OP Malhotra Class 12 Maths Solutions Chapter 6 Matrices Ex 6(h)

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S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(h)

Question 1.
Show that A is a singular matrix if
(i) A = \(\left[\begin{array}{ll}
3 & 6 \\
2 & 4
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{rrr}
1 & 1 & 3 \\
2 & 2 & 6 \\
2 & -3 & 1
\end{array}\right]\)
Solution:
(i) Given A = \(\left[\begin{array}{ll}
3 & 6 \\
2 & 4
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{ll}
3 & 6 \\
2 & 4
\end{array}\right|\) = 12 – 12 = 0
∴ A be a singular matrix

(ii) Given A = \(\left[\begin{array}{rrr}
1 & 1 & 3 \\
2 & 2 & 6 \\
2 & -3 & 1
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{rrr}
1 & 1 & 3 \\
2 & 2 & 6 \\
2 & -3 & 1
\end{array}\right]\)
Expanding along R₁
= 1(2 + 18) – 1(2 – 12) + 3(- 6 – 4)
= 20 + 10 – 30 = 0
Thus A be a singular matrix.

Question 2.
Find x if \(\left[\begin{array}{rrr}
8 & -6 & 2 \\
-6 & 7 & -4 \\
2 & -4 & x
\end{array}\right]\) is a singular matrix.
Solution:
Let A = \(\left[\begin{array}{rrr}
8 & -6 & 2 \\
-6 & 7 & -4 \\
2 & -4 & x
\end{array}\right]\)
since A be a singular matrix ∴ |A| = 0
\(\left[\begin{array}{rrr}
8 & -6 & 2 \\
-6 & 7 & -4 \\
2 & -4 & x
\end{array}\right]\) = 0; Expanding along R₁
⇒ 8(7x – 16) + 6(-6x + 8) + 2(24 – 14) = 0
⇒ 56x – 128 – 36x + 48 + 48 – 28 = 0
⇒ 20x – 60 = 0
⇒ x = 3

Question 3.
For each of the following matrices, determine wheter the inverse exists. If it exists, find it.
(i) \(\left[\begin{array}{ll}
4 & 5 \\
2 & 3
\end{array}\right]\)
(ii) \(\left[\begin{array}{rr}
2 & -6 \\
-1 & 3
\end{array}\right]\)
(iii) \(\left[\begin{array}{ll}
2 & 6 \\
1 & 1
\end{array}\right]\)
(iv) \(\left[\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\)
Solution:
(i) Let A = \(\left[\begin{array}{ll}
4 & 5 \\
2 & 3
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
4 & 5 \\
2 & 3
\end{array}\right|\) = 12 – 10 = 2 ≠ 0
Thus A^-1 exists ∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
∴ The cofactors of R₁ are ; 3 ; – 2
The cofactors of R₂ are ; – 5 ; 4
∴ adj A = \(\left[\begin{array}{cc}
3 & -2 \\
-5 & 4
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
3 & -5 \\
-2 & 4
\end{array}\right]\)
Thus A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A = \(\frac{1}{2}\left[\begin{array}{cc}
3 & -5 \\
-2 & 4
\end{array}\right]\)

(ii) Let A = \(\left[\begin{array}{rr}
2 & -6 \\
-1 & 3
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{rr}
2 & -6 \\
-1 & 3
\end{array}\right]\)
Thus A is a singular matrix
∴ A^-1 does not exists.

