2x + y + 3z = 1.

Solution

The given equations are

x – y + z = 4 …(1)

x – 2y – 2z = 9 …(2)

2x + y + 3z = 1 …(3)

Let A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
1 & -2 & -2 \\
2 & 1 & 3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and C = \(\left[\begin{array}{l}
4 \\
9 \\
1
\end{array}\right]\).

Then, the given system of equations is AX = C.

Now, AB = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
1 & -2 & -2 \\
2 & 1 & 3
\end{array}\right]\left[\begin{array}{rrr}
-4 & 4 & 4 \\
-7 & 1 & 3 \\
5 & -3 & -1
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
-4+7+5 & 4-1-3 & 4-3-1 \\
-4+14-10 & 4-2+6 & 4-6+2 \\
-8-7+15 & 8+1-9 & 8+3-3
\end{array}\right]=\left[\begin{array}{lll}
8 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 8
\end{array}\right]\)

= 8I

⇒ \(A \cdot\left(\frac{1}{8} B\right)=I\)

⇒ \(A^{-1}=\frac{1}{8} B=\frac{1}{8} \cdot\left[\begin{array}{rrr}
-4 & 4 & 4 \\
-7 & 1 & 3 \\
5 & -3 & -1
\end{array}\right]\)

Now, AX = C

⇒ X = A-1C

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{8} \cdot\left[\begin{array}{rrr}
-4 & 4 & 4 \\
-7 & 1 & 3 \\
5 & -3 & -1
\end{array}\right]\left[\begin{array}{l}
4 \\
9 \\
1
\end{array}\right]\)

= \(\frac{1}{8} \cdot\left[\begin{array}{r}
-16+36+4 \\
-28+9+3 \\
20-27-1
\end{array}\right]=\frac{1}{8} \cdot\left[\begin{array}{r}
24 \\
-16 \\
-8
\end{array}\right]=\left[\begin{array}{r}
3 \\
-2 \\
-1
\end{array}\right]\)

⇒ x = 3, y = -2 and z = -1.

Hence, x = 3, y = -2 and z = -1.

Example 9 The sum of three numbers is 6. Twice the third number when added to the first number gives 7. On adding the sum of the second and third numbers to thrice the first number, we get 12. Find the numbers, using matrix method.

Solution

Let the first, second and third numbers be x,y,z respectively. Then,

x + y + z = 6 …(1)

x + 2z = 7 …(2)

3x + y + z = 12 …(3)

Let A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 0 & 2 \\
3 & 1 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{r}
6 \\
7 \\
12
\end{array}\right]\)

Then, the given system in matrix form is AX = B.

Now, |A| = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 0 & 2 \\
3 & 1 & 1
\end{array}\right|=\left|\begin{array}{rrr}
1 & 1 & 1 \\
0 & -1 & 1 \\
0 & -2 & -2
\end{array}\right|\)

\(\left[\begin{array}{l}
R_2 \rightarrow R_2-R_1 ; \\
R_3 \rightarrow R_3-3 R_1
\end{array}\right]\)

= 1.(2+2) = 4 ≠ 0.

∴ A is invertible.

So, the given system has a unique solution, X = A-1B.

The minors of the elements of |A| are

M11 = -2, M12 = -5, M13 = 1; M21 = 0, M22 = -2, M23 = -2; M31 = 2, M32 = 1, M33 = -1.

The cofactors of the elements of |A| are

A11 = -2, A12 = 5, A13 = 1; A21 = 0, A22 = -2, A23 = 2; A31 = 2, A32 = -1, A33 = -1.

∴ (adj A) = \(\left[\begin{array}{rrr}
-2 & 5 & 1 \\
0 & -2 & 2 \\
2 & -1 & -1
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
-2 & 0 & 2 \\
5 & -2 & -1 \\
1 & 2 & -1
\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|} \cdot({adj} A)=\frac{1}{4} \cdot\left[\begin{array}{rrr}
-2 & 0 & 2 \\
5 & -2 & -1 \\
1 & 2 & -1
\end{array}\right]\)

⇒ X = A-1B

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{4} \cdot\left[\begin{array}{rrr}
-2 & 0 & 2 \\
5 & -2 & -1 \\
1 & 2 & -1
\end{array}\right]\left[\begin{array}{r}
6 \\
7 \\
12
\end{array}\right]\)

= \(\frac{1}{4} \cdot\left[\begin{array}{r}
-12+0+24 \\
30-14-12 \\
6+14-12
\end{array}\right]=\frac{1}{4} \cdot\left[\begin{array}{r}
12 \\
4 \\
8
\end{array}\right]=\left[\begin{array}{l}
3 \\
1 \\
2
\end{array}\right]\)

⇒ x = 3, y = 1, z = 2.

