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

The availability of ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(b) encourages students to tackle difficult exercises.

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

Question 1.
Write each sum or difference as a single matrix.
(i) \(\left[\begin{array}{rr}
-3 & -2 \\
2 & 0
\end{array}\right]+\left[\begin{array}{rr}
3 & 2 \\
-2 & 0
\end{array}\right]\)
(ii) \(\left[\begin{array}{rrr}
1 & 2 & -1 \\
3 & 1 & 4 \\
3 & 1 & -6
\end{array}\right]+\left[\begin{array}{rrr}
-1 & -2 & 1 \\
-3 & 0 & -4 \\
-2 & -1 & 6
\end{array}\right]\)
(iii) \(\left[\begin{array}{rr}
\cos ^2 x & \sin ^2 x \\
\sin x & \cos ^2 x
\end{array}\right]+\left[\begin{array}{rr}
\sin ^2 x & \cos ^2 x \\
-\cos x & -\sin ^2 x
\end{array}\right]\)
Solution:
(i) \(\left[\begin{array}{rr}
-3 & -2 \\
2 & 0
\end{array}\right]+\left[\begin{array}{rr}
3 & 2 \\
-2 & 0
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-3+3 & -2+2 \\
2-2 & 0+0
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

(ii) \(\left[\begin{array}{rrr}
1 & 2 & -1 \\
3 & 1 & 4 \\
3 & 1 & -6
\end{array}\right]+\left[\begin{array}{rrr}
-1 & -2 & 1 \\
-3 & 0 & -4 \\
-2 & -1 & 6
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
1-1 & 2-2 & -1+1 \\
3-3 & 1+0 & 4-4 \\
3-2 & 1-1 & -6+6
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\)

(iii) \(\left[\begin{array}{rr}
\cos ^2 x & \sin ^2 x \\
\sin x & \cos ^2 x
\end{array}\right]+\left[\begin{array}{rr}
\sin ^2 x & \cos ^2 x \\
-\cos x & -\sin ^2 x
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos ^2 x+\sin ^2 x & \sin ^2 x+\cos ^2 x \\
\sin x-\cos x & \cos ^2 x-\sin ^2 x
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1 & 1 \\
\sin x-\cos x & \cos 2 x
\end{array}\right]\)

Question 2.
(i) Given A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4 \\
2 & 6
\end{array}\right]\), B = \(\left[\begin{array}{rr}
2 & -1 \\
3 & -2 \\
0 & 1
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
4 & 2 \\
1 & 0 \\
-2 & -4
\end{array}\right]\) compute the following
(a) A + B
(b) (A + B) + C
(c) A + (B + C)
(d) A – B
(e) (A – B) + C
(f) B – A
(ii) Consider the answers to part (b) and (c), what law is illustrated?
(iii) Consider the parts (d) and (f), what conclusion can be drawn?
Solution:
(i) Given A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4 \\
2 & 6
\end{array}\right]\), B = \(\left[\begin{array}{rr}
2 & -1 \\
3 & -2 \\
0 & 1
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
4 & 2 \\
1 & 0 \\
-2 & -4
\end{array}\right]\)

(ii) From (b) & (c), we have (A + B) + C = A + (B + C)
∴ association law holds under matrix addition.

(iii) From part (d) & part (f), we have A – B = – (B – A)
∴ Commutative law does not holds under matrix subtraction.

Question 3.
Solve the equation
X + \(\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]=\left[\begin{array}{lll}
2 & 1 & 2 \\
3 & 2 & 3 \\
4 & 3 & 4
\end{array}\right]\) for the 3 x 3 matrix X.
Solution:
Given X + \(\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]=\left[\begin{array}{lll}
2 & 1 & 2 \\
3 & 2 & 3 \\
4 & 3 & 4
\end{array}\right]\)
∴ X = \(\left[\begin{array}{lll}
2 & 1 & 2 \\
3 & 2 & 3 \\
4 & 3 & 4
\end{array}\right]-\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
2-0 & 1-0 & 2-1 \\
3-0 & 2-1 & 3-0 \\
4-1 & 3-0 & 4-0
\end{array}\right]\)
= \(\left[\begin{array}{lll}
2 & 1 & 1 \\
3 & 1 & 3 \\
3 & 3 & 4
\end{array}\right]\)