(iii) Let A = \(\left[\begin{array}{ll}
2 & 6 \\
1 & 1
\end{array}\right]\);
Here |A| = \(\left|\begin{array}{ll}
2 & 6 \\
1 & 1
\end{array}\right|\) = 2 – 6 = – 4 ≠ 0
Thus A^-1 exists
∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A … (i)
The cofactor of R₁ are ; 1 ; – 1
The cofactor of R₂ are ; – 6; 2
∴ adj A = \(\left[\begin{array}{cc}
1 & -1 \\
-6 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
1 & -6 \\
-1 & 2
\end{array}\right]\)
Thus from (1) ; we have
A^-1 = \(\frac{1}{-4}\left[\begin{array}{cc}
1 & -6 \\
-1 & 2
\end{array}\right]=\frac{1}{4}\left[\begin{array}{cc}
-1 & 6 \\
1 & -2
\end{array}\right]\)

(iv) Let A = \(\left[\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\)

Question 4.
(a) If A = \(\left[\begin{array}{ll}
2 & x \\
4 & 2
\end{array}\right]\), x ≠ 1, calculate
(i) A²
(ii) (A²)^-1
(b) Find the sum of \(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\) multiplicative inverse.
Solution:
(a) Given A = \(\left[\begin{array}{ll}
2 & x \\
4 & 2
\end{array}\right]\)
∴ A² = A.A = \(\left[\begin{array}{ll}
2 & x \\
4 & 2
\end{array}\right]\left[\begin{array}{ll}
2 & x \\
4 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4+4 x & 2 x+2 x \\
8+8 & 4 x+4
\end{array}\right]\)
∴ A² = \(\left[\begin{array}{cc}
4+4 x & 4 x \\
16 & 4 x+4
\end{array}\right]\)
∴ |B| = \(\left|\begin{array}{cc}
4+4 x & 4 x \\
16 & 4 x+4
\end{array}\right|\)
= (4 + 4x)² – 64x = (4 – 4x)²
= 16(1 – x)²
= 16(x – 1)² ≠ 0 [ ∵ x ≠ 1]
The cofactor of R₁ are ; + 4x + 4 ; – 16
The cofactor of R₂ are ; – 4x ; 4x + 4

The cofactor of R₁ are ; 7 ; -5
The cofactor of R₂ are ; – 3 ; 2

Question 5.
(i) Find a matrix X for which
X\(\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\). Also find the inverse of \(\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\).
(ii) Find 2 x 2 matrix M such that
\(\left[\begin{array}{rr}
-5 & 2 \\
15 & -7
\end{array}\right] M=\left[\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right]\).
Solution:
Given X\(\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\)
∴ X = \(\left[\begin{array}{rr}
4 & 1 \\
2 & 3
\end{array}\right]\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]^{-1}\) … (1)
Let A = \(\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]\)
∴\(|\mathrm{A}|=\left|\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right|\)
= – 3 – 2 = – 5 ≠ 0
Thus A^-1 exists
∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactor of R₁ are ; – 1 ; – 1
The cofactor of R₂ are ; – 2 ; 3

Thus B^-1 exists & B^-1 = \(\frac{1}{|\mathrm{~B}|}\) adj B
The cofactor of R₁ are ; 3 ; – 2
The cofactor of R₂ are ; – 1 ; 4

The cofactor of R₁ are ; -7 ; -15
The cofactor of R₂ are ; -2 ; -15

Question 6.
Find the inverse of the following matrices and verify your result.
(i) \(\left[\begin{array}{rr}
2 & 5 \\
-3 & 1
\end{array}\right]\)
(ii) \(\left[\begin{array}{rr}
-4 & 3 \\
5 & -5
\end{array}\right]\)
Solution:
(i) Let A = \(\left[\begin{array}{rr}
2 & 5 \\
-3 & 1
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{rr}
2 & 5 \\
-3 & 1
\end{array}\right|\)
Thus, A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactor of R₁ are ; 1 ; 3
The cofactor of R₂ are ; – 5 ; 2

Question 7.
(a) Find the inverse of each of the following matrices and verify your result.
(i) \(\left[\begin{array}{rrr}
1 & 4 & 3 \\
2 & 5 & 4 \\
1 & -3 & -2
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccr}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin a & -\cos \alpha
\end{array}\right]r\)
(b) Verify that AA^-1 = A^-1A = I, if
\(\left[\begin{array}{rrr}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\)
(c) If A = \(\left[\begin{array}{rrr}
3 & 0 & 2 \\
1 & 5 & 9 \\
-6 & 4 & 7
\end{array}\right]\) and AB = BA = I, find B.
Solution:

(ii) Let A = \(\left[\begin{array}{ccr}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin a & -\cos \alpha
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin \alpha & -\cos \alpha
\end{array}\right|\);
Expanding along R₁ = – cos²α – sin²α = – 1 ≠ 0
∴ A^-1 exists
Thus, A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A … (1)
The cofactor of R₁ are ;

Expanding along C₁
= 1(3 – 0) +1(2 – 4) = 3 – 2 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactor of R₁ are ;
\(\left|\begin{array}{rr}
3 & 0 \\
-2 & -1
\end{array}\right| ;-\left|\begin{array}{rr}
-1 & 0 \\
0 & 1
\end{array}\right| ;\left|\begin{array}{rr}
-1 & 3 \\
0 & -2
\end{array}\right|\)
i.e. 3 ; 1 ; 2
The cofactor of R₂ are ;
\(-\left|\begin{array}{rr}
2 & -2 \\
-2 & 1
\end{array}\right| ;\left|\begin{array}{rr}
1 & -2 \\
0 & 1
\end{array}\right| ;-\left|\begin{array}{rr}
1 & 2 \\
0 & -2
\end{array}\right|\)
i.e. 2 ; 1 ; 2
The cofactor of R₃ are ;

(c) Let A = \(\left[\begin{array}{rrr}
3 & 0 & 2 \\
1 & 5 & 9 \\
-6 & 4 & 7
\end{array}\right]\)
since AB = BA = I,
Also, AA’ = A^-1A = I
∴ B = A^-1
Now, |A| = \(\left|\begin{array}{rrr}
3 & 0 & 2 \\
1 & 5 & 9 \\
-6 & 4 & 7
\end{array}\right|\)
Expanding along R₁
= 3(35 – 36) + 0 + 2(4 + 30)
= – 3 + 68 = 65 ≠ 0
∴ A^-1 exists
& A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactor of R₁ are ;
\(\left|\begin{array}{ll}
5 & 9 \\
4 & 7
\end{array}\right| ;-\left|\begin{array}{cc}
1 & 9 \\
-6 & 7
\end{array}\right| ;\left|\begin{array}{cc}
1 & 5 \\
-6 & 4
\end{array}\right|\)
The cofactor of R₂ are
\(-\left|\begin{array}{cc}
0 & 2 \\
4 & 7
\end{array}\right| ;\left|\begin{array}{cc}
3 & 2 \\
-6 & 7
\end{array}\right| ;-\left|\begin{array}{cc}
3 & 0 \\
-6 & 4
\end{array}\right|\)
i.e. 8 ; 33 ; – 12
The cofactor of R₃ are
\(\left|\begin{array}{ll}
0 & 2 \\
5 & 9
\end{array}\right| ;-\left|\begin{array}{ll}
3 & 2 \\
1 & 9
\end{array}\right| ;\left|\begin{array}{ll}
3 & 0 \\
1 & 5
\end{array}\right|\)
i.e. -10 ; -25 ; 15
∴ adj A = \(\left[\begin{array}{ccc}
-1 & -61 & 34 \\
8 & 33 & -12 \\
-10 & -25 & 15
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-1 & 8 & -10 \\
-61 & 33 & -25 \\
34 & -12 & 15
\end{array}\right]\)
Thus, B = A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
= \(\frac{1}{65}\left[\begin{array}{ccc}
-1 & 8 & -10 \\
-61 & 33 & -25 \\
34 & -12 & 15
\end{array}\right]\)

