ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices Ex 3.4

Well-structured ML Aggarwal Class 12 Solutions Chapter 3 Matrices Ex 3.4 facilitate a deeper understanding of mathematical principles.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices Ex 3.4

Question 1.
If A = \(\left[\begin{array}{rrr}
2 & -3 & 0 \\
-1 & 4 & 5
\end{array}\right]\), then find (3A)’.
Solution:
Given A = \(\left[\begin{array}{rrr}
2 & -3 & 0 \\
-1 & 4 & 5
\end{array}\right]\)
∴ 3A = 3 \(\left[\begin{array}{rrr}
2 & -3 & 0 \\
-1 & 4 & 5
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
6 & -9 & 0 \\
-3 & 12 & 15
\end{array}\right]\)

⇒ (3A)’ = \(\left[\begin{array}{rrr}
6 & -9 & 0 \\
-3 & 12 & 15
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{rr}
6 & -3 \\
-9 & 12 \\
0 & 15
\end{array}\right]\)

Question 2.
If A = \(\left[\begin{array}{rrr}
2 & -1 & 5 \\
4 & 0 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
-2 & 3 & 1 \\
-1 & 2 & -3
\end{array}\right]\), find A’ B’.
Solution:
Given A = \(\left[\begin{array}{rrr}
2 & -1 & 5 \\
4 & 0 & 3
\end{array}\right]\)
and B = \(\left[\begin{array}{rrr}
-2 & 3 & 1 \\
-1 & 2 & -3
\end{array}\right]\)

∴ A’ = \(\left[\begin{array}{rr}
2 & 4 \\
-1 & 0 \\
5 & 3
\end{array}\right]\)
and B’ = \(\left[\begin{array}{rr}
-2 & -1 \\
3 & 2 \\
1 & -3
\end{array}\right]\)

Thus A’ + B’ = \(\left[\begin{array}{rr}
2 & 4 \\
-1 & 0 \\
5 & 3
\end{array}\right]+\left[\begin{array}{rr}
-2 & -1 \\
3 & 2 \\
1 & -3
\end{array}\right]\)
= \(\left[\begin{array}{ll}
0 & 3 \\
2 & 2 \\
6 & 0
\end{array}\right]\)

Question 3.
If A = \(\left[\begin{array}{rr}
\sin x & -\cos x \\
\cos x & \sin x
\end{array}\right]\), 0 < x < \(\frac{\pi}{2}\) and A + A’ = I where I is unity matrix, find the value of x.
Solution:
Given A = \(\left[\begin{array}{rr}
\sin x & -\cos x \\
\cos x & \sin x
\end{array}\right]\) ; 0 < x < \(\frac{\pi}{2}\)
Given A + A’ = I
⇒ \(\left[\begin{array}{rr}
\sin x & -\cos x \\
\cos x & \sin x
\end{array}\right]\) + \(\left[\begin{array}{rr}
\sin x & \cos x \\
-\cos x & \sin x
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

⇒ \(\left[\begin{array}{cc}
2 \sin x & 0 \\
0 & 2 \sin x
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
∴ 2 sin x = 1
⇒ sin x = \(\frac{1}{2}\) = sin 30°
⇒ x = nπ + (- 1)ⁿ \(\frac{\pi}{6}\) ; n ∈ I
But 0 < x < \(\frac{\pi}{2}\)
∴ x = \(\frac{\pi}{6}\) [for n = 0]

Question 4.
If A^T = \(\left[\begin{array}{rr}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\), then find A^T – B^T.
Solution:
Given A^T = \(\left[\begin{array}{rr}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]\)
and B = \(\left[\begin{array}{rrr}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\)
∴ A^T – B^T = \(\left[\begin{array}{rr}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]-\left[\begin{array}{rr}
-1 & 1 \\
2 & 2 \\
1 & 3
\end{array}\right]\)
= \(\left[\begin{array}{rr}
4 & 3 \\
-3 & 0 \\
-1 & -2
\end{array}\right]\)

Question 5.
If A = \(\left[\begin{array}{rr}
-1 & 5 \\
3 & 7
\end{array}\right]\), determine whether A + A’ is symmetric or skew-symmetric.
Solution:
Given A = \(\left[\begin{array}{rr}
-1 & 5 \\
3 & 7
\end{array}\right]\)
⇒ A’ = \(\left[\begin{array}{rr}
-1 & 3 \\
5 & 7
\end{array}\right]\)
∴ A + A’ = \(\left[\begin{array}{rr}
-1 & 5 \\
3 & 7
\end{array}\right]+\left[\begin{array}{rr}
-1 & 3 \\
5 & 7
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-2 & 8 \\
8 & 14
\end{array}\right]\)
Thus (A + A’)’ = \(\left[\begin{array}{rr}
-2 & 8 \\
8 & 14
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{rr}
-2 & 8 \\
8 & 14
\end{array}\right]^{\prime}\)
= A + A’
∴ A + A’ is symmetric.

