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

Practicing ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(g) is the ultimate need for students who intend to score good marks in examinations.

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

Question 1.
Let A be a square matrix of order 3. Write the value of |2A|, where |A| = 4.
Solution:
Given A be a square matrix of order 3
s.t. |A| = 4
∴ |2A| = 2²|A|
= 8 x 4 = 32
[∵ |KA| = Kⁿ|A| where A be a square matrix of order n]

Question 2.
If A is a square matrix of order 3 such that |adj A| = 64, find |A|.
Solution:
Given |adj A| = 64
Also A be a square matrix of order 3
|adj A| = |A|^3-1 = |A|²
∴ |A|² = 64
⇒ |A| = ± 8

Question 3.
Find the adjoint of the following matrices :
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
(ii) \(\left[\begin{array}{rrr}
-1 & -2 & 3 \\
-2 & 1 & 1 \\
-4 & -5 & 2
\end{array}\right]\)
Solution:
(i) Let 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 = \(\left[\begin{array}{cc}
4 & -3 \\
-2 & 1
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]\)

(ii) Let A = \(\left[\begin{array}{rrr}
-1 & -2 & 3 \\
-2 & 1 & 1 \\
-4 & -5 & 2
\end{array}\right]\)
The cofactors of R₁ are ;
\(\left|\begin{array}{cc}
1 & 1 \\
-5 & 2
\end{array}\right| ;-\left|\begin{array}{cc}
-2 & 1 \\
-4 & 2
\end{array}\right| ;\left|\begin{array}{cc}
-2 & 1 \\
-4 & -5
\end{array}\right|\)
i.e. (2 + 5) ; -(-4 + 4); (10 + 4)
i.e 7 ; 0; 14
The cofactors of R₁ are ;
– \(\left|\begin{array}{ll}
-2 & 3 \\
-5 & 2
\end{array}\right| ;\left|\begin{array}{ll}
-1 & 3 \\
-4 & 2
\end{array}\right| ;-\left|\begin{array}{ll}
-1 & -2 \\
-4 & -5
\end{array}\right|\)
i.e. – (-4 + 15) ; (- 2 + 12) ; – (5 – 8)
i.e. – 11 ; 10 ; 3
The cofactors of R₃ are ;
\(\left|\begin{array}{cc}
-2 & 3 \\
1 & 1
\end{array}\right| ;-\left|\begin{array}{ll}
-1 & 3 \\
-2 & 1
\end{array}\right| ;\left|\begin{array}{cc}
-1 & -2 \\
-2 & 1
\end{array}\right|\)
i.e. (- 2 – 3) ; – (- 1 + 6) ; (- 1 – 4)
i.e. – 5 ; – 5 ; – 5
∴ adj A = \(\left[\begin{array}{rrr}
7 & 0 & 14 \\
-11 & +10 & 3 \\
-5 & -5 & -5
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
7 & -11 & -5 \\
0 & 10 & -5 \\
14 & 3 & -5
\end{array}\right]^{\prime}\)

Question 4.
Find the adjoint of the matrix A and verify
A (adj. A) = (adj. A) A = | A | I₂.
(i) A = \(\left[\begin{array}{ll}
3 & 4 \\
5 & 7
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{ll}
5 & -2 \\
3 & -2
\end{array}\right]\)
Solution:
(i) Given A = \(\left[\begin{array}{ll}
3 & 4 \\
5 & 7
\end{array}\right]\)
∴ |A| = 21 – 20 = 1 ≠ 0
The cofactors of R₁ are ; 7 ; – 5
The cofactors of R₂ are; – 4 ; 3

(ii) Let A = \(\left[\begin{array}{ll}
5 & -2 \\
3 & -2
\end{array}\right]\)
∴ |A| = – 10 + 6 = – 4
The cofactors of R₁ are ; – 2 ; – 3
The cofactors of R₂ are ; 2 ; 5

Question 5.
For the matrix A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
2 & 3 & 0 \\
18 & 2 & 10
\end{array}\right]\), Prove that A (adj. A) = 0.
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
2 & 3 & 0 \\
18 & 2 & 10
\end{array}\right]\)
The cofactors of R₂ are
i.e. (30 – 0) ; – (20 – 0) ; (4 – 54)
i.e. 30 ; – 20 ; – 50
The cofactors of R₂ are ;
\(-\left|\begin{array}{cc}
-1 & 1 \\
2 & 10
\end{array}\right| ;\left|\begin{array}{cc}
1 & 1 \\
18 & 10
\end{array}\right| ;-\left|\begin{array}{cc}
1 & -1 \\
18 & 2
\end{array}\right|\)
i.e. -(-10 – 2) ; (10 – 18) ; -(2 + 18)
i.e. 12 ; – 8 ; – 20
The cofactors of R₃ are ;
\(\left|\begin{array}{cc}
-1 & 1 \\
3 & 0
\end{array}\right| ;-\left|\begin{array}{cc}
1 & 1 \\
2 & 0
\end{array}\right| ; \quad\left|\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right|\)
i.e. – 3 ; 2 ; 5

Question 6.
Find A (adj. A) for the matrix A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
0 & 2 & -1 \\
-4 & 5 & 2
\end{array}\right]\)
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
0 & 2 & -1 \\
-4 & 5 & 2
\end{array}\right]\)
The cofactors of R₁ are ;
\(\left|\begin{array}{cc}
2 & -1 \\
5 & 2
\end{array}\right| ;-\left|\begin{array}{cc}
0 & -1 \\
-4 & 2
\end{array}\right| ;\left|\begin{array}{cc}
0 & 2 \\
-4 & 5
\end{array}\right|\)
i.e. (4 + 5) ; – (0 – 4) ; (0 + 8)
i.e. 9 ; 4 ; 8
The cofactors of R₂ are ;
\(-\left|\begin{array}{cc}
-2 & 3 \\
5 & 2
\end{array}\right| ;\left|\begin{array}{cc}
1 & 3 \\
-4 & 2
\end{array}\right| ;-\left|\begin{array}{cc}
1 & -2 \\
-4 & 5
\end{array}\right|\)
i.e. 19, 14, 3
The cofactors of R₃ are ;

Question 7.
Find the adjoint of the matrix
A = \(\left[\begin{array}{rrr}
-1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right]\) and hence show that
A (adj. A) = | A | I₃.
Solution:

Question 8.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2 \\
1 & 0 & 3
\end{array}\right]\), find the value of A (adj A) without finding Adj. A.
Solution:
Here |A| = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2 \\
1 & 0 & 3
\end{array}\right]\);
Expanding along R₁.
= 1(3 – 0) – 2(9 – 2) + 3(0 – 1)
= 3 – 14 – 3 = – 14
we know that,
A(adj A) = (adj A) A = |A|I
∴ A adj A = |A|I₃
= – 14\(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-14 & 0 & 0 \\
0 & -14 & 0 \\
0 & 0 & -14
\end{array}\right]\)

Question 9.
If A = \(\left[\begin{array}{rrr}
-4 & -3 & -3 \\
1 & 0 & 1 \\
4 & 4 & 3
\end{array}\right]\) show that adj A = A.
Solution:

Question 10.
For the matrix A = \(\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\), verify the theorem
A (adj A) = (adj A) A = | A | I.
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\)
∴ |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 – 4)
= 3 – 9 – 5 = – 11
The cofactors of R₁ are ;
\(\left|\begin{array}{cc}
2 & -3 \\
-1 & 3
\end{array}\right| ;-\left|\begin{array}{cc}
1 & -3 \\
2 & 3
\end{array}\right| ;\left|\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right|\)
i.e. 3 ; – 9 ; – 5
The cofactors of R₂ are ;

Question 11.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right]\), prove that adj AB = (adj B) (adj A).
Solution:
Given A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\)
∴ The cofactor of R₁ are ; 1 ; – 2
The cofactor of R₂ are ; – 2 ; 1
∴ adj A = \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 1
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
1 & -2 \\
-2 & 1
\end{array}\right]\)
Also, B = \(\left[\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right]\)
∴ The cofactor of R₁ are ; 1 ; – 2
The cofactor of R₂ are ; – 1 ; 1

Question 12.
If A = \(\left[\begin{array}{rrr}
-1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right]\), show that adj A = 3A’.
Solution:

Question 13.
If A = \(\left[\begin{array}{rrr}
1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right]\) find a non-zero unit matrix B such that AB = BA.
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right]\)
we know that A(adj A) = (adj A) A
also given AB = BA
∴ adj A = B
The cofactors of R₁ are ;

Question 14.
Prove that
|adj AB| = |adj A| |adj B|.
Solution:
First of all we prove that
adj (A B) = (adj B) (adj A)
where A & B are non-singular matrices of order n.
we know that,
AB(adj AB) = |AB|I = (adj AB)AB (AB) (adj B) (adj A)
= A(B adj B) adj A = A |B|I adj A
= |B|(A adj A)I = |B||A|I = |A||B||I
= |AB|I = AB (adj AB)
⇒ adj AB = (adj B)(adj A)
[if A, B & C are non-singular matrices s.t. AB = AC ⇒ B = C]
∴ |adj AB| = |(adj BXadj A)| = |adj B| |adj A| [∵ |AB| = |A||B|]