Question 8.
Verify that (AB)^-1 = B^-1A^-1 for the matrices A and B where
(i) A = \(\left[\begin{array}{ll}
2 & 1 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
4 & 5 \\
3 & 4
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
4 & 6 \\
3 & 2
\end{array}\right]\)
Solution:
(i) Given A = \(\left[\begin{array}{ll}
2 & 1 \\
5 & 3
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{ll}
2 & 1 \\
5 & 3
\end{array}\right]\) = 6 – 5 = 1 ≠ 0
Thus, A^-1 exists and A^-1 =\(\frac{1}{|\mathrm{~A}|}\) adj A …(1)
The cofactor of R₁ are ; 3 ; – 5
The cofactor of R₂ are ; – 1 ; 2
∴ adj A = \(\left[\begin{array}{cc}
3 & -5 \\
-1 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
3 & -1 \\
-5 & 2
\end{array}\right]\)
Thus, from (1);
A^-1 = \(\frac{1}{1}\left[\begin{array}{cc}
3 & -1 \\
-5 & 2
\end{array}\right]=\left[\begin{array}{cc}
3 & -1 \\
-5 & 2
\end{array}\right]\)
& B = \(\left[\begin{array}{ll}
4 & 5 \\
3 & 4
\end{array}\right]\)
∴ |B| = \(\left|\begin{array}{ll}
4 & 5 \\
3 & 4
\end{array}\right|\)
= 16 – 15 = 1 ≠ 0
Thus, B^-1 exists and B^-1 =\(\frac{1}{|\mathrm{~A}|}\) adj A …(2)
The cofactor of R₁ are ; 4 ; – 3
The cofactor of R₂ are ; – 5 ; 4

The cofactor of R₁ are ; 37 ; – 29
The cofactor of R₂ are ; – 14 ; 11
∴ adj (AB)
= \(\left[\begin{array}{cc}
37 & -29 \\
-14 & 11
\end{array}\right]=\left[\begin{array}{cc}
37 & -14 \\
-29 & 11
\end{array}\right]\)
Thus (AB)^-1 = \(\frac{1}{|\mathrm{AB}|}\)adj (AB)
= \(\left[\begin{array}{cc}
37 & -14 \\
-29 & 11
\end{array}\right]\) … (2)
[ ∵ |AB| = \(\left|\begin{array}{ll}
11 & 14 \\
29 & 37
\end{array}\right|=407-406=1 \neq 0\)
∴ (AB)^-1 exists]
from (1) & (2); we have
B^-1A^-1 = (AB)^-1

(ii) Given A = \(\left[\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right|=15-14=1 \neq 0\)
∴ A^-1 exists & A^-1 = \(\frac { 1 }{ |A| }\)adj A
The cofactor of R₁ are ; 5 ; – 7
The cofactor of R₂ are ; – 2 ; 3

The cofactor of R₁ are ; 2 ; – 3
The cofactor of R₂ are ; – 6 ; 4

The cofactor of R₁ are ; 52 ; – 43
The cofactor of R₂ are ; – 22 ; 18
∴ adj (AB)
= \(\left[\begin{array}{cc}
52 & -43 \\
-22 & 18
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
52 & -22 \\
-43 & 18
\end{array}\right]\)
Thus, (AB)^-1 = \(\frac{1}{|\mathrm{AB}|}\)adj (AB)
= \(\frac{-1}{10}\left[\begin{array}{cc}
52 & -22 \\
-43 & 18
\end{array}\right]=\frac{1}{10}\left[\begin{array}{cc}
-52 & 22 \\
43 & -18
\end{array}\right]\)
from (1) & (2) ; we have
(AB)^-1 = B^-1A^-1

Question 9.
Given A = \(\left[\begin{array}{rr}
2 & -3 \\
-4 & 7
\end{array}\right]\), compute A^-1 and show that 2A^-1 = 9I – A.
Solution:
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 & A^-1 = \(\frac { 1 }{ |A| }\)adj A … (1)
The cofactor of R₁ are ; 7 ; 4
The cofactor of R₂ are ; 3 ; 2

Question 10.
(i) If A = \(\left[\begin{array}{rrr}
2 & 4 & -1 \\
-1 & 0 & 2
\end{array}\right]\), B = \(\left[\begin{array}{rr}
3 & 4 \\
-1 & 2 \\
2 & 1
\end{array}\right]\), find (AB)^-1.
(ii) If A = \(\left[\begin{array}{lll}
5 & 0 & 4 \\
2 & 3 & 2 \\
1 & 2 & 1
\end{array}\right]\), B^-1 = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\), find (AB)^-1.
(iii) If A^-1 = \(\frac{1}{10}\left[\begin{array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\).
Solution:

(iii) Given A^-1 = \(\frac{1}{10}\left[\begin{array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right] \quad \& B=\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\)

Question 11.
If A = \(\left[\begin{array}{rr}
2 & -3 \\
4 & 6
\end{array}\right]\), verily that (adj. A)^-1 = adj. (A^-1).
Solution:
Given A = \(\left[\begin{array}{rr}
2 & -3 \\
4 & 6
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{rr}
2 & -3 \\
4 & 6
\end{array}\right|\) = 12 + 12 = 24 ≠ 0
Thus A^-1 exists & A^-1 = \(\frac { 1 }{ |A| }\)adj A
The cofactor of R₁ are ; 6 ; – 4
The cofactor of R₂ are ; 3 ; 2

The cofactor of R₁ are ; 2 ; 4
The cofactor of R₂ are ; – 3 ; 6
∴ adj B = \(\left[\begin{array}{cc}
2 & 4 \\
-3 & 6
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right]\)
Thus B^-1 = \(\frac{1}{|B|} \text { adj } B=\frac{1}{24}\left[\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right]\)
⇒ (adj A)^-1 \(\frac{1}{24}\left[\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right]\) … (2)
from eqn. (1) & eqn. (2) ; we have
(adj A)^-1 = adj A^-1

Question 12.
If A = \(\left[\begin{array}{rr}
-1 & -1 \\
2 & -2
\end{array}\right]\), show that A² + 3A + 4I = 0. Hence find A^-1.
Solution:
Given A = \(\left[\begin{array}{rr}
-1 & -1 \\
2 & -2
\end{array}\right]\)

∴ 4I = – 3A – A² ; pre multiplying both sides by A^-1 ; we have
∴ 4A^-1I = – 3A^-1 A – (A^-1A)A
∴ 4A^-1 = – 31 – IA [∵ A^-1A = I = AA^-1]
⇒ A^-1 = \(\frac { 1 }{ 4 }\)[- 3I – A]
= \(\frac{1}{4}\left[-3\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
-1 & -1 \\
2 & -2
\end{array}\right]\right]\)
= \(\frac{1}{4}\left[\left[\begin{array}{cc}
-3 & 0 \\
0 & -3
\end{array}\right]-\left[\begin{array}{cc}
-1 & -1 \\
2 & -2
\end{array}\right]\right]\)
= \(\frac{1}{4}\left[\begin{array}{cc}
-3+1 & 0+1 \\
0-2 & -3+2
\end{array}\right]\)
= \(\frac{1}{4}\left[\begin{array}{cc}
-2 & 1 \\
-2 & -1
\end{array}\right]=-\frac{1}{4}\left[\begin{array}{cc}
2 & -1 \\
2 & 1
\end{array}\right]\)

Question 13.
If A² – A + I = 0, then show that A^-1 = I – A.
Solution:
Given A² – A + I = 0 … (1)
pre-multiplying both sides of en. (1) by A^-1, we have
⇒ A^-1 A² – A^-1 A + A^-1I = A^-10
⇒ (A^-1A)A – I + A^-1 = 0
[∵ A^-1 A = I = AA^-1]
⇒ IA – I + A^-1 = 0
⇒ A – I + A^-1 = 0 [∵ IA = A = AI]
⇒ A^-1 = I – A.

