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## ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.4

Question 1.
(i) For what value of x, is the following matrix singular? $$\left[\begin{array}{cc} 3-2 x & x+1 \\ 2 & 4 \end{array}\right]$$
(ii) If 0 < x < π and the matrix $$\left[\begin{array}{cc} \cos x & 1 \\ 3 & 4 \cos x \end{array}\right]$$ is singular, find the values of x.
Solution:
(i) Let A = $$\left[\begin{array}{cc} 3-2 x & x+1 \\ 2 & 4 \end{array}\right]$$ and A is given to be singular.
∴ |A| = 0
⇒ $$\left|\begin{array}{cc} 3-2 x & x+1 \\ 2 & 4 \end{array}\right|$$ = 0
⇒ 4 (3 – 2x) – 2 (x + 1) = 0
⇒ – 10x + 10 = 0
⇒ x = $$\frac{10}{10}$$ = 1.

(ii) Let A = $$\left[\begin{array}{cc} \cos x & 1 \\ 3 & 4 \cos x \end{array}\right]$$
and A is given to be singular.
∴ |A| = 0
⇒ $$\left|\begin{array}{cc} \cos x & 1 \\ 3 & 4 \cos x \end{array}\right|$$ = 0
⇒ 4 cos2 x – 3 = 0
⇒ cos2 x = $$\left(\frac{\sqrt{3}}{2}\right)^2$$
= cos2 $$\frac{\pi}{6}$$
⇒ x = nπ ± $$\frac{\pi}{6}$$ ∀ n ∈ I
Since 0 < x < π
∴ x = $$\frac{\pi}{6}, \frac{5 \pi}{6}$$.

Question 1 (old).
If the matrix $$\left[\begin{array}{cc} 5-x & x+1 \\ 2 & 4 \end{array}\right]$$ is singular, find x.
Solution:
Given A = $$\left[\begin{array}{cc} 5-x & x+1 \\ 2 & 4 \end{array}\right]$$ is singular
∴ |A| = 0
⇒ $$\left|\begin{array}{cc} 5-x & x+1 \\ 2 & 4 \end{array}\right|$$ = 0
⇒ 4 (5 – x) – 2 (x + 1) = 0
⇒ 20 – 4x – x – 2 = 0
⇒ – 6x + 18 = 0
⇒ x = 3

Question 2.
Which of the following matrices are non-singular?
(i) $$\left[\begin{array}{rrr} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right]$$
(ii) $$\left[\begin{array}{rrr} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right]$$
Solution:
(i) Let A = $$\left[\begin{array}{rrr} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rrr} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|$$ ;
expanding along R1
= 6 (- 2 – 10) + 3 (4 + 20) + 2 (10 – 10)
= – 72 + 72 + 0 = 0
Thus, matrix A is singular.

(ii) Let |A| = $$\left|\begin{array}{ccc} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right|$$ ;
expanding along R1
= 1 (- 2 – 10) + 3 (8 – 6) + 2 (20 + 3)
= – 12 + 6 + 46 = 40 ≠ 0
Thus A is non-singular matrix.

Question 3.
Write the adjoint of the matrices :
(i) $$\left[\begin{array}{rr} 2 & -1 \\ 4 & 3 \end{array}\right]$$
(ii) $$\left[\begin{array}{cc} 3 & 0 \\ -2 & -5 \end{array}\right]$$.
Solution:
(i) Let A = $$\left[\begin{array}{rr} 2 & -1 \\ 4 & 3 \end{array}\right]$$
∴ Cofactors of R1 are : 3, – 4
Cofactors of R2 are : 1, 2
∴ adj A = $$\left[\begin{array}{rr} 3 & -4 \\ 1 & 2 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rr} 3 & 1 \\ -4 & 2 \end{array}\right]$$

(ii) Let A = $$\left[\begin{array}{rr} 3 & 0 \\ -2 & -5 \end{array}\right]$$
∴ Cofactors of R1 are : – 5, 2
Cofactors of R2 are : 0, 3
∴ adj A = $$\left[\begin{array}{rr} -5 & 2 \\ 0 & 3 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rr} -5 & 0 \\ 2 & 3 \end{array}\right]$$

Question 4.
(i) For what values of k, the matrix $$\left[\begin{array}{ll} 2 & k \\ 3 & 5 \end{array}\right]$$ has no inverse?
(ii) If A = $$\left[\begin{array}{rr} 3 & 10 \\ 2 & 7 \end{array}\right]$$, then write A-1.
(iii) Write the inverse of the matrix $$\left[\begin{array}{rr} 3 & -1 \\ 1 & 2 \end{array}\right]$$.
Solution:
(i) Let A = $$\left[\begin{array}{ll} 2 & k \\ 3 & 5 \end{array}\right]$$
Since matrix A has no inverse
∴ A is not invertible.
∴ A is singular.
∴ |A| = 0
⇒ $$\left|\begin{array}{ll} 2 & k \\ 3 & 5 \end{array}\right|$$ = 0
⇒ 10 – 3k = 0
⇒ k = $$\frac{10}{3}$$.

(ii) Given A = $$\left[\begin{array}{rr} 3 & 10 \\ 2 & 7 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rr} 3 & 10 \\ 2 & 7 \end{array}\right|$$
= 21 – 20
= 1 ≠ 0
Thus A-1 exists
and A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A ……….(1)
∴ Cofactors of R1 are ; 7 , – 2
Cofactors of R2 are ; – 10, 3
∴ adj A = $$\left[\begin{array}{rr} 7 & -2 \\ -10 & 3 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rr} 7 & -10 \\ -2 & 3 \end{array}\right]$$
from (1) ;
A = $$\frac{1}{1}\left[\begin{array}{rr} 7 & -10 \\ -2 & 3 \end{array}\right]$$
= $$\left[\begin{array}{rr} 7 & -10 \\ -2 & 3 \end{array}\right]$$

(iii) Given A = $$\left[\begin{array}{rr} 3 & -1 \\ 1 & 2 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rr} 3 & -1 \\ 1 & 2 \end{array}\right|$$
= 6 + 1
= 7 ≠ 0
∴ A-1 exists
and A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A ………………(1)
∴ Cofactors of R1 are ; 2, – 1
Cofactors of R1 are ; 1, 3
∴ adj A = $$\left[\begin{array}{rr} 2 & -1 \\ 1 & 3 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rr} 2 & 1 \\ -1 & 3 \end{array}\right]$$
∴ from (1);
A-1 = $$\frac{1}{7}\left[\begin{array}{rr} 2 & 1 \\ -1 & 3 \end{array}\right]$$.