Hence, the required numbers are 3,1,2.

Cramer’s Rule

System of Linear Equations In Two Unknowns

Theorem 1 The solution of the system of equations

a1x + b1y = c1 …(1)

a2x + b2y = c2 …(2)

is given by: x = \(\frac{\Delta_1}{\Delta}\) and y = \(\frac{\Delta_2}{\Delta}\), where

Δ = \(\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|\), Δ1 = \(\left|\begin{array}{ll}
c_1 & b_1 \\
c_2 & b_2
\end{array}\right|\) and Δ2 = \(\left|\begin{array}{ll}
a_1 & c_1 \\
a_2 & c_2
\end{array}\right|\) and Δ ≠ 0.

Proof

On multiplying (1) by b2 and (2) by b1 and subtracting, we get:

(a1b2 – a2b1)x = (b2c1 – b1c2)

⇒ \(x=\frac{\left(b_2 c_1-b_1 c_2\right)}{\left(a_1 b_2-a_2 b_1\right)}=\frac{\left|\begin{array}{ll}
c_1 & b_1 \\
c_2 & b_2
\end{array}\right|}{\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|}=\frac{\Delta_1}{\Delta} .\)

Again, multiplying (1) by a2 and (2) by b1 and subtracting, we get: (a2b1 – a1b2)y = (a2c1 – a1c2)

⇒ \(y=\frac{\left(a_2 c_1-a_1 c_2\right)}{\left(a_2 b_1-a_1 b_2\right)}=\frac{\left|\begin{array}{ll}
a_1 & c_1 \\
a_2 & c_2
\end{array}\right|}{\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|}=\frac{\Delta_2}{\Delta} .\)

Hence, x = \(\frac{\Delta_1}{\Delta}\) and y = \(\frac{\Delta_2}{\Delta}\).

Remark Here Δ = \(\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|\) = determinant of coefficients of x and y.

To obtain Δ1, we replace a1, a2 by c1, c2 respectively.

To obtain Δ2, we replace b1, b2 by c1, c2 respectively.

Example 1 Solve the following system of linear equations using Cramer’s rule: 3x + 4y = 5, x – y + 3 = 0.

Solution

The given equations are:

3x + 4y = 5 …(1)

x – y = -3 …(2)

We have

Δ = \(\left|\begin{array}{rr}
3 & 4 \\
1 & -1
\end{array}\right|\) = (-3-4) = -7;

Δ1 = \(\left|\begin{array}{rr}
5 & 4 \\
-3 & -1
\end{array}\right|\) = (-5+12) = 7

and Δ2 = \(\left|\begin{array}{rr}
3 & 5 \\
1 & -3
\end{array}\right|\) = (-9-5) = -14.

∴ \(x=\frac{\Delta_1}{\Delta}=\frac{7}{-7}=-1 \text { and } y=\frac{\Delta_2}{\Delta}=\frac{-14}{-7}=2 \text {. }\)

Hence x = -1 and y – 2.

System of Linear Equations In Three Unknowns

Theorem 2 The solution of the system of equations

a1x + b1y + c1z = d1 …(1)

a2x + b2y + c2z = d2 …(2)

a3x + b3y + c3z = d3 …(3)

is given by: x = \(\frac{\Delta_1}{\Delta}\), y = \(\frac{\Delta_2}{\Delta}\) and z = \(\frac{\Delta_3}{\Delta}\), where

\(\Delta=\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right| \neq 0, \quad \Delta_1=\left|\begin{array}{lll}
d_1 & b_1 & c_1 \\
d_2 & b_2 & c_2 \\
d_3 & b_3 & c_3
\end{array}\right| \neq 0 ;\) \(\Delta_2=\left|\begin{array}{lll}
a_1 & d_1 & c_1 \\
a_2 & d_2 & c_2 \\
a_3 & d_3 & c_3
\end{array}\right| \text { and } \Delta=\left|\begin{array}{lll}
a_1 & b_1 & d_1 \\
a_2 & b_2 & d_2 \\
a_3 & b_3 & d_3
\end{array}\right| \text {. }\)

Solution

We have:

xΔ = \(\left|\begin{array}{lll}
a_1 x & b_1 & c_1 \\
a_2 x & b_2 & c_2 \\
a_3 x & b_3 & c_3
\end{array}\right|\)