Question 14.
For the matrix A = \(\left[\begin{array}{rr}
2 & -3 \\
3 & 4
\end{array}\right]\), show that A² – 6A + 17I = O. Hence find A^-1.
Solution:

Question 15.
Show that A = \(\left[\begin{array}{rrr}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right]\) satisfies the equation A³ – A² – 3A – I = 0. Hence, find A^-1.
Solution:
Here, A² = A.A

I₃ = A₃ – A₂ – 3A
pre-multiplying both sides by A^-1 ; we have
[Here, |A| = \(\left|\begin{array}{rrr}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right|\); Expanding along R₁
= 1(-1 – 8) + 0 – 2(- 8 + 3) = – 9 + 0 + 10 = 1 ≠ 0 ∴ A^-1 exists.]
⇒ A^-1I₃ = (A^-11 A)A² – (A^-1A)A – 3(A^-1A)
⇒ A^-1 = IA² – IA – 3I [∵ A^-1 A = I = AA^-1]
⇒ A^-1 = A2 – A – 3I [∵IA = A = AI]
= \(\left[\begin{array}{ccc}
-5 & -8 & -4 \\
6 & 9 & 4 \\
-2 & 0 & 3
\end{array}\right]-\left[\begin{array}{rrr}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right]-3\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-1-1-3 & -8-0-0 & -4+2+0 \\
6+2+0 & 9+1-3 & 4-2+0 \\
-2-3+0 & 0-4+0 & 3-1-3
\end{array}\right]=\left[\begin{array}{ccc}
-9 & -8 & -2 \\
8 & 7 & 2 \\
-5 & -4 & -1
\end{array}\right]\)

Question 16.
Let A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\), Prove that
A² – 4A – 5I = 0. Hence obtain A^-1.
Solution:
Given A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\)

Here |A| = \(\left|\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right|\); Expanding along R₁
= 1(1 – 4) -2(2 – 4) + 2(4 – 2) = – 3 + 4 + 4 = 5 ≠ 0
∴ A^-1 exists, pre-multiplying eqn. (1) both sides by A^-1 we have
(A^-1A)A – 4A^-1A – 5A^-1I = A^-10
⇒ IA – 4I – 5 A^-1 = O
⇒ 5A^-1 = A – 4I [∵ A^-1A = I]
⇒ 5A^-1 = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-\left[\begin{array}{lll}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{array}\right]=\left[\begin{array}{ccc}
-3 & 2 & 2 \\
2 & -3 & 2 \\
2 & 2 & -3
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{5}\left[\begin{array}{ccc}
-3 & 2 & 2 \\
2 & -3 & 2 \\
2 & 2 & -3
\end{array}\right]=\left[\begin{array}{ccc}
-3 / 5 & 2 / 5 & 2 / 5 \\
2 / 5 & -3 / 5 & 2 / 5 \\
2 / 5 & 2 / 5 & -3 / 5
\end{array}\right]\)

Question 17.
If A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\) , find (A’)^-1
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\)
∴ A^-1 = \(\left[\begin{array}{rrr}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right]\)
∴ |A’| = \(\left|\begin{array}{rrr}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right|\); Expanding along R₁.
= 1(- 1 – 8) + 0 – 2(- 8 + 3)
= – 9 + 10 = 1 ≠ 0.
∴ (A^-1)^-1 exists.
The cofactor of R₁ are ;
\(\left|\begin{array}{cc}
-1 & 2 \\
4 & 1
\end{array}\right| ;-\left|\begin{array}{ll}
-2 & 2 \\
+3 & 1
\end{array}\right| ;\left|\begin{array}{cc}
-2 & -1 \\
+3 & 4
\end{array}\right|\)
i.e. – 9 ; + 8 ; – 5
The cofactor of R₂ are ;
\(-\left|\begin{array}{cc}
+0 & -2 \\
4 & 1
\end{array}\right| ;\left|\begin{array}{cc}
1 & -2 \\
+3 & 1
\end{array}\right| ;-\left|\begin{array}{cc}
1 & 0 \\
3 & 4
\end{array}\right|\)
i.e. – 8 ; 7 ; – 4
The cofactor of R₃ are ;
\(\left|\begin{array}{cc}
0 & -2 \\
-1 & 2
\end{array}\right| ;-\left|\begin{array}{cc}
1 & -2 \\
-2 & 2
\end{array}\right| ; \quad\left|\begin{array}{cc}
1 & 0 \\
-2 & -1
\end{array}\right|\)
i.e. – 2 ; 2 ; – 1
Thus, adj (A’) = \(\left[\begin{array}{ccc}
-9 & +8 & -5 \\
-8 & 7 & -4 \\
-2 & 2 & -1
\end{array}\right]=\left[\begin{array}{ccc}
-9 & -8 & -2 \\
+8 & 7 & 2 \\
-5 & -4 & -1
\end{array}\right]\)
∴ (A’)^-1) = \(\frac{1}{|A|} \text { adj } A^{\prime}=\left[\begin{array}{ccc}
-9 & 8 & -2 \\
-8 & 7 & 2 \\
-5 & -4 & -1
\end{array}\right]\)