Question 4 (Old).
Write A-1 for A = $$\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]$$.
Solution:
Given A = $$\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]$$
∴ Cofactors of R1 are ; – 5 , 2
Cofactors of R2 are ; 0, 3
∴ adj A = $$\left[\begin{array}{rr} -5 & 2 \\ 0 & 3 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rr} -5 & 0 \\ 2 & 3 \end{array}\right]$$

Question 5.
(i) If A = $$\left[\begin{array}{ll} 8 & 2 \\ 3 & 2 \end{array}\right]$$, then find |adj A|.
(ii) If A is a square matrix of order 3 and |A| = 5, then find |adj A|.
(iii) If A is a non-singular square matrix of order 3 such that |adj A| = 64, find |A|.
(iv) If A is a square matrix of order 3 such that |A| = – 2, then find |3 adj A|.
(v) If |A-1| = 5, then find the value of |A|.
(vi) If |A| = – 4 and A-1 = $$\left[\begin{array}{rr} -2 & 1 \\ \frac{5}{4} & -\frac{1}{2} \end{array}\right]$$, then write adj A.
(vi) If |A| = – 4 and A-1 = $$\left[\begin{array}{rr} -2 & 1 \\ \frac{5}{4} & -\frac{1}{2} \end{array}\right]$$, then write adj A.
(vii) If A = $$\left[\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right]$$ be such that A-1 = kA, then find the value of k.
(viii) If A (adj A) = $$\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]$$, then find the value of |adj A|.
(ix) If the value of a determinant of third order is 12, then what is the value of the determinant formed by replacing each element by its co-factor?
Solution:
(i) Given A = $$\left[\begin{array}{ll} 8 & 2 \\ 3 & 2 \end{array}\right]$$
∴ |A| = 16 – 6 = 10
We know that,
Given A be a matrix of order 2
∴ |adj A| = |a| = 10

= |A|2
= 52
= 25

(iii) Given A be a non-singular matrix of order 3
∴ |A| ≠ 0
and n = 3
⇒ |A|n-1 = 64
⇒ |A|3-1 = 64
⇒ |A|2 = 64
⇒ |A| = ± 8.

(iv) If A be a matrix of order n.
Then |kA| = kn |A|
Given A be a matrix of order 3
then adj A be a matrix of order 3.
= 27 |A|2
= 27 (- 2)2 = 108.

(v) Given |A-1| = 5
|A|-1 = 5
⇒ $$\frac{1}{|A|}$$ = 5
⇒ |A| = $$\frac{1}{5}$$

(vi) We know that
A-1 = $$\frac{1}{|A|}$$ adj A
⇒ adj A = A-1 |A|
= – 4 $$\left[\begin{array}{rr} -2 & 1 \\ \frac{5}{4} & -\frac{1}{2} \end{array}\right]$$
= $$\left[\begin{array}{rr} 8 & -4 \\ -5 & 2 \end{array}\right]$$

(vii) Given A = $$\left[\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right|$$
= – 4 – 15
= – 19 ≠ 0
Given A-1 = kA
⇒ |A-1| = |kA|
⇒ |A|-1 = k2 |A|
[∵ A be a matrix of order 2]
⇒ k2 = $$\frac{1}{(-19)^2}$$
⇒ k = ± $$\frac{1}{19}$$

(viii) Given A adj A = $$\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]$$ = 3I
Also A adj A = |A| I
∴ |A| = 3
= 32 = 9
[If A be a matrix of order n then |adj A| = |A|n -1]

(ix) Let A be the given matrix of order 3.
Then |A| = 12
Thus required det = |adj A|
= |A|3-1
= |A|2
= 122
= 144.

Question 5 (old).
(i) If A is a square matrix of order 3 and |A| = – 5 then find |adj A|.
(ii) If A is a non-singular square matrix of order 3 such that |adj A| = 64, find |A|.
Solution:
(i) We know that if A be a square matrix of order n.
Since A be a square matrix of order 3
∴ n = 3
= (- 5)3-1 = 25
[∵ |A| = – 5]

(ii) Given A be a non-singular square matrix of order 3
∴ |A| ≠ 0
and n = 3
⇒ |A|n-1 = 64
⇒ |A|3-1 = 64
⇒ |A| 2 = 64
⇒ |A| = ± 8

Question 6.
(i) A = $$\left[\begin{array}{rr} 1 & 3 \\ -1 & 4 \end{array}\right]$$
(ii) A = $$\left[\begin{array}{rrr} -2 & 0 & 0 \\ 3 & 4 & 0 \\ 10 & -7 & 3 \end{array}\right]$$
(iii) A = $$\left[\begin{array}{rrr} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right]$$
Solution:
We know that if A be a square matrix of order n,
Then |adj A| = |A|n-1 …………..(1)

(i) Given A = $$\left[\begin{array}{rr} 1 & 3 \\ -1 & 4 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rr} 1 & 3 \\ -1 & 4 \end{array}\right|$$
= 4 + 3 = 7
Here A be a square matrix of order 2 × 2
∴ n = 2
= 72-1
= 71 = 7.

(ii) Given A = $$\left[\begin{array}{rrr} -2 & 0 & 0 \\ 3 & 4 & 0 \\ 10 & -7 & 3 \end{array}\right]_{3 \times 3}$$
Here A be a square matrix of order 3 × 3
∴ |A| = $$\left|\begin{array}{rrr} -2 & 0 & 0 \\ 3 & 4 & 0 \\ 10 & -7 & 3 \end{array}\right|$$ ;
expanding along R1
= – 2 (12 – 0)
= – 24
= (- 24)3-1
= (- 24)2
= 576.

(iii) Given A = $$\left[\begin{array}{rrr} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right]$$
Here A be a square matrix of order 3 × 3
∴ n = 3
∴ |A| = $$\left|\begin{array}{rrr} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right|$$ ;
Expanding along R1
= 2 (0 – 5) + 1 (0 + 3) – 2 (0 – 6)
= – 10 + 3 + 12
= 5
∴ |adj A| = |A|n – 1
= 53-1
= 52
= 25.

Question 7.
(i) If A = $$\left[\begin{array}{ll} 2 & 1 \\ 7 & 5 \end{array}\right]$$, find |A adj A|.
(ii) If A = $$\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]$$, find |A adj A|.
(iii) If A = $$\left[\begin{array}{rr} 3 & -1 \\ 4 & 5 \end{array}\right]$$, find adj (adj A).
Solution:
(i) Given A = $$\left[\begin{array}{ll} 2 & 1 \\ 7 & 5 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ll} 2 & 1 \\ 7 & 5 \end{array}\right|$$
= 10 – 7 = 3
= |A| |A|n-1
= |A|n
= 32 = 9
[∵ A be a square matrix of order 2 × 2]

(ii) Given A = $$\left[\begin{array}{ccc} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]_{3 \times 3}$$
i.e., A be a matrix of order 3 × 3.
∴ |A| = $$\left|\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right|$$ ;
Expanding along R1
= a3
[∵ |AB| = |A| |B|]
= |A| |A|n-1
= |A|n
= (a3)3
= a9.

(iii) We kbnow that for a non-singular matrix B of order n, we have
B adj B = |B| I …………..(1)
replacing B by adj A in eqn. (1) ; we have
[∵ A . I = A]
⇒ adj (adj A) = $$\frac{\mathrm{A}|\mathrm{A}|^{n-1}}{|\mathrm{~A}|}$$
[∵ |A| ≠ 0]
Given A = $$\left[\begin{array}{rr} 3 & -1 \\ 4 & 5 \end{array}\right]$$
∴ |A| = 15 + 4
= 19 ≠ 0
∴ from (2) ; we have
adj (adj A) = (19)2-2 $$\left[\begin{array}{rr} 3 & -1 \\ 4 & 5 \end{array}\right]=\left[\begin{array}{rr} 3 & -1 \\ 4 & 5 \end{array}\right]$$.

Question 8.
(i) If A is a non-singular matrix, then show that |A-1| = $$\frac{1}{|A|}$$.
(ii) If A is a matrix such that A3 = I, then show that A is inertibIe.
(iii) If A is a matrix such that A4 = I, then show that A-1 = A3.
Solution:
(i) Given A be a non-singular matrix
∴ |A| ≠ 0
Thus AA-1 = In = A-1A
⇒ |AA-1| = |In|
[Taking determinants on bothsides]
⇒ |A| |A-1| = 1
[∵ |AB| = |A| |B| and |In| = 1]
⇒ |A-1| = $$\frac{1}{|\mathrm{~A}|}$$
[∵ |A| ≠ 0]

(ii) Given, A3 = I
|∵ A3| = |I|
⇒ |A|3 = 1
[∵ |An| = |A|n]
⇒ |A| = 1 ≠ 0
∴ A be non-singular matrix.
Thus A be invertible.

(iii) Given, A4 = 1
⇒ |A4| = |1|
⇒ |A|4 = 1
⇒ |A| = ± 1 ≠ 0
∴ A is non-singular
⇒ A is invertible
Hence A-1 exists.
again A4 = I
pre multiplying bothsides by A-1 ; we have
(A-1A) A3 = A-1 I
⇒ IA3 = A-1
⇒ A3 = A-1
[∵ A-1 A = I = AA-1]

Question 9.
If A = $$\left[\begin{array}{rr} 3 & -5 \\ 4 & 2 \end{array}\right]$$, find A (adj A).
Solution:
Given, A = $$\left[\begin{array}{rr} 3 & -5 \\ 4 & 2 \end{array}\right]$$
The Cofactors of R1 are; 2, – 4
Cofactors of R2 are ; 5, 3
∴ adj A = $$\left[\begin{array}{rr} 2 & -4 \\ 5 & 3 \end{array}\right]^{\mathrm{T}}$$
= $$\left[\begin{array}{rr} 2 & 5 \\ -4 & 3 \end{array}\right]$$

Thus A adj A = $$\left[\begin{array}{rr} 3 & -5 \\ 4 & 2 \end{array}\right]\left[\begin{array}{rr} 2 & 5 \\ -4 & 3 \end{array}\right]$$
= $$\left[\begin{array}{rr} 26 & 0 \\ 0 & 26 \end{array}\right]$$

Aliter:
|A| = $$\left|\begin{array}{rr} 3 & -5 \\ 4 & 2 \end{array}\right|$$
= 6 + 20
= 26 ≠ 0
∴ A-1 exists
Thus A adj A = |A| I2
= 26 $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{ll} 26 & 0 \\ 0 & 26 \end{array}\right]$$.

Question 10.
If the matrix $$\left[\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right]$$ in singular, find x.
Solution:
Let A = $$\left[\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|$$ ;
operate C1 → C1 + C2 + C3
= $$\left|\begin{array}{ccc} 3 x+4 & x & x \\ 3 x+4 & x+4 & x \\ 3 x+4 & x & x+4 \end{array}\right|$$ ;
Taking (3x + 4) common from C1
= (3x + 4) $$\left|\begin{array}{ccc} 1 & x & x \\ 1 & x+4 & x \\ 1 & x & x+4 \end{array}\right|$$ ;
operate R2 → R2 – R1
R3 → R3 – R1
= (3x + 4) $$\left|\begin{array}{lll} 1 & x & x \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right|$$
Expanding along C1
= 16 (3x + 4)
Given matrix A is to be singular.
∴ |A| = 0
⇒ 16 (3x + 4) = 0
⇒ x = – $$\frac{4}{3}$$

Question 10 (old).
(i) If the matrix A = $$\left[\begin{array}{rrr} 1 & -2 & 3 \\ 1 & 2 & 1 \\ x & 2 & -3 \end{array}\right]$$ is singular, find x.
Solution:
(i) Given A = $$\left[\begin{array}{rrr} 1 & -2 & 3 \\ 1 & 2 & 1 \\ x & 2 & -3 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rrr} 1 & -2 & 3 \\ 1 & 2 & 1 \\ x & 2 & -3 \end{array}\right|$$ ;
Expanding along R1
= 1 (- 6 – 2) + 2 (- 3 – x) + 3 (2 – 2x)
= – 8 – 6 – 2x + 6 – 6x
= – 8 – 8x
Since A be singular.
∴ |A| = 0
⇒ – 8 – 8x = 0
⇒ x = – 1.

Question 11.
Find the adjoint of the following matrices:
(i) A = $$\left[\begin{array}{cc} -2 & 1 \\ 3 & 4 \end{array}\right]$$
(ii) A = $$\left[\begin{array}{rr} 2 & 3 \\ -4 & -6 \end{array}\right]$$
Also verify that
Solution:
(i) $$\left[\begin{array}{cc} -2 & 1 \\ 3 & 4 \end{array}\right]$$
The cofactors of R1 are ; 4, – 3
The cofactors of R2 are ; – 1, – 2
∴ adj. A = $$\left[\begin{array}{cc} 4 & -3 \\ -1 & -2 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{cc} 4 & -1 \\ -3 & -2 \end{array}\right]$$

(ii) Given A = $$\left[\begin{array}{rr} 2 & 3 \\ -4 & -6 \end{array}\right]$$
The cofactors of R1 are ; – 6 , 4
The cofactors of R1 are ; – 3 , 2
∴ adj A = $$\left[\begin{array}{ll} -6 & 4 \\ -3 & 2 \end{array}\right]^{\mathrm{T}}$$
= $$\left[\begin{array}{rr} -6 & -3 \\ 4 & 2 \end{array}\right]$$
Now |A| = $$\left|\begin{array}{rr} 2 & 3 \\ -4 & -6 \end{array}\right|$$
= – 12 + 12 = 0
Now A adj A = $$\left[\begin{array}{rr} 2 & 3 \\ -4 & -6 \end{array}\right]\left[\begin{array}{rr} -6 & -3 \\ 4 & 2 \end{array}\right]$$
= $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$
= O
= 0 . $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
⇒ A adj A = |A| I2 ………………(1)
(adj A) A = $$\left[\begin{array}{rr} -6 & -3 \\ +4 & 2 \end{array}\right]\left[\begin{array}{rr} 2 & 3 \\ -4 & -6 \end{array}\right]$$
= $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$
= 0 $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= |A| I2 ………………….(2)
Using (1) and (2) ; we have
A adj A = |A| I2.

Question 11 (old).
Find the adjoint of the following matrices:
(i) A = $$\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]$$
Solution:
Given A = $$\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]$$
The cofactors of R1 are ; 4, – 3
Cofactors of R2 are ; – 2, 1
∴ adj A = $$\left[\begin{array}{rr} 4 & -3 \\ -2 & 1 \end{array}\right]^{\prime}=\left[\begin{array}{rr} 4 & -2 \\ -3 & 1 \end{array}\right]$$
= $$\left|\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right|$$
Now |A| = $$\left|\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right|$$
= 4 – 6
= – 2 ≠ 0
∴ A-1 exists.
Now A adj A = $$\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{rr} 4 & -2 \\ -3 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rr} -2 & 0 \\ 0 & -2 \end{array}\right]$$
= – 2 $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= – 2I2
= |A| I2 ………………(1)
(adj A) A = $$\left[\begin{array}{rr} 4 & -2 \\ -3 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]$$
= $$\left[\begin{array}{rr} -2 & 0 \\ 0 & -2 \end{array}\right]$$
= – 2 $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= |A| I2 …………….(2)
From (1) and (2) ; we have

Question 12.
If A = $$\left[\begin{array}{ccc} 3 & 1 & 2 \\ 2 & -3 & -1 \\ 1 & 2 & 1 \end{array}\right]$$, find A (adj A) without calculating adj A.
Solution:
Given A = $$\left[\begin{array}{ccc} 3 & 1 & 2 \\ 2 & -3 & -1 \\ 1 & 2 & 1 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rrr} 3 & 1 & 2 \\ 2 & -3 & -1 \\ 1 & 2 & 1 \end{array}\right|$$ ;
expanding along R1
= 3 (- 3 + 2) – 1 (2 + 1) + 2 (4 + 3)
= – 3 – 3 + 14
= 8 ≠ 0
We know that,
A adj A = |A| I3
⇒ A adj A = 8 $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}\right]$$

Question 13.
Find the adjoint of the following matrices:
(i) A = $$\left[\begin{array}{rrr} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{array}\right]$$
(ii) A = $$\left[\begin{array}{rrr} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right]$$
Also verify that A (adj A) = |A| I3 = (adj A) A. (NCERT)
Solution:
(i) Given A = $$\left[\begin{array}{rrr} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{array}\right]$$
The cofactors of R1 are :

Expanding along R1
= 1 (3 – 0) + 1 (2 + 10) + 2 (0 + 6)
= 3 + 12 + 12
= 27 ≠ 0
∴ A adj A = $$\left[\begin{array}{rrr} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{array}\right]\left[\begin{array}{rrr} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{array}\right]$$
= 27 $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
= |A| I3 ………….(1)
= $$\left[\begin{array}{rrr} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{array}\right]\left[\begin{array}{rrr} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{array}\right]$$
= 27 $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
= |A| I3 ………….(2)
From (1) and (2) ; we have
A (adj A) = |A| I3

(ii) Given A = $$\left[\begin{array}{rrr} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right]$$
The cofactors of R1 are :

From (1) and (2) ; we have
= |A| I3

Question 14.
If A = $$\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]$$, find adj A and verify that A (adj A) = (adj A) A = |A| I3.
Solution:
Given A = $$\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right|$$ ;
Expanding along R3
= cos2 α + sin2 α
= 1 ≠ 0

Question 15.
For the matrix A = $$\left[\begin{array}{rrr} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right]$$, verify that A (adj A) = O.
Solution:
Given A = $$\left[\begin{array}{rrr} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right]$$
The cofactors are R1 are ;
$$\left|\begin{array}{rr} 2 & 5 \\ 4 & -1 \end{array}\right| ;-\left|\begin{array}{rr} 4 & 5 \\ 0 & -1 \end{array}\right| ;+\left|\begin{array}{ll} 4 & 2 \\ 0 & 4 \end{array}\right|$$
i.e., – 22 ; 4 ; 16

The cofactors of R2 are ;
$$-\left|\begin{array}{rr} -1 & 3 \\ 4 & -1 \end{array}\right| ;\left|\begin{array}{rr} 2 & 3 \\ 0 & -1 \end{array}\right| ;-\left|\begin{array}{rr} 2 & -1 \\ 0 & 4 \end{array}\right|$$
i.e., – 11 ; 2 ; 8

The cofactors of R3 are ;
$$\left|\begin{array}{rr} -1 & 3 \\ 2 & 5 \end{array}\right| ;-\left|\begin{array}{rr} 2 & 3 \\ 4 & 5 \end{array}\right| ;\left|\begin{array}{rr} 2 & -1 \\ 4 & 2 \end{array}\right|$$
i.e., – 11 ; 2 ; 8

∴ adj A = $$\left[\begin{array}{rrr} -22 & 4 & 16 \\ 11 & -2 & -8 \\ -11 & 2 & 8 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rrr} -22 & 11 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8 \end{array}\right]$$

Thus, A adj A = $$\left[\begin{array}{rrr} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right]\left[\begin{array}{rrr} -22 & 11 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8 \end{array}\right]$$
= $$\left[\begin{array}{rrr} -44-4+48 & 22+2-24 & -2-2+24 \\ -88+8+80 & 44-4-40 & -4+4+40 \\ 0+16-16 & 0-8+8 & 0+8-8 \end{array}\right]$$
= $$\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$$
= O.

Question 16.
For the matrix A = $$\left[\begin{array}{rr} 5 & 2 \\ -3 & -1 \end{array}\right]$$, verify that adj A’ = (adj A)’.
Solution:
Given A = $$\left[\begin{array}{rr} 5 & 2 \\ -3 & -1 \end{array}\right]$$
The cofactors of R1 are ; – 1 ; 3
The cofactors of R2 are ; – 2 ; 5
∴ adj A = $$\left[\begin{array}{ll} -1 & 3 \\ -2 & 5 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rr} -1 & -2 \\ 3 & 5 \end{array}\right]$$
Thus, (adj A) = $$\left[\begin{array}{rr} -1 & -2 \\ 3 & 5 \end{array}\right]^{\mathrm{T}}$$
= $$\left[\begin{array}{ll} -1 & 3 \\ -2 & 5 \end{array}\right]$$ …………. (1)
∴ A’ = B = $$\left[\begin{array}{ll} 5 & -3 \\ 2 & -1 \end{array}\right]$$
The cofactors of R1 are ; – 1 ; – 2
The cofactors of R2 are ; 3 ; 5
∴ adj B = $$\left[\begin{array}{rr} -1 & -2 \\ 3 & 5 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{ll} -1 & 3 \\ -2 & 5 \end{array}\right]$$
⇒ adj A’ = $$\left[\begin{array}{ll} -1 & 3 \\ -2 & 5 \end{array}\right]$$ ………….(2)
From eqn. (1) and (2) ; we have

Question 17.
(i) Given A = $$\left[\begin{array}{rr} 2 & -3 \\ -4 & 7 \end{array}\right]$$, compute A-1 and show that 2A-1 = 9I – A.
(ii) If A = $$\left[\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right]$$, show that A – 3I = 2 (I + 3A-1)
Solution:
(i) Given A = $$\left[\begin{array}{rr} 2 & -3 \\ -4 & 7 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rr} 2 & -3 \\ -4 & 7 \end{array}\right|$$
= 14 – 12
= 2 ≠ 0
∴ A-1 exists
and A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A ……………(1)
The cofactors of R1 are ; 7 ; 4
Cofactors of R2 are ; 3 ; 2
∴ adj A = $$\left[\begin{array}{ll} 7 & 4 \\ 3 & 2 \end{array}\right]^{\mathrm{T}}$$
= $$\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]$$
∴ from (1) ;
A-1 = $$\frac{1}{2}\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]$$
⇒ 2A-1 = $$\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]$$ ………….(2)
∴ 9I – A = $$9\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{rr} 2 & -3 \\ -4 & 7 \end{array}\right]$$
= $$\left[\begin{array}{ll} 9 & 0 \\ 0 & 9 \end{array}\right]-\left[\begin{array}{rr} 2 & -3 \\ -4 & 7 \end{array}\right]$$
= $$\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]$$ ……………..(3)
From (2) and (3) ; we have
2A-1 = 9I – A

(ii) Given A = $$\left[\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right|$$
= 4 – 10
= – 6 ≠ 0
∴ A-1 exists and
A-1 = $$\frac{1}{|A|}$$ adj A ……………(1)
The cofactors of R1 are ; 1 ; – 2
The cofactors of R2 are ; – 5 ; 4
Thus from (1) ; we have

From (3) and (4) ; we have
L.H.S. = R.H.S.
Thus A – 3I = 2 (I + 3A-1).

Question 18.
Find the inverse of the matrix A = $$\left[\begin{array}{cc} a & b \\ c & \frac{1+b c}{a} \end{array}\right]$$ and show that aA-1 = (a2 + bc + 1) I – aA.
Solution:
Let A = $$\left[\begin{array}{cc} a & b \\ c & \frac{1+b c}{a} \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{cc} a & b \\ c & \frac{1+b c}{a} \end{array}\right|$$
= 1 + bc – bc
= 1 ≠ 0
∴ A-1 exists and
A-1 = $$\frac{1}{|A|}$$ adj A ……………(1)
The cofactors of R1 are ; $$\frac{1+bc}{a}$$ ; – c
The cofactors of R2 are ; – b ; a

From (1) and (2) ;
we have L.H.S. = R.H.S.
⇒ aA-1 = (a2 + bc + 1) I – aA.

Question 18 (old).
Given the matrix A = $$\left[\begin{array}{rrr} 1 & 0 & 2 \\ -2 & 1 & 0 \\ 0 & -1 & 2 \end{array}\right]$$, compute A-1.
Solution:
Given A = $$\left[\begin{array}{rrr} 1 & 0 & 2 \\ -2 & 1 & 0 \\ 0 & -1 & 2 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rrr} 1 & 0 & 2 \\ -2 & 1 & 0 \\ 0 & -1 & 2 \end{array}\right|$$ ;
expanding along R1
= 1 (2 – 0) + 0 + 2 (2 – 0)
= 2 + 4
= 6 ≠ 0
∴ A-1 exists and
A-1 = $$\frac{1}{|A|}$$ adj A ……………(1)
The cofactors of R1 are ;

Question 19.
If A = $$\left[\begin{array}{lll} 0 & 1 & 3 \\ 1 & 2 & x \\ 2 & 3 & 1 \end{array}\right]$$ and A-1 = $$\frac{1}{2}\left[\begin{array}{rrr} 1 & -8 & 5 \\ -1 & 6 & -3 \\ 1 & 2 y & 1 \end{array}\right]$$, find the value(s) of x and y.
Solution:
We know that A-1A = I
⇒ $$\frac{1}{2}\left[\begin{array}{rrr} 1 & -8 & 5 \\ -1 & 6 & -3 \\ 1 & 2 y & 1 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 3 \\ 1 & 2 & x \\ 2 & 3 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right]$$

⇒ $$\left[\begin{array}{ccc} 2 & 0 & 8-8 x \\ 0 & 2 & -6+6 x \\ 2 y+2 & 4+4 y & 4+2 x y \end{array}\right]=\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]$$

Thus their corresponding entries are equal.
∴ 8 – 8x = 0
⇒ x = 1
2y + 2 = 0
⇒ y = – 1
Now 4 + 2xy = 4 + 2 × 1 × (- 1) = 2, which is true.

Question 20.
If A = $$\left[\begin{array}{rr} 2 & -3 \\ 5 & 1 \end{array}\right]$$ and B = $$\left[\begin{array}{rr} -1 & 3 \\ 4 & 0 \end{array}\right]$$, verify that adj (AB) = (adj B) (adj A).
Solution:
Given A = $$\left[\begin{array}{rr} 2 & -3 \\ 5 & 1 \end{array}\right]$$
∴ The cofactors of R1 are ; 1 ; – 5
The cofactors of R2 are ; 3 ; 2
∴ adj A = $$\left[\begin{array}{rr} 1 & -5 \\ 3 & 2 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rr} 1 & 3 \\ -5 & 2 \end{array}\right]$$
Given B = $$\left[\begin{array}{rr} -1 & 3 \\ 4 & 0 \end{array}\right]$$
∴ The cofactors of R1 are ; 0 ; – 4
The cofactors of R2 are ; – 3 ; – 1
∴ adj B = $$\left[\begin{array}{rr} 0 & -4 \\ -3 & -1 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rr} 0 & -3 \\ -4 & -1 \end{array}\right]$$
Now AB = $$\left[\begin{array}{rr} 2 & -3 \\ 5 & 1 \end{array}\right]\left[\begin{array}{rr} -1 & 3 \\ 4 & 0 \end{array}\right]$$
= $$\left[\begin{array}{rr} -14 & 6 \\ -1 & 15 \end{array}\right]$$
∴ The cofactors of R1 are ; 15 ; 1
The cofactors of R2 are ; – 6 ; – 14
∴ adj (AB) = $$\left[\begin{array}{rr} 15 & 1 \\ -6 & -14 \end{array}\right]^{\prime}$$
= $$\left[\begin{array}{rr} 15 & -6 \\ 1 & -14 \end{array}\right]$$
Now, (adj B) (adj A) = $$\left[\begin{array}{rr} 0 & -3 \\ -4 & -1 \end{array}\right]\left[\begin{array}{rr} 1 & 3 \\ -5 & 2 \end{array}\right]$$
= $$\left[\begin{array}{rr} 15 & -6 \\ 1 & -14 \end{array}\right]$$

Question 21.
Find the inverse of each of the following matrices:
(i) $$\left[\begin{array}{rrr} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{array}\right]$$ (NCERT)
(ii) $$\left[\begin{array}{rrr} -1 & 1 & 2 \\ 3 & -1 & 1 \\ -1 & 3 & 4 \end{array}\right]$$
Also verify that A-1 A = I3.
Solution:
(i) |A| = $$=\left|\begin{array}{rrr} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{array}\right|$$
= 1 (- 3)
= – 3 ≠ 0
[Expanding along R1]
∴ A-1 exists and
A-1 = $$\frac{1}{|A|}$$ adj A ……………(1)
Cofactors of 1st row are,
$$\left|\begin{array}{rr} 3 & 0 \\ 2 & -1 \end{array}\right| ;-\left|\begin{array}{rr} 3 & 0 \\ 5 & -1 \end{array}\right| ;\left|\begin{array}{rr} 3 & 3 \\ 5 & 2 \end{array}\right|$$
i.e., – 3 ; 3 ; – 9

Cofactors of 2nd row are,
$$-\left|\begin{array}{rr} 0 & 0 \\ 2 & -1 \end{array}\right| ;\left|\begin{array}{rr} 1 & 0 \\ 5 & -1 \end{array}\right| ;-\left|\begin{array}{rr} 1 & 0 \\ 5 & 2 \end{array}\right|$$
i.e., 0 ; – 1 ; – 2

Cofactors of 3rd row are,
$$\left|\begin{array}{cc} 0 & 0 \\ 3 & 0 \end{array}\right| ;-\left|\begin{array}{cc} 1 & 0 \\ 3 & 0 \end{array}\right| ;\left|\begin{array}{cc} 1 & 0 \\ 3 & 3 \end{array}\right|$$
i.e., 0 ; 0 ; 3

∴ adj A = $$\left[\begin{array}{rrr} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{array}\right]$$
From (1) ;
A-1 = $$\left[\begin{array}{rrr} +1 & 0 & 0 \\ -1 & 1 / 3 & 0 \\ 3 & 2 / 3 & -1 \end{array}\right]$$
Now A-1 A = $$\left[\begin{array}{rrr} 1 & 0 & 0 \\ -1 & 1 / 3 & 0 \\ 3 & 2 / 3 & -1 \end{array}\right]\left[\begin{array}{rrr} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{array}\right]$$
= $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
= I3

(ii) Given A = $$=\left[\begin{array}{rrr} -1 & 1 & 2 \\ 3 & -1 & 1 \\ -1 & 3 & 4 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rrr} -1 & 1 & 2 \\ 3 & -1 & 1 \\ -1 & 3 & 4 \end{array}\right|$$ ;
expanding along R1
= – 1 (- 4 – 3) – 1 (12 + 1) + 2 (9 – 1)
= 7 – 13 + 16
= 10 ≠ 0
∴ A-1 exists and
A-1 = $$\frac{1}{|A|}$$ adj A ……………(1)

The cofactors of R1 are ;
$$\left|\begin{array}{rr} -1 & 1 \\ 3 & 4 \end{array}\right| ;-\left|\begin{array}{rr} 3 & 1 \\ -1 & 4 \end{array}\right| ;\left|\begin{array}{rr} 3 & -1 \\ -1 & 3 \end{array}\right|$$
i.e., – 7 ; – 13 ; 8

The cofactors of R2 are ;
$$-\left|\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right| ;+\left|\begin{array}{ll} -1 & 2 \\ -1 & 4 \end{array}\right| ;-\left|\begin{array}{ll} -1 & 1 \\ -1 & 3 \end{array}\right|$$
i.e., 2 ; – 2 ; 2

The cofactors of R3 are ;
$$\left|\begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array}\right| ;-\left|\begin{array}{rr} -1 & 2 \\ 3 & 1 \end{array}\right| ;\left|\begin{array}{rr} -1 & 1 \\ 3 & -1 \end{array}\right|$$
i.e., 3 ; 7 ; – 2

Question 22.
Given A = $$\left[\begin{array}{lll} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{array}\right]$$, B-1 = $$\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]$$. Compute (AB)-1.
Solution:
Given A = $$\left[\begin{array}{lll} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{array}\right]$$
and B-1 = $$\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]$$
Now |A| = $$\left|\begin{array}{lll} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{array}\right|$$
= 5 (- 1) + 4 (1)
= – 5 + 4
= – 1 ≠ 0
∴ A-1 exists
and A-1 = $$\frac{1}{|A|}$$ adj A ……………(1)

Cofactors of 1st row are,
$$\left|\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}\right| ;-\left|\begin{array}{cc} 2 & 2 \\ 1 & 1 \end{array}\right| ;\left|\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right|$$
i.e., – 1 ; 0 ; 1

Cofactors of 2nd row are,
$$-\left|\begin{array}{cc} 0 & 4 \\ 2 & 1 \end{array}\right| ;+\left|\begin{array}{cc} 5 & 4 \\ 1 & 1 \end{array}\right| ;-\left|\begin{array}{cc} 5 & 0 \\ 1 & 2 \end{array}\right|$$
i.e., 8 ; 1 ; – 10

Cofactors of 3rd row are,
$$\left|\begin{array}{ll} 0 & 4 \\ 3 & 2 \end{array}\right| ;-\left|\begin{array}{cc} 5 & 4 \\ 2 & 2 \end{array}\right| ;\left|\begin{array}{cc} 5 & 0 \\ 2 & 3 \end{array}\right|$$
i.e., – 12 ; – 2 ; 15

∴ adj A = $$\left[\begin{array}{rrr} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{array}\right]$$

∴ from (1) ;
A-1 = $$\left[\begin{array}{rrr} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{array}\right]$$

∴ (AB)-1 = B-1A-1
= $$\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]\left[\begin{array}{rrr} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{array}\right]$$
= $$\left[\begin{array}{lll} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{array}\right]$$

Question 23.
(i) Show that A = $$\left[\begin{array}{rr} 5 & 3 \\ -1 & -2 \end{array}\right]$$ satisfies the equation x2 – 3x – 7 = 0 and hence, find A-1.
(ii) If A = $$\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]$$ and I is the identity matrix of order 2, then show that A2 = 4A – 3I.
(iii) If A = , then find the value of λ so that A2 = λA – 2I. Hence, find A-1.
Solution:
(i) Given A = $$\left[\begin{array}{rr} 5 & 3 \\ -1 & -2 \end{array}\right]$$
satisfies the equation x2 – 3x – 7 = 0
if A2 – 3A – 7I = O
Here, A2 – 3A – 7I = $$\left[\begin{array}{rr} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{rr} 5 & 3 \\ -1 & -2 \end{array}\right]-3\left[\begin{array}{rr} 5 & 3 \\ -1 & -2 \end{array}\right]-7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{ll} 22 & 9 \\ -3 & 1 \end{array}\right]+\left[\begin{array}{rr} -15 & -9 \\ 3 & 6 \end{array}\right]+\left[\begin{array}{rr} -7 & 0 \\ 0 & -7 \end{array}\right]$$
= $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$
= $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$
= O

Hence A satisfies the equation x2 – 3x – 7 = 0
Now, A2 – 3A – 7I = 0
⇒ 7I = A2 – 3A
pre multiplying both sides by A-1 ; we get
7A-1 I = (A-1 A)A – 3 (A-1 A)
7A-1 = IA – 3I
= A – 3I
⇒ 7A-1 = $$\left[\begin{array}{rr} 5 & 3 \\ -1 & -2 \end{array}\right]-\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]$$
= $$\left[\begin{array}{rr} 2 & 3 \\ -1 & -5 \end{array}\right]$$
⇒ A-1 = $$\frac{1}{7}\left[\begin{array}{rr} 2 & 3 \\ -1 & -5 \end{array}\right]$$

(ii) Given A = $$\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]$$
∴ |A| = 4 – 1
= 3 ≠ 0
∴ A-1 exists.
∴ A2 = A . A
= $$\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]$$
= $$\left[\begin{array}{rr} 5 & -4 \\ -4 & 5 \end{array}\right]$$

and 4A – 3I = $$4\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rr} 8 & -4 \\ -4 & 8 \end{array}\right]-\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]$$
= $$\left[\begin{array}{rr} 5 & -4 \\ -4 & 5 \end{array}\right]$$
∴ A2 = 4A – 3I ………….(1)
pre-multiplying eqn. (1) by A-1 ; we have
(A-1 A) A = 4 (A-1 A) – 3 A-1 I
⇒ IA = 4I – 3A-1
[∵ A-1 A = I = AA-1]
⇒ A = 4I – 3A-1
3A-1 = 4I – A
⇒ 3A-1 = $$4\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]$$
= $$\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right]-\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]$$
⇒ 3A-1 = $$\left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right]$$
⇒ A-1 = $$\frac{1}{3}\left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right]$$

(iii) Given A2 = λA – 2I ………………(1)
⇒ $$\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]=\lambda\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
⇒ $$\left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right]=\left[\begin{array}{cc} 3 \lambda-2 & -2 \lambda \\ 4 \lambda & -2 \lambda-2 \end{array}\right]$$
∴ 1 = 3λ – 2
⇒ 3λ = 3
⇒ λ = 1
Thus eqn. (1) becomes ;
A2 = A – 2I

Here |A| = $$\left|\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right|$$
= – 6 + 8
= 2 ≠ 0
∴ A-1 exists
∴ (A-1A) A = A-1A – 2A-1I
⇒ IA = I – 2A-1
⇒ A = I – 2A-1
⇒ 2A-1 = I – A
= $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]$$
= $$\left[\begin{array}{ll} -2 & 2 \\ -4 & 3 \end{array}\right]$$
⇒ A-1 = $$\frac{1}{2}\left[\begin{array}{ll} -2 & 2 \\ -4 & 3 \end{array}\right]$$