= \(\left|\begin{array}{lll}
a_1 x+b_1 y+c_1 z & b_1 & c_1 \\
a_2 x+b_2 y+c_2 z & b_2 & c_2 \\
a_3 x+b_3 y+c_3 z & b_3 & c_3
\end{array}\right| \quad\left[C_1 \rightarrow C_1+y C_2+z C_3\right]\)

= \(\left|\begin{array}{lll}
d_1 & b_1 & c_1 \\
d_2 & b_2 & c_2 \\
d_3 & b_3 & c_3
\end{array}\right|=\Delta_1\) [using (1), (2) and (3)].

Similarly, yΔ = Δ2 and zΔ = Δ3.

∴ x = \(\frac{\Delta_1}{\Delta}\), y = \(\frac{\Delta_2}{\Delta}\) and z = \(\frac{\Delta_3}{\Delta}\).

Remark Here Δ = \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\).

To obtain Δ1 replace a1, a2, a3 by d1, d2, d3 respectively in Δ.

To obtain Δ2 replace b1, b2, b3 by d1, d2, d3 respectively in Δ.

To obtain Δ3 replace c1, c2, c3 by d1, d2, d3 respectively in Δ.

Example 2 Solve the following system of linear equations by Cramer’s rule: x – 3y + 2z = 8, 3x + 4y + z = 5, 4x – 2y + 9z = -2.

Solution

We have

Δ = \(\left|\begin{array}{rrr}
1 & -3 & 2 \\
3 & 4 & 1 \\
4 & -2 & 9
\end{array}\right|=1 \cdot(36+2)+3 \cdot(27-4)+2 \cdot(-6-16)\)

= 38 + 69 – 44 = 63.

Δ1 = \(\left|\begin{array}{rrr}
8 & -3 & 2 \\
5 & 4 & 1 \\
-2 & -2 & 9
\end{array}\right|=8 \cdot(36+2)+3 \cdot(45+2)+2 \cdot(-10+8)\)

= 304 + 141 – 4 = 441.

Δ2 = \(\left|\begin{array}{rrr}
1 & 8 & 2 \\
3 & 5 & 1 \\
4 & -2 & 9
\end{array}\right|=1 \cdot(45+2)-8 \cdot(27-4)+2 \cdot(-6-20)\)

= 47 – 184 – 52 = -189.

Δ3 = \(\left|\begin{array}{rrr}
1 & -3 & 8 \\
3 & 4 & 5 \\
4 & -2 & -2
\end{array}\right|=1 \cdot(-8+10)+3 \cdot(-6-20)+8 \cdot(-6-16)\)

= 2 – 78 – 176 = -252.

∴ \(x=\frac{\Delta_1}{\Delta}=\frac{441}{63}=7, y=\frac{\Delta_2}{\Delta}=\frac{-189}{63}=-3 \text { and } z=\frac{\Delta_3}{\Delta}=\frac{-252}{63}=-4 \text {. }\)

Hence, x = 7, y = -3 and z = -4.

Consistent and Inconsistent Systems of Equations

A system of equations is said to be consistent or inconsistent according as its solution exists or does not exist respectively.

Various Conditions For consistency

For equations in two unknowns, we have

x = \(\frac{\Delta_1}{\Delta}\) and y = \(\frac{\Delta_2}{\Delta}\).

In this case, we have the following conditions:

(1) If Δ ≠ 0, then the given system has a unique solution.

(2) If Δ = 0, Δ1 ≠ 0 and Δ3 ≠ 0, then the given system has infinitely many solutions.

(3) If Δ = 0, Δ1 ≠ 0 or Δ2 ≠ 0, then the system is inconsistent.

For equations in three unknowns, we have

x = \(\frac{\Delta_1}{\Delta}\), y = \(\frac{\Delta_2}{\Delta}\) and z = \(\frac{\Delta_3}{\Delta}\).

In this case, we have the following conditions.

(1) If Δ ≠ 0, then the given system has a unique solution.

(2) If Δ = 0 and at least one of Δ1, Δ2, Δ3 is nonzero, then the given system is inconsistent.

(3) If Δ = Δ1 = Δ2 = Δ3 = 0, then the system may or may not be consistent.

Example 3 Show that the following system of equations is inconsistent: x + 3y = 2, 2x + 6y = 7.