Question 6.
If A = \(\left[\begin{array}{rr}
5 & -1 \\
-2 & 6
\end{array}\right]\), determine whether A – A^T is symmetric ao skew-symmetric.
Solution:
Given A = \(\left[\begin{array}{rr}
5 & -1 \\
-2 & 6
\end{array}\right]\)
⇒ A^T = \(\left[\begin{array}{rr}
5 & -2 \\
-1 & 6
\end{array}\right]\)
∴ A – A^T = \(\left[\begin{array}{rr}
5 & -1 \\
-2 & 6
\end{array}\right]-\left[\begin{array}{rr}
5 & -2 \\
-1 & 6
\end{array}\right]\)
= \(\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\)
Let P = A – A^T
= \(\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\)
P^T = \(\left[\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right]\)
= – \(\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\)
= – P
Thus P = A – A^T is skew-symmetric.

Question 7.
(i) A = \(\left[\begin{array}{ll}
5 & a \\
b & 0
\end{array}\right]\) and A is symmetric matrix, show that a = b. (ISC 2018)
(ii) If the matrix \(\left[\begin{array}{cc}
6 & -x^2 \\
2 x-15 & 10
\end{array}\right]\) is symmetric, find the value(s) of x. (ISC 2017)
Solution:
(i) Given A = \(\left[\begin{array}{ll}
5 & a \\
b & 0
\end{array}\right]\)
since A is symmetric.
∴ A’ = A
⇒ \(\left[\begin{array}{ll}
5 & a \\
b & 0
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
5 & a \\
b & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
5 & b \\
a & 0
\end{array}\right]=\left[\begin{array}{ll}
5 & a \\
b & 0
\end{array}\right]\)
Thus their corresponding elements are equal.
∴ a = b.

(ii) Let A = \(\left[\begin{array}{cc}
6 & -x^2 \\
2 x-15 & 10
\end{array}\right]\) and A is symmetric.
∴ A’ = A
⇒ \(\left[\begin{array}{cc}
6 & 2 x-15 \\
-x^2 & 10
\end{array}\right]=\left[\begin{array}{cc}
6 & -x^2 \\
2 x+15 & 10
\end{array}\right]\)
∴ 2x – 15 = – x²
⇒ x² + 2x – 15 = 0
⇒ x + 5x – 3x – 15 = 0
⇒ x (x + 5) – 3 (x + 5) = 0
⇒ (x + 5) (x – 3) = 0
⇒ x = – 5, 3.

Question 8.
If A is a square matri%, ¡wove that A’A is symmetric.
Solution:
Let P = A’A
⇒ P’ = (A’A)’ = A'(A’)
[∵ (AB)’ = B’A’]
= A’A = P
[∵ (A’)’ = A]
Thus P is symmetric.
∴ A’A is symmetric.

Question 9.
If A, B are symmetric matrices and AB = BA then show that AB is symmetric.
Solution:
Given AB = BA …………..(1)
A and B are symmetric matrices
∴ A^T = A
and B^T = B
Now (AB)^T = B^TA^T = BA = AB
[using (1) and (2)]
∴ AB is symmetric.

Question 10.
If A, B and AB are aN symmetric matrices, then show that AB = BA.
Solution:
Given A and B are symmetric
∴ A^T = A and B^T = B …………..(1)
Also, AB is symmetric
∴ (AB)^T = AB
⇒ B^TA^T = AB
⇒ BA = AB [using (1)]

Question 11.
If A, B are skew-symmetric matrices and AB = BA, then show that AB is symmetric.
Solution:
Given A and B are skew symmetric matrices
∴ A^T = – A
and B^T = – B …………..(1)
also AB = BA …………….(2)
Now (AB)^T = B^TA^T [reversal law]
= (- B) (- A) [using (1)]
= BA = AB [using (2)]
Thus AB is symmetric.

Question 12.
If A, B are skew-symmetric matrices of same order and AB is symmetric, then show that AB = BA.
Solution:
Given A and B are skew symmetric matrices
∴ A^T = – A
and B^T = – B ………..(1)
(AB)^T = AB [∵ AB is symmetric]
⇒ B^T A^T = AB [using reversal law]
⇒ (- B) (- A) = AB [using (1)]
⇒ BA = AB.

Question 13.
Let A = \(\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right]\), verify that
(i) (A’)’ = A
(ii) (3A)’ = 3A’
(iii) (A – B)’ = A’ – B’
(iv) (AB)’ = B’A’
Solution:
Given A = \(\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]\)
and B = \(\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right]\)

(i) A’ = \(\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{rr}
5 & 6 \\
-1 & 7
\end{array}\right]\)
⇒ (A’)’ = \(\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]\)
= A

(ii) 3A = 3 \(\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]\)
= \(\left[\begin{array}{ll}
15 & -3 \\
18 & 21
\end{array}\right]\)
⇒ (3A)’ = \(\left[\begin{array}{rr}
15 & 18 \\
-3 & 21
\end{array}\right]\)
⇒ (3A)’ = 3 \(\left[\begin{array}{rr}
5 & 6 \\
-1 & 7
\end{array}\right]\) = 3A’

(iii) A – B = \(\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]-\left[\begin{array}{rr}
2 & 1 \\
3 & 4
\end{array}\right]\)
= \(\left[\begin{array}{rr}
3 & -2 \\
3 & 3
\end{array}\right]\)
∴ (A – B)’ = \(\left[\begin{array}{rr}
3 & -2 \\
3 & 3
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{rr}
3 & 3 \\
-2 & 3
\end{array}\right]\) …………….(1)
A’ – B’ = \(\left[\begin{array}{rr}
5 & 6 \\
-1 & 7
\end{array}\right]-\left[\begin{array}{ll}
2 & 3 \\
1 & 4
\end{array}\right]\)
= \(\left[\begin{array}{rr}
3 & 3 \\
-2 & 3
\end{array}\right]\) ………………(2)
From (1) and (2) ; we have
(A – B)’ = A’ – B’

(iv) AB = \(\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]\left[\begin{array}{rr}
2 & 1 \\
3 & 4
\end{array}\right]\)
= \(\left[\begin{array}{rr}
10-3 & 5-4 \\
12+21 & 6+28
\end{array}\right]\)
= \(\left[\begin{array}{rr}
7 & 1 \\
33 & 34
\end{array}\right]\)
∴ (AB)’ = \(\left[\begin{array}{rr}
7 & 1 \\
33 & 34
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{ll}
7 & 33 \\
1 & 34
\end{array}\right]\) ………….(1)
and B’A’ = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 4
\end{array}\right]\left[\begin{array}{rr}
5 & 6 \\
-1 & 7
\end{array}\right]\)
= \(\left[\begin{array}{rr}
10-3 & 12+21 \\
5-4 & 6+28
\end{array}\right]\)
= \(\left[\begin{array}{ll}
7 & 33 \\
1 & 34
\end{array}\right]\) …………..(2)
From (1) and (2) ; we have
(AB)’ = B’A’.

Question 14.
If A = \(\left[\begin{array}{rrr}
3 & \sqrt{3} & 2 \\
4 & 2 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
2 & -1 & 2 \\
1 & 2 & 4
\end{array}\right]\), then verify that
(i) (A’)’ = A
(ii) (A + B)’ = A’ + B’
(iii) (kB)’ = kB’, where k is any real number. (NCERT)
Solution:
Given A = \(\left[\begin{array}{rrr}
3 & \sqrt{3} & 2 \\
4 & 2 & 0
\end{array}\right]\)
and B = \(\left[\begin{array}{rrr}
2 & -1 & 2 \\
1 & 2 & 4
\end{array}\right]\)

(i) ∴ A’ = \(\left[\begin{array}{rrr}
3 & \sqrt{3} & 2 \\
4 & 2 & 0
\end{array}\right]^{\mathrm{T}}\)
= \(\left[\begin{array}{rr}
3 & 4 \\
\sqrt{3} & 2 \\
2 & 0
\end{array}\right]\)
∴ (A’)’ = \(\left[\begin{array}{rr}
3 & 4 \\
\sqrt{3} & 2 \\
2 & 0
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{rrr}
3 & \sqrt{3} & 2 \\
4 & 2 & 0
\end{array}\right]\)
= A

(ii) A + B = \(\left[\begin{array}{rrr}
3 & \sqrt{3} & 2 \\
4 & 2 & 0
\end{array}\right]+\left[\begin{array}{rrr}
2 & -1 & 2 \\
1 & 2 & 4
\end{array}\right]\)
= \(=\left[\begin{array}{ccc}
5 & \sqrt{3}-1 & 4 \\
5 & 4 & 4
\end{array}\right]\)

∴ (A + B)’ = \(\left[\begin{array}{cc}
5 & 5 \\
\sqrt{3}-1 & 4 \\
4 & 4
\end{array}\right]\) …………..(1)

Now A’ + B’ = \(\left[\begin{array}{rr}
3 & 4 \\
\sqrt{3} & 2 \\
2 & 0
\end{array}\right]+\left[\begin{array}{rr}
2 & 1 \\
-1 & 2 \\
2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
5 & 5 \\
\sqrt{3}-1 & 4 \\
4 & 4
\end{array}\right]\) …………..(2)
From (1) and (2) ; we have
(A + B)’ = A’ + B’

(iii) kB = k \(\left[\begin{array}{rrr}
2 & -1 & 2 \\
1 & 2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
2 k & -k & 2 k \\
k & 2 k & 4 k
\end{array}\right]\)

∴ (kB)’ = \(\left[\begin{array}{rrr}
2 k & -k & 2 k \\
k & 2 k & 4 k
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{rr}
2 k & k \\
-k & 2 k \\
2 k & 4 k
\end{array}\right]\) …………..(1)

and kB’ = k \(\left[\begin{array}{rr}
2 & 1 \\
-1 & 2 \\
2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{rr}
2 k & k \\
-k & 2 k \\
2 k & 4 k
\end{array}\right]\) …………..(2)

From (1) and (2) ; we have
(KB)’ = KB’

Question 15.
If A’ = \(\left[\begin{array}{rr}
-2 & 3 \\
1 & 2
\end{array}\right]\) and B = \(=\left[\begin{array}{rr}
-1 & 0 \\
1 & 2
\end{array}\right]\), then find (A + 2B)’. (NCERT)
Solution:
We have,
A’ = \(\left[\begin{array}{rr}
-2 & 3 \\
1 & 2
\end{array}\right]\) ;
B = \(=\left[\begin{array}{rr}
-1 & 0 \\
1 & 2
\end{array}\right]\)
∴ (A’)’ = A
= \(\left[\begin{array}{rr}
-2 & 1 \\
3 & 2
\end{array}\right]\)

Now A + 2B = \(\left[\begin{array}{rr}
-2 & 1 \\
3 & 2
\end{array}\right]+\left[\begin{array}{rr}
-2 & 0 \\
2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-4 & 1 \\
5 & 6
\end{array}\right]\)

∴ (A + 2B)’ = \(\left[\begin{array}{rr}
-4 & 5 \\
1 & 6
\end{array}\right]\)

Question 16.
If A = \(\left[\begin{array}{r}
1 \\
-4 \\
3
\end{array}\right]\) and B = [- 1 2 1], verify that (AB)’ = B’A’. (NCERT)
Solution:
Given, A = \(\left[\begin{array}{r}
1 \\
-4 \\
3
\end{array}\right]\)
and B = [- 1 2 1]
∴ A’ = [1 -4 3]
and B’ = \(\left[\begin{array}{r}
-1 \\
2 \\
1
\end{array}\right]\)
Here AB and B’A’ both are defined.
Now, AB = \(\left[\begin{array}{r}
1 \\
-4 \\
3
\end{array}\right]\) [- 1 2 1]
= \(\left[\begin{array}{rrr}
-1 & 2 & 1 \\
4 & -8 & -4 \\
-3 & 6 & 3
\end{array}\right]\)

∴ (AB)’ = \(\left[\begin{array}{rrr}
-1 & 4 & -3 \\
2 & -8 & 6 \\
1 & -4 & 3
\end{array}\right]\)
and B’A’ = \(\left[\begin{array}{r}
-1 \\
2 \\
1
\end{array}\right]\) [1 – 4 3]
= \(\left[\begin{array}{rrr}
-1 & 4 & -3 \\
2 & -8 & 6 \\
1 & -4 & 3
\end{array}\right]\)
Thus (AB)’ = B’A’.

Question 17.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 1 \\
5 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 2 \\
6 & 4 \\
7 & 3
\end{array}\right]\) then verify that
(i) (2A + B)’ = 2A’ + B’
(ii) (A – B)’ = A’ – B’
Solution:
Given A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 1 \\
5 & 6
\end{array}\right]\)
and B = \(\left[\begin{array}{ll}
1 & 2 \\
6 & 4 \\
7 & 3
\end{array}\right]\)

(i) 2A + B = \(2\left[\begin{array}{ll}
1 & 2 \\
4 & 1 \\
5 & 6
\end{array}\right]+\left[\begin{array}{ll}
1 & 2 \\
6 & 4 \\
7 & 3
\end{array}\right]\)
= \(\left[\begin{array}{rr}
2 & 4 \\
8 & 2 \\
10 & 12
\end{array}\right]+\left[\begin{array}{rr}
1 & 2 \\
6 & 4 \\
7 & 3
\end{array}\right]\)
= \(\left[\begin{array}{rr}
3 & 6 \\
14 & 6 \\
17 & 15
\end{array}\right]\)

∴ (2A + B)’ = \(2\left[\begin{array}{lll}
1 & 4 & 5 \\
2 & 1 & 6
\end{array}\right]+\left[\begin{array}{lll}
1 & 6 & 7 \\
2 & 4 & 3
\end{array}\right]\)

2A’ + B’ = \(2\left[\begin{array}{lll}
1 & 4 & 5 \\
2 & 1 & 6
\end{array}\right]+\left[\begin{array}{lll}
1 & 6 & 7 \\
2 & 4 & 3
\end{array}\right]\)

= \(\left[\begin{array}{lll}
2 & 8 & 10 \\
4 & 2 & 12
\end{array}\right]+\left[\begin{array}{lll}
1 & 6 & 7 \\
2 & 4 & 3
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
3 & 14 & 17 \\
6 & 6 & 15
\end{array}\right]\)
∴ (2A + B)’ = 2A’ + B’

(ii) A – B = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 1 \\
5 & 6
\end{array}\right]-\left[\begin{array}{ll}
1 & 2 \\
6 & 4 \\
7 & 3
\end{array}\right]\)
= \(\left[\begin{array}{rr}
0 & 0 \\
-2 & -3 \\
-2 & 3
\end{array}\right]\)

∴ (A – B)’ = \(\left[\begin{array}{rrr}
0 & -2 & -2 \\
0 & -3 & 3
\end{array}\right]\) ……………..(1)
and A’ – B’ = \(\left[\begin{array}{lll}
1 & 4 & 5 \\
2 & 1 & 6
\end{array}\right]-\left[\begin{array}{lll}
1 & 6 & 7 \\
2 & 4 & 3
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
0 & -2 & -2 \\
0 & -3 & 3
\end{array}\right]\) ……………(2)
From (1) and (2) ; we have
(A – B)’ = A – B’

Question 18.
If A = \(\left[\begin{array}{lll}
2 & 4 & 0 \\
3 & 9 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 8 \\
1 & 3
\end{array}\right]\) verify that (AB)’ = B’A’. (NCERT Exampler)
Solution:
Given A = \(\left[\begin{array}{lll}
2 & 4 & 0 \\
3 & 9 & 6
\end{array}\right]_{2 \times 3}\)
and B = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 8 \\
1 & 3
\end{array}\right]_{3 \times 2}\)

Since No. of columns in A = no. of rows in B = 3
∴ AB exists.
AB = \(\left[\begin{array}{lll}
2 & 4 & 0 \\
3 & 9 & 6
\end{array}\right]\left[\begin{array}{ll}
1 & 4 \\
2 & 8 \\
1 & 3
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2 \times 1+4 \times 2+0 \times 1 & 2 \times 4+4 \times 8+0 \times 3 \\
3 \times 1+9 \times 2+6 \times 1 & 3 \times 4+9 \times 8+6 \times 3
\end{array}\right]\)
= \(\left[\begin{array}{rr}
10 & 40 \\
27 & 112
\end{array}\right]\)

∴ (AB)’ = \(\left[\begin{array}{rr}
10 & 27 \\
40 & 112
\end{array}\right]\) ………………..(1)
and B’A’ = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
4 & 8 & 3
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
4 & 9 \\
0 & 6
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1 \times 2+2 \times 4+1 \times 0 & 1 \times 3+2 \times 9+1 \times 6 \\
4 \times 2+8 \times 4+3 \times 0 & 4 \times 3+8 \times 9+3 \times 6
\end{array}\right]\)
= \(\left[\begin{array}{rr}
10 & 27 \\
40 & 112
\end{array}\right]\) …………..(2)
From (1) and (2) ; we have
(AB)’ = B’A’.

Question 19.
If A = \(\left[\begin{array}{rr}
3 & 2 \\
-1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
-1 & 0 \\
2 & 5 \\
3 & 4
\end{array}\right]\), find (BA)’.
Solution:
Given A = \(\left[\begin{array}{rr}
3 & 2 \\
-1 & 1
\end{array}\right]\)
and B =\(\left[\begin{array}{rr}
-1 & 0 \\
2 & 5 \\
3 & 4
\end{array}\right]\)

Since No. of columns in B = no. of rows in A = 2
∴ BA exists.
∴ BA = \(\left[\begin{array}{rr}
-1 & 0 \\
2 & 5 \\
3 & 4
\end{array}\right]\left[\begin{array}{rr}
3 & 2 \\
-1 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-1 \times 3+0 \times(-1) & -1 \times 2+0 \times 1 \\
2 \times 3+5 \times(-1) & 2 \times 2+5 \times 1 \\
3 \times 3-4(1) & 3 \times 2+4 \times 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-3 & -2 \\
1 & 9 \\
5 & 10
\end{array}\right]\)

Thus (BA)’ = \(\left[\begin{array}{rrr}
-3 & 1 & 5 \\
-2 & 9 & 10
\end{array}\right]\).

Question 20.
If A = \(\left[\begin{array}{rr}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right]\), then verify that A’A = I. (NCERT)
Solution:
Given A = \(\left[\begin{array}{rr}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right]\)

∴ A’ = \(\left[\begin{array}{rr}
\sin \alpha & -\cos \alpha \\
\cos \alpha & \sin \alpha
\end{array}\right]\)

∴ A’A = \(\left[\begin{array}{rr}
\sin \alpha & -\cos \alpha \\
\cos \alpha & \sin \alpha
\end{array}\right]\left[\begin{array}{rr}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right]\)

= \(\left[\begin{array}{cc}
\sin ^2 \alpha+\cos ^2 \alpha & \sin \alpha \cos \alpha-\cos \alpha \sin \alpha \\
\cos \alpha \sin \alpha-\sin \alpha \cos \alpha & \cos ^2 \alpha+\sin ^2 \alpha
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= I₂

Question 21.
If A = \(\frac{1}{3}\left[\begin{array}{rrr}
1 & -2 & 2 \\
-2 & 1 & 2 \\
-2 & -2 & -1
\end{array}\right]\), show that AA’ = I.
Solution:
Given A = \(\frac{1}{3}\left[\begin{array}{rrr}
1 & -2 & 2 \\
-2 & 1 & 2 \\
-2 & -2 & -1
\end{array}\right]\)

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

Thus, AA’ = \(\frac{1}{9}\left[\begin{array}{rrr}
1 & -2 & 2 \\
-2 & 1 & 2 \\
-2 & -2 & -1
\end{array}\right]\left[\begin{array}{rrr}
1 & -2 & -2 \\
-2 & 1 & -2 \\
2 & 2 & -1
\end{array}\right]\)

∴ AA’ = \(\frac{1}{9}\left[\begin{array}{ccc}
1 \times 1-2 \times(-2)+2 \times 2 & -2-2+4 & -2+4-2 \\
-2-2+4 & 4+1+4 & 4-2-2 \\
-2+4-2 & 4-2-2 & 4+4+1
\end{array}\right]\)
= \(\frac{1}{9}\left[\begin{array}{lll}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right]\)
= \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
= 1

Question 22.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & x \\
-2 & 2 & -1
\end{array}\right]\) is a matrix satisfying AA’ = 9I₃, find x.
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & x \\
-2 & 2 & -1
\end{array}\right]\)
∴ A’ = \(\left[\begin{array}{rrr}
1 & 2 & -2 \\
2 & 1 & 2 \\
2 & x & -1
\end{array}\right]\)
Since AA’ = 9I₃

⇒ \(\left[\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & x \\
-2 & 2 & -1
\end{array}\right]\left[\begin{array}{rrr}
1 & 2 & -2 \\
2 & 1 & 2 \\
2 & x & -1
\end{array}\right]=9\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
\(\left[\begin{array}{rrr}
1+4+4 & 2+2+2 x & -2+4-2 \\
2+2+2 x & 4+1+x^2 & -4+2-x \\
-2+4-2 & -4+2-x & 4+4+1
\end{array}\right]\) = 9 \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)

⇒ \(\left[\begin{array}{rrr}
9 & 4+2 x & 0 \\
4+2 x & 5+x^2 & -2-x \\
0 & -2-x & 9
\end{array}\right]=\left[\begin{array}{lll}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right]\)

Thus their corresponding elements are equal.
∴ 4 + 2x = 0
⇒ x = – 2;
5 + x² = 9
⇒ x = ± 2
– 2 – x = 0
⇒ x = – 2
Thus, the common value of x be – 2.

Question 23.
Find the untegral value of x if [x 4 -1] \(\left[\begin{array}{rrr}
2 & 1 & -1 \\
1 & 0 & 0 \\
2 & 2 & 4
\end{array}\right]\) [x 4 – 1]^t = O.
Solution:
Given [x 4 -1] \(\left[\begin{array}{rrr}
2 & 1 & -1 \\
1 & 0 & 0 \\
2 & 2 & 4
\end{array}\right]\) [x 4 – 1]^t = O
⇒ [x 4 – 1] \(\left[\begin{array}{rrr}
2 & 1 & -1 \\
1 & 0 & 0 \\
2 & 2 & 4
\end{array}\right]\left[\begin{array}{r}
x \\
4 \\
-1
\end{array}\right]\) = O
⇒ [x 4 -1] \(\left[\begin{array}{c}
2 x+4+1 \\
x+0+0 \\
2 x+8-4
\end{array}\right]\) = O

⇒ [x 4 – 1] \(\left[\begin{array}{c}
2 x+5 \\
x \\
2 x+4
\end{array}\right]\) = O
⇒ x (2x + 5) + 4x – 1 (2x + 4) = 0
⇒ 2x² + 7x – 4 = 0
2x² + 8x – x – 4 = 0
2x (x + 4) – 1 (x + 4) = 0
(2x – 1) (x + 4) = 0
x = \(\frac{1}{2}\), – 4.

Question 24.
Show that the matrix A, where A = \(\left[\begin{array}{rrr}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\), is a symmetric matrix. (NCERT)
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\)
⇒ A’ = \(\left[\begin{array}{rrr}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\) = A
∴ A is symmetric matrix.

Question 24 (old).
Define a symmetric matrix. Prove that for A = \(\left[\begin{array}{ll}
2 & 4 \\
5 & 6
\end{array}\right]\), A + A^t is a symmetric matrix where A^t is a symmetric matrix where A^t
Solution:
Let A be a square matrix of order n.
Then A is symmetric iff a_ij = a_ji ∀ i ≠ j
i.e., A is symmetric iff A’ = A
Given A = \(\left[\begin{array}{ll}
2 & 4 \\
5 & 6
\end{array}\right]\)
A^t = \(\left[\begin{array}{ll}
2 & 5 \\
4 & 6
\end{array}\right]\)
Thus P = A + A^t
= \(\left[\begin{array}{ll}
2 & 4 \\
5 & 6
\end{array}\right]+\left[\begin{array}{ll}
2 & 5 \\
4 & 6
\end{array}\right]\)
= \(\left[\begin{array}{rr}
4 & 9 \\
9 & 12
\end{array}\right]\)
∴ P’ = \(\left[\begin{array}{rr}
4 & 9 \\
9 & 12
\end{array}\right]\) = P
∴ P is symmetric.
Thus A + A^t is symmetric.

Question 25.
If the matrix \(\left[\begin{array}{rrr}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & -1
\end{array}\right]\) is symmetric, find the value of a and b.
Solution:
Let A = \(\left[\begin{array}{rrr}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & -1
\end{array}\right]\)
and A is given to be symmetric.
∴ A = A’
⇒ \(\left[\begin{array}{rrr}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & -1
\end{array}\right]=\left[\begin{array}{rrr}
0 & 3 & 3 a \\
2 b & 1 & 3 \\
-2 & 3 & -1
\end{array}\right]\)
Thus corresponding elements are equal.
∴ 2b = 3
⇒ b = \(\frac{3}{2}\) and – 2 = 3a
⇒ a = – \(\frac{2}{3}\)

Question 25 (old).
If A = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right]\), prove that A – A^T is a skew-symmetric matrix.
Solution:
Given A = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right]\)
∴ A’ = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 5
\end{array}\right]\)
Thus P = A – A’
= \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right]-\left[\begin{array}{ll}
2 & 4 \\
3 & 5
\end{array}\right]\)
= \(\left[\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right]\)

∴ P’ = \(\left[\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\)
= \(-\left[\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right]\)
= – P
Thus P i.e., A – A’ is skew symmetric matrix.

Question 26.
If A and B are symmetric matrices of the same order, prove that AB + BA is symmetric.
Solution:
Given A and B are symmetric matrices of same order.
∴ A’ = A
and B’ = B …………….(1)
Now, (AB + BA)’ = (AB)’ + (BA)’
[∵ (A + B)’ = A’ + B’]
= B’A’ + A’B’ [reversal law]
= BA + AB [using (1)]
= AB + BA [∵ A + B = B + A]
Thus, AB + BA is symmetric.

Question 27.
Express each of the following matrices as the sum of a symmetric and a skew-symmetric matrices:
(i) \(\left[\begin{array}{rr}
3 & 5 \\
1 & -1
\end{array}\right]\) (NCERT)
(ii) \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rr}
3 & 5 \\
1 & -1
\end{array}\right]\)
∴ A’ = \(\left[\begin{array}{rr}
3 & 1 \\
5 & -1
\end{array}\right]\)

∴ A + A’ = \(\left[\begin{array}{rr}
3 & 5 \\
1 & -1
\end{array}\right]+\left[\begin{array}{rr}
3 & 1 \\
5 & -1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
6 & 6 \\
6 & -2
\end{array}\right]\)

⇒ \(\frac{1}{2}\) (A + A’) = \(\left[\begin{array}{rr}
3 & 3 \\
3 & -1
\end{array}\right]\)
B = \(\frac{1}{2}\) (A + A’)
= \(\left[\begin{array}{rr}
3 & 3 \\
3 & -1
\end{array}\right]\)
∴ B’ = B
⇒ B is symmetric.
and A – A’ = \(\left[\begin{array}{rr}
0 & 2 \\
-2 & 0
\end{array}\right]\)
⇒ \(\frac{1}{2}\) (A – A’) = \(\left[\begin{array}{rr}
0 & 2 \\
-2 & 0
\end{array}\right]\)
∴ C = \(\frac{1}{2}\) (A – A’)
= \(\left[\begin{array}{rr}
0 & 2 \\
-2 & 0
\end{array}\right]\)
⇒ C’ = \(\left[\begin{array}{rr}
0 & -2 \\
+2 & 0
\end{array}\right]\)
= – \(\left[\begin{array}{rr}
0 & +2 \\
-2 & 0
\end{array}\right]\)
= – C
Thus C is skew-symmetric.
Now, A = \(\frac{1}{2}\) (2A)
= \(\frac{1}{2}\) [(A + A’) + (A – A’)]
∴ A = \(\frac{1}{2}\) (A + A’) + \(\frac{1}{2}\) (A – A’)
= B + C
= \(\left[\begin{array}{rr}
3 & 3 \\
3 & -1
\end{array}\right]+\left[\begin{array}{rr}
0 & 2 \\
-2 & 0
\end{array}\right]\)

Question 28.
Express the following matrices as the sum of symmetric and skew-symmetric matrices:
(i) \(\left[\begin{array}{lll}
3 & 2 & 5 \\
4 & 1 & 3 \\
0 & 6 & 7
\end{array}\right]\)
(ii) \(\left[\begin{array}{rrr}
2 & 4 & -6 \\
7 & 3 & 5 \\
1 & -2 & 4
\end{array}\right]\) (NCERT Exampler)
Solution:
(i) Let A = \(\left[\begin{array}{lll}
3 & 2 & 5 \\
4 & 1 & 3 \\
0 & 6 & 7
\end{array}\right]\)
∴ A’ = \(\left[\begin{array}{lll}
3 & 4 & 0 \\
2 & 1 & 6 \\
5 & 3 & 7
\end{array}\right]\)
∴ A + A’ = \(\left[\begin{array}{rrr}
6 & 6 & 5 \\
6 & 2 & 9 \\
5 & 9 & 14
\end{array}\right]\)
⇒ \(\frac{1}{2}\) (A + A’) = \(\left[\begin{array}{rrr}
3 & 3 & 5 / 2 \\
3 & 1 & 9 / 2 \\
5 / 2 & 9 / 2 & 7
\end{array}\right]\)
Let P = \(\frac{1}{2}\) (A + A’)
= \(\left[\begin{array}{rrr}
3 & 3 & 5 / 2 \\
3 & 1 & 9 / 2 \\
5 / 2 & 9 / 2 & 7
\end{array}\right]\)

∴ P’ = \(\left[\begin{array}{rrr}
3 & 3 & 5 / 2 \\
3 & 1 & 9 / 2 \\
5 / 2 & 9 / 2 & 7
\end{array}\right]\) = P
Thus P is symmetric.

Now A – A’ = \(\left[\begin{array}{rrr}
0 & -2 & 5 \\
2 & 0 & -3 \\
-5 & 3 & 0
\end{array}\right]\)
⇒ \(\frac{1}{2}\) (A – A’) = \(\left[\begin{array}{rrr}
0 & -1 & 5 / 2 \\
+1 & 0 & -3 / 2 \\
-5 / 2 & 3 / 2 & 0
\end{array}\right]\)
Let Q = \(\frac{1}{2}\) (A – A’)
= \(\left[\begin{array}{rrr}
0 & -1 & 5 / 2 \\
1 & 0 & -3 / 2 \\
-5 / 2 & 3 / 2 & 0
\end{array}\right]\)
∴ Q’ = \(\left[\begin{array}{rrr}
0 & 1 & -5 / 2 \\
-1 & 0 & 3 / 2 \\
5 / 2 & -3 / 2 & 0
\end{array}\right]\)
= – Q
which is skew symmetric.
Since A = \(\frac{1}{2}\) (2A)
⇒ A = \(\frac{1}{2}\) [A + A’ + A -A’]
⇒ A = P + Q
Thus A can be expressed as sum of symmetric and skew symmetric matrix.

(ii) Let A = \(\left[\begin{array}{rrr}
2 & 4 & -6 \\
7 & 3 & 5 \\
1 & -2 & 4
\end{array}\right]\)
∴ A’ = \(\left[\begin{array}{rrr}
2 & 7 & 1 \\
4 & 3 & -2 \\
-6 & 5 & 4
\end{array}\right]\)

Thus Q is skew-symmetric matrix.
Since A = \(\frac{1}{2}\) (A + A’ + A – A’)
= \(\frac{1}{2}\) (A + A’) + \(\frac{1}{2}\) (A – A’)
⇒ A = P + Q
Thus, A can be expressible as sum of a symmetric matrix and a skew symmetric matrix.