Question 4.
If \(\left[\begin{array}{rr}
2 & -3 \\
4 & 0
\end{array}\right]-\left[\begin{array}{ll}
x_1 & x_2 \\
y_1 & y_2
\end{array}\right]=\left[\begin{array}{rr}
-3 & 4 \\
5 & -1
\end{array}\right]\) determine x₁, x₂, y₁ and y₂.
Solution:
Given \(\left[\begin{array}{rr}
2 & -3 \\
4 & 0
\end{array}\right]-\left[\begin{array}{ll}
x_1 & x_2 \\
y_1 & y_2
\end{array}\right]=\left[\begin{array}{rr}
-3 & 4 \\
5 & -1
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2-x_1 & -3-x_2 \\
4-y_1 & 0-y_2
\end{array}\right]=\left[\begin{array}{cc}
-3 & 4 \\
5 & -1
\end{array}\right]\)
Thus, their corresponding element are equal.
∴ ⇒ 2 – x₁ = – 3 ⇒ x₁ = 5
– y₂ = – 1 ⇒ y₂ = 1
– 3 – x₂ = 4 ⇒ x₂ = – 7
4 – y₁ = 5 ⇒ y₁ = – 1

Question 5.
If A = \(\left[\begin{array}{rr}
1 & 2 \\
-1 & 8 \\
4 & 9
\end{array}\right]\), construct a matrix X such that X + A = 0.
Solution:
A = \(\left[\begin{array}{rr}
1 & 2 \\
-1 & 8 \\
4 & 9
\end{array}\right]\) we want to find matrix x
s.t. A + X = 0 ⇒ X = O – A
= – A = – \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 8 \\
4 & 9
\end{array}\right]=\left[\begin{array}{cc}
-1 & -2 \\
1 & -8 \\
-4 & -9
\end{array}\right]\)

Question 6.
For A = \(\left[\begin{array}{rrr}
2 & 2 & 2 \\
2 & 1 & -3 \\
1 & 0 & 4
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
3 & 3 & 3 \\
3 & 0 & 5 \\
6 & 9 & -1
\end{array}\right]\), C = \(\left[\begin{array}{rrr}
4 & 4 & 4 \\
5 & -1 & 4 \\
7 & 8 & -1
\end{array}\right]\)
compute
(a) 3A – 6B + 9C
(b) 7A – 2B – C
Solution:

Question 7.
For A = \(\left[\begin{array}{rr}
1 & -2 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
2 & 1 \\
-1 & 1
\end{array}\right]\), solve each over S_2×2:
(i) X + 2A = B
(ii) X – A = 3B
(iii) 2X – 3A = 2B – X
Solution:
Given A = \(\left[\begin{array}{rr}
1 & -2 \\
1 & 0
\end{array}\right]\) & B = \(\left[\begin{array}{rr}
2 & 1 \\
-1 & 1
\end{array}\right]\)

Question 8.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
-2 & 5 & 7
\end{array}\right]\) and 2A – 3B = \(\left[\begin{array}{rrr}
4 & 5 & -9 \\
1 & 2 & 3
\end{array}\right]\), find B.
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
-2 & 5 & 7
\end{array}\right]\)
Also, 2A – 3B = \(\left[\begin{array}{rrr}
4 & 5 & -9 \\
1 & 2 & 3
\end{array}\right]\)
⇒ 3B = 2\(\left[\begin{array}{rrr}
1 & 2 & 3 \\
-2 & 5 & 7
\end{array}\right]-\left[\begin{array}{rrr}
4 & 5 & -9 \\
1 & 2 & 3
\end{array}\right]\)
⇒ 3B = \(\left[\begin{array}{rrr}
2 & 4 & 6 \\
-4 & 10 & 14
\end{array}\right]-\left[\begin{array}{rrr}
4 & 5 & -9 \\
1 & 2 & 3
\end{array}\right]\)
⇒ 3B = \(\left[\begin{array}{rrr}
2-4 & 4-5 & 6+9 \\
-4-1 & 10-2 & 14-3
\end{array}\right]\)
⇒ B = \(\frac{1}{3}\left[\begin{array}{rrr}
-2 & -1 & 15 \\
-5 & 8 & 11
\end{array}\right]\)

Question 9.
If A = \(\left[\begin{array}{rrr}
1 & -3 & 2 \\
2 & 0 & 2
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
2 & -1 & -1 \\
1 & 0 & -1
\end{array}\right]\), find the matrix C such that A + B + C ius a zero matrix.
Solution:
Given A + B + C = O

Question 10.
Find the value of k, a non-zero scalar, if
\(2