Question 18.
Show that A = \(\left[\begin{array}{rr}
-8 & 5 \\
2 & 4
\end{array}\right]\) satisfies the equation x² + 4x – 42 = 0. Hence find A^-1.
Solution:
Given A = \(\left[\begin{array}{rr}
-8 & 5 \\
2 & 4
\end{array}\right]\)
Now A satisfies the eqn. x² + 4x – 42 = 0 if A² + 4A – 42I = 0
Here, A² + 4A – 42I

Question 19.
If A = \(\left[\begin{array}{ll}
4 & 3 \\
2 & 5
\end{array}\right]\), find x and y such that A² – xA + yI = 0.
Solution:
Given A = \(\left[\begin{array}{ll}
4 & 3 \\
2 & 5
\end{array}\right]\) & A² – xA + yI = 0
⇒ \(\left[\begin{array}{ll}
4 & 3 \\
2 & 5
\end{array}\right]\left[\begin{array}{ll}
4 & 3 \\
2 & 5
\end{array}\right]-x\left[\begin{array}{ll}
4 & 3 \\
2 & 5
\end{array}\right]+y\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
16+6 & 12+15 \\
8+10 & 6+25
\end{array}\right]-\left[\begin{array}{ll}
4 x & 3 x \\
2 x & 5 x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
0 & y
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
22-4 x+y & 27-3 x \\
18-2 x & 31-5 x+y
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
Thus, their corresponding elements are equal.
∴ 22 – 4x + y = 0;
27 – 3x = 0 ⇒ x = 9 … (1)
Also 18 – 2x = 0 ⇒ x = 9
& 31 – 5x + y = 0 … (2)
from (1) ; 22 – 36 + y = 0
⇒ y= 14
Also both values of x & y satisfies eqn. (2).
Thus, x = 9 and y = 14.

Question 20.
If A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
2 & -1 & 0 \\
1 & 0 & 0
\end{array}\right]\), find A² and show that A² – A^-1.
Solution:

Question 21.
If A = \(\left[\begin{array}{lll}
1 & 1 & 2 \\
1 & 9 & 3 \\
1 & 4 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
2 & 3 & -1 \\
1 & -1 & 3
\end{array}\right]\), Verify that (AB)^-1 = B^-1A^-1.
Solution:
Given A = \(\left[\begin{array}{lll}
1 & 1 & 2 \\
1 & 9 & 3 \\
1 & 4 & 2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
1 & 1 & 2 \\
1 & 9 & 3 \\
1 & 4 & 2
\end{array}\right|\); Expanding along R₁
⇒ |A| = 1(18 – 12) -1(2 – 3) + 2(4 – 9) = 6 + 1 – 10 = – 3 ≠ 0
∴ A^-1 exists & A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
The cofactors of R₁ are ;

∴ |AB| = 5(52 – 0) -3(44 – 0) + 5(264 – 286) = 260 – 132 – 110 = 18 ≠ 0
Thus, (AB)^-1 exists & (AB)^-1 = \(\frac{1}{|\mathrm{AB}|}\) adj AB
The cofactors of R₁ are ;

From eqn. (1) & eqn. (2) ; we have (AB)^-1 = B^-1.

Question 22.
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’.
Solution: