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

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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,
|adj A| = |A|^n-1
Given A be a matrix of order 2
∴ |adj A| = |a| = 10

(ii) |adj A| = |A|^3-1
= |A|²
= 5²
= 25

(iii) Given A be a non-singular matrix of order 3
∴ |A| ≠ 0
and n = 3
Also, |adj A| = 64
⇒ |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.
∴ |3 adj A| = 3³ |adj A|
= 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
Thus, |adj A| = |A|²
= 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.
Then |adj A| = |A|^n-1
Since A be a square matrix of order 3
∴ n = 3
∴ |adj A| = |A|^n-1
= (- 5)^3-1 = 25
[∵ |A| = – 5]

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

Question 6.
Without computing adj A, find |adj A| if
(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
∴ |adj A| = |A|^n-1
= 7^2-1
= 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
∴ |adj A| = |A|^n-1
= (- 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}
= 5^3-1
= 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 adj A| = |A| |adj A|
= |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
Now |A adj A| = |A| |adj A|
[∵ |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
adj A adj (adj A) = A |adj A| I
⇒ (A adj A) adj (adj A) = A |adj A| I
⇒ |A| I adj (adj A) = A |adj A|
[∵ A . I = A]
⇒ adj (adj A) = \(\frac{\mathrm{A}|\mathrm{A}|^{n-1}}{|\mathrm{~A}|}\)
[∵ |A| ≠ 0]
⇒ adj (adj A) = |A|^n-2 A …………..(2)
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 = I_n = A^-1A
⇒ |AA^-1| = |I_n|
[Taking determinants on bothsides]
⇒ |A| |A^-1| = 1
[∵ |AB| = |A| |B| and |I_n| = 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
A (adj A) = |A| I₂ = (adj A) A.
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
A adj A = |A| I₂ = (adj A) A

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)
and (adj A) A
= \(\left[\begin{array}{rrr}
3 & 1 & -11 \\
-12 & 5 & -1 \\
6 & 2 & 5
\end{array}\right]\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 3 & 5 \\
-2 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
27 & 0 & 0 \\
0 & 27 & 0 \\
0 & 0 & 27
\end{array}\right]\)
= 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
= (adj A) A

(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 (adj A) = (adj A) A
= |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
adj A’ = (adj A)’.

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) = \(\left[\begin{array}{rr}
0 & -3 \\
-4 & -1
\end{array}\right]\left[\begin{array}{rr}
1 & 3 \\
-5 & 2
\end{array}\right]\)
= \(\left[\begin{array}{rr}
15 & -6 \\
1 & -14
\end{array}\right]\)
= adj (AB)

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, A^T = \(\frac{1}{9}\left[\begin{array}{rrr}
-8 & 4 & 1 \\
1 & 4 & -8 \\
4 & 7 & 4
\end{array}\right]\) ………….(3)
∴ A^-1 = A^T [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 = \(\left[\begin{array}{rr}
2 & -1 \\
-3 & 2
\end{array}\right]\left[\begin{array}{rr}
13 & 8 \\
1 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
25 & 15 \\
-37 & -22
\end{array}\right]\)