Solution

The given equations are:

x + 3y = 2 …(1)

2x + 6y = 7 …(2)

∴ Δ = \(\left|\begin{array}{ll}
1 & 3 \\
2 & 6
\end{array}\right|\) = (6-6) = 0

and Δ1 = \(\left|\begin{array}{ll}
2 & 3 \\
7 & 6
\end{array}\right|\) = (12-21) = -9 ≠ 0.

Thus, Δ = 0 and Δ1 ≠ 0.

Hence, the given system of equations is inconsistent.

Example 4 Solve the following system of equations using determinants: 2x + 5y = 8, 4x + 10y = 16.

Solution

The given equations are:

2x + 5y = 8 …(1)

4x + 10y = 16 …(2)

∴ Δ = \(\left|\begin{array}{rr}
2 & 5 \\
4 & 10
\end{array}\right|\) = (20 – 20) = 0,

Δ1 = \(\left|\begin{array}{rr}
8 & 5 \\
16 & 10
\end{array}\right|\) = (80-80) = 0

and Δ2 = \(\left|\begin{array}{rr}
2 & 8 \\
4 & 16
\end{array}\right|\) = (32-32) = 0.

Thus, Δ = 0, Δ1 = 0 and Δ2 = 0.

So, the given system has an infinite number of solutions.

Let y = k. Then, 2x + 5k = 8 ⇒ x = \(\frac{1}{2}\)(8-5k).

Hence, x = \(\frac{1}{2}\)(8-5k) and y = k is the required solution, where k is arbitraty.

Example 5 Show that the following system of equations is inconsistent: 2x – y + z = 4, x + 3y + 2z = 12, 3x + 2y + 3z = 10.

Solution

We have

Δ = \(\left|\begin{array}{rrr}
2 & -1 & 1 \\
1 & 3 & 2 \\
3 & 2 & 3
\end{array}\right|=2(9-4)+1 \cdot(3-6)+1 \cdot(2-9)\)

= 10 – 3 – 7 = 0

and Δ1 = \(\left|\begin{array}{rrr}
4 & -1 & 1 \\
12 & 3 & 2 \\
10 & 2 & 3
\end{array}\right|=4(9-4)+1 \cdot(36-20)+1 \cdot(24-30)\)

= 20 + 16 – 6 = 30 ≠ 0.

Thus, Δ = 0 and Δ1 ≠ 0.

Hence, the given system is inconsistent.

Example 6 Determine whether the following system of equations is inconsistent: 2x + y – 2z = 4, x – 2y + z = -2, 5x – 5y + z = -2. If so, find its solution.

Solution

We leave it to the reader to verify that:

Δ = 0, Δ1 = 0, Δ2 = 0 and Δ3 = 0.

So, we cannot conclude whether the given system is consistent or inconsistent.

Let us put z = k, where k is an arbitrary constant.

Then, the given equations become:

2x + y = (4+2k) …(1)

x – 2y = (-2-k) …(2)

5x – 5y = (-2-k) …(3)

Solving (1) and (2) by Cramer’s rule in x and y, we get

Δ = \(\left|\begin{array}{rr}
2 & 1 \\
1 & -2
\end{array}\right|\) = (-4-1) = -5 ≠ 0.

Δ1 = \(\left|\begin{array}{cc}
4+2 k & 1 \\
-2-k & -2
\end{array}\right|\) = (-8-4k) + (2+k) = (-6-3k)

and Δ2 = \(\left|\begin{array}{cc}
2 & 4+2 k \\
1 & -2-k
\end{array}\right|\) = (-4-2k) – (4+2k) = (-8-4k).

∴ \(x=\frac{\Delta_1}{\Delta}=\frac{(-6-3 k)}{-5}=\frac{(6+3 k)}{5} \text { and } y=\frac{\Delta_2}{\Delta}=\frac{(-8-4 k)}{-5}=\frac{(8+4 k)}{5} \text {. }\)

Thus, \(x=\frac{6+3 k}{5} \text { and } y=\frac{8+4 k}{5} \text {. }\)

Putting these values of x and y in (3), we get

LHS = \(\left.5\left(\frac{6+3 k}{5}\right)-5\left(\frac{8+4 k}{5}\right)=(6+3 k)-8+4 k\right)=(-2-k)\) = RHS.

So, the above values of x and y satisfy the third equation.

∴ \(x=\frac{6+3 k}{5}, y=\frac{8+4 k}{5} \text { and } z=k \text {. }\)

By giving arbitrary values to k, we find that the given system of equations has an infinite number of solutions.

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