Question 24.
(i) A = $$\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]$$, prove that A2 – 4A – 5I = O. Hence find A-1.
(ii) Find A-1 if A = $$\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$$. Also show that A-1 = $$\frac{A^2-3 I}{2}$$. (NCERT Exampler)
Solution:
(i) Given A = $$\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]$$
Here |A| = 1 (1 – 4) – 2 (2 – 4) + 2 (4 – 2)
= – 3 + 4 + 4
= 5 ≠ 0
∴ A-1 exists
and A-1 = $$\frac{1}{|A|}$$ adj A ………..(1)

(ii) Given A = $$\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right|$$
= 0 – 1 (0 – 1) + 1 (1 – 0)
= 2 ≠ 0
[Expanding along R1]
∴ A-1 exists
and A-1 = $$\frac{1}{|A|}$$ adj A ………..(1)
The cofactors of R1 are :
$$\left|\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right| ;-\left|\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right| ;\left|\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right|$$
i.e., – 1 ; 1 ; 1

The cofactors of R2 are :
$$-\left|\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array}\right| ;\left|\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right| ;-\left|\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\right|$$
i.e., 1 ; – 1 ; 1

The cofactors of R3 are :
$$\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right| ;-\left|\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\right| ;\left|\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right|$$
i.e., 1 ; 1 ; – 1

∴ from eqn. (1) ; we have

Question 25.
(i) If A = $$\frac{1}{9}\left[\begin{array}{rrr} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{array}\right]$$, prove that A-1 = A’.
(ii) If A = $$\left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right]$$, prove that A-1 = A2.
(iii) If F(x) = $$\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]$$, prove that (F(x))-1 = F(- x).
Solution:
(i) Given A = $$\frac{1}{9}\left[\begin{array}{rrr} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{array}\right]_{3 \times 3}$$
= $$\left[\begin{array}{rrr} -8 / 9 & 1 / 9 & 4 / 9 \\ 4 / 9 & 4 / 9 & 7 / 9 \\ 1 / 9 & -8 / 9 & 4 / 9 \end{array}\right]$$
∴ |A| = $$\left(\frac{1}{9}\right)^3\left|\begin{array}{rrr} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{array}\right|$$
= $$\frac{1}{729}$$ [- 8 (16 + 56) – 1 (16 – 7) + 4 (- 32 – 4)]
= $$\frac{1}{729}$$ [- 576 – 9 – 144]
= – $$\frac{729}{729}$$
= – 1 ≠ 0
∴ A-1 exists
and A-1 = $$\frac{1}{|A|}$$ adj A ………..(1)

Now, AT = $$\frac{1}{9}\left[\begin{array}{rrr} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{array}\right]$$ ………….(3)
∴ A-1 = AT [using (2) and (3)]

(ii) Here |A| = $$\left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right]$$ ;
Expanding along C3
C3 = 1 (0 + 1)
= 1 ≠ 0
∴ A-1 exists
and A-1 = $$\frac{1}{|A|}$$ adj A ………..(1)
The cofactors R1 are ;
$$\left|\begin{array}{cc} -1 & 0 \\ 0 & 0 \end{array}\right| ;-\left|\begin{array}{cc} 2 & 0 \\ 1 & 0 \end{array}\right| ;\left|\begin{array}{cc} 2 & -1 \\ 1 & 0 \end{array}\right|$$
i.e., 0 : 0 : 1

The cofactors of R2 are ;
$$-\left|\begin{array}{cc} -1 & 1 \\ 0 & 0 \end{array}\right| ;\left|\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right| ;-\left|\begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array}\right|$$
i.e., 0 ; – 1 ; – 1

The cofactors R3 are ;
$$\left|\begin{array}{ll} -1 & 1 \\ -1 & 0 \end{array}\right| ;-\left|\begin{array}{ll} 1 & 1 \\ 2 & 0 \end{array}\right| ;\left|\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right|$$
i.e., 1 ; 2 ; 1

∴ adj A-1 = $$\left[\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & -1 \\ 1 & 2 & 1 \end{array}\right]$$
= $$\left[\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right]$$
∴ A = $$\left[\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right]$$ …………..(2)
and A2 = A . A
= $$\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right]\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right]$$
= $$\left[\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right]$$ …………..(3)
From eqn. (2) and (3) ;
We have
A2 = A-1.

(iii) Given F(x) = $$\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]$$

(i) |F(x)| = $$\left|\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right|$$ ;
expanding along C3
= cos2 x + sin2 x
= 1 ≠ 0
∴ [F(x)]-1 exists
and [F(x)]-1 = $$\frac{1}{|\mathrm{~F}(x)|}$$ adj (F(x)) …………….(1)
Cofactors of R1 are ;
cos x, – sin x , 0
Cofactors of R2 are ;
sin x ; cos x ; 0
Cofactors of R3 are ;
0 ; 0 ; 1

∴ adj (F(x)) = $$\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]$$
∴ From (1) ;
we have
[F(x)]-1 = $$\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]$$
and F(- x) = $$=\left[\begin{array}{ccc} \cos (-x) & -\sin (-x) & 0 \\ \sin (-x) & \cos (-x) & 0 \\ 0 & 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]$$
∴[F(x)]-1 = F (- x).

Question 26.
Show that for the matrix A = $$\left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]$$, A3 – 6A2 + 5A + 11I = O. Hence, find A-1.
Solution:
Here |A| = $$\left|\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right|$$ ;
Expanding along R1
= 1 (6 – 3) – 1 (3 + 6) + 1 (- 1 – 4)
= 3 – 9 – 5
= – 11 ≠ 0
∴ A-1 exists

⇒ 11I = – A3 + 6A2 – 5A
Some A-1 exists, pre-multiplying eqn. (1) by A-1;
we have
11A-1I = – (A-1 A) A2 + 6 (A-1 A) A – 5 (A-1 A)
⇒ 11A-1 = – IA2 + 61A – 5I
= – A2 + 6A – 5I
= $$-\left[\begin{array}{rrr} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]+6\left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]-5\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rrr} -4+6-5 & -2+6+0 & -1+6+0 \\ 3+6+0 & -8+12-5 & 14-18+0 \\ -7+12+0 & 3-6+0 & -14+18-6 \end{array}\right]$$
= $$\left[\begin{array}{rrr} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{array}\right]$$
∴ A-1 = $$\frac{1}{11}\left[\begin{array}{rrr} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{array}\right]$$.

Question 27.
(i) Find a 2 × 2 matrix B such that $$B\left[\begin{array}{rr} 1 & -2 \\ 1 & 4 \end{array}\right]=\left[\begin{array}{ll} 6 & 0 \\ 0 & 6 \end{array}\right]$$.
(ii) Solve the matrix equation $$\left[\begin{array}{ll} 5 & 4 \\ 1 & 1 \end{array}\right] A=\left[\begin{array}{rr} 1 & -2 \\ 1 & 3 \end{array}\right]$$.
Solution:
(i) Let A = $$\left[\begin{array}{rr} 1 & -2 \\ 1 & 4 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{rr} 1 & -2 \\ 1 & 4 \end{array}\right|$$
= 4 + 2
= 6
∴ A-1 exists
and A-1 = $$\frac{1}{|A|}$$ adj A ………..(1)
The cofactors of R1 are ; 4 ; – 1
The cofactors of R2 are ; 2 ; 1
∴ adj A = $$\left[\begin{array}{rr} 4 & -1 \\ 2 & 1 \end{array}\right]^{]}$$
= $$\left[\begin{array}{rr} 4 & 2 \\ -1 & 1 \end{array}\right]$$
∴ from (1) ;
A-1 = $$\left[\begin{array}{rr} 4 & 2 \\ -1 & 1 \end{array}\right]$$
∴ Given eqn. becomes BA = 6 $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
= 6I2 ………..(1)
post multiplying eqn. (1) by A-1 ; we have
BAA-1 = 6I2A-1
⇒ B (AA-1) = 6A-1
⇒ BI = 6A-1
⇒ B = $$\frac{6}{6}\left[\begin{array}{rr} 4 & 2 \\ -1 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rr} 4 & 2 \\ -1 & 1 \end{array}\right]$$

(ii) Given matrix eqn. be $$\left[\begin{array}{ll} 5 & 4 \\ 1 & 1 \end{array}\right] X=\left[\begin{array}{rr} 1 & -2 \\ 1 & 3 \end{array}\right]$$
i.e., AX = B …………..(1)
where A = $$\left[\begin{array}{ll} 5 & 4 \\ 1 & 1 \end{array}\right]$$
and B = $$\left[\begin{array}{rr} 1 & -2 \\ 1 & 3 \end{array}\right]$$
Here, |A| = 5 – 4
= 1 ≠ 0
∴ A-1 exists
∴ from (1) ;
A-1 (AX) = A-1B
(A-1 A) X = A-1 B
IX = A-1B
⇒ X = A-1 B …………….(2)
The cofactors of R1 are ; 1 ; – 1
cofactors of R2 are ; – 4 ; 5
∴ adj A = $$\left[\begin{array}{rr} 1 & -1 \\ -4 & 5 \end{array}\right]^{\mathrm{T}}$$
= $$\left[\begin{array}{rr} +1 & -4 \\ -1 & 5 \end{array}\right]$$
∴ A-1 = $$\frac{1}{|\mathrm{~A}|}$$ adj A
= $$\left[\begin{array}{rr} 1 & -4 \\ -1 & 5 \end{array}\right]$$
Thus from (2) ; we have
X = $$\left[\begin{array}{rr} 1 & -4 \\ -1 & 5 \end{array}\right]\left[\begin{array}{rr} 1 & -2 \\ 1 & 3 \end{array}\right]$$
= $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

Question 28.
(i) Find the matrix A satisfying the matrix equation $$\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right] A\left[\begin{array}{ll} 4 & 7 \\ 3 & 5 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$.
(ii) Find the matrix P satisfying the matrix equation $$\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] P\left[\begin{array}{rr} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right]$$.
Solution:
(i) Let B = $$\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right]$$
and C = $$\left[\begin{array}{ll} 4 & 7 \\ 3 & 5 \end{array}\right]$$
∴ |B| = $$\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|$$
= 3 – 4
= – 1 ≠ 0
∴ B-1 exists
and B-1 = $$\frac{1}{|B|}$$ adj B ……………(1)
The cofactors of R1 are ; 3 ; – 2
The cofactors of R2 are ; – 2 ; 1
∴ adj B = $$\left[\begin{array}{rr} 3 & -2 \\ -2 & 1 \end{array}\right]^{\mathrm{T}}$$
= $$\left[\begin{array}{rr} 3 & -2 \\ -2 & 1 \end{array}\right]$$
∴ from (1) ;
B-1 = $$\frac{1}{-1}\left[\begin{array}{rr} 3 & -2 \\ -2 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rr} -3 & 2 \\ 2 & -1 \end{array}\right]$$ ……………(2)
Now |C| = $$\mid \begin{array}{ll} 4 & 7 \\ 3 & 5 \end{array}$$
= 20 – 21
= – 1 ≠ 0
∴ C-1 exists
and C-1 = $$\frac{1}{|C|}$$ adj C ……………(1)
The cofactors of R1 are ; 5 ; – 3
The cofactors of R2 are ; – 7 ; 4
∴ adj C = $$=\left[\begin{array}{rr} 5 & -3 \\ -7 & 4 \end{array}\right]^{\mathrm{T}}$$
= $$\left[\begin{array}{rr} 5 & -7 \\ -3 & 4 \end{array}\right]$$
∴ from (3) ;
C-1 = $$\frac{1}{-1}\left[\begin{array}{rr} 5 & -7 \\ -3 & 4 \end{array}\right]$$
= $$\left[\begin{array}{rr} -5 & 7 \\ 3 & -4 \end{array}\right]$$ …………(4)
Thus, given eqn. becomes ;
BAC = I2 …………..(5)
pre multiplying both sides of eqn. (5) by B-1; we have
(B-1 B) AC = B-1 I
⇒ IAC = B-1
⇒ AC = B-1 …………(6)
Now post multiplying both sides of eqn. (6) by C-1 ; we have
A (CC-1) = B-1C-1
⇒ AI = B-1C-1
⇒ A = B-1C-1
= $$\left[\begin{array}{rr} -3 & 2 \\ 2 & -1 \end{array}\right]\left[\begin{array}{rr} -5 & 7 \\ 3 & -4 \end{array}\right]$$
= $$\left[\begin{array}{rr} 21 & -29 \\ -13 & 18 \end{array}\right]$$

(ii) Let B = $$\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]$$
and C = $$\left[\begin{array}{rr} -3 & 2 \\ 5 & -3 \end{array}\right]$$
∴ |B| = $$\left|\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right|$$
= 4 – 3
= 1 ≠ 0
∴ B-1 exists
and B-1 = $$\frac{1}{|B|}$$ adj B ……………(1)
The cofactors of R1 are ; 2 ; – 3
The cofactors of R2 are ; – 1 ; 2
∴ B-1 = $$\left[\begin{array}{rr} 2 & -3 \\ -1 & 2 \end{array}\right]^{\mathrm{T}}$$
= $$\left[\begin{array}{rr} 2 & -1 \\ -3 & 2 \end{array}\right]$$ [using (1)]
Now |C| = $$\left|\begin{array}{rr} -3 & 2 \\ 5 & -3 \end{array}\right|$$
= 9 – 10
= – 1 ≠ 0
∴ C-1 exists
and C-1 = $$\frac{1}{|C|}$$ adj C ……………(1)
The cofactors of R1 are ; – 3 ; – 5
The cofactors of R2 are ; – 2 ; – 3
∴adj C = $$\left[\begin{array}{ll} -3 & -2 \\ -5 & -3 \end{array}\right]$$
∴ from (2) ;
C-1 = $$-\left[\begin{array}{ll} -3 & -2 \\ -5 & -3 \end{array}\right]$$
= $$\left[\begin{array}{ll} 3 & 2 \\ 5 & 3 \end{array}\right]$$
Thus given eqn. becomes BPC
= $$\left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right]$$ …………….(3)
pre-multiplying both sides of eqn. (4) by C-1; we get
P(CC-1) = B-1 $$\left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right]$$ C-1
PI = B-1 $$\left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right]$$ C-1
⇒ P = $$\left[\begin{array}{rr} 2 & -1 \\ -3 & 2 \end{array}\right]\left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right]\left[\begin{array}{ll} 3 & 2 \\ 5 & 3 \end{array}\right]$$
⇒ P = $$\left[\begin{array}{rr} 2 & -1 \\ -3 & 2 \end{array}\right]\left[\begin{array}{rr} 13 & 8 \\ 1 & 1 \end{array}\right]$$
= $$\left[\begin{array}{rr} 25 & 15 \\ -37 & -22 \end{array}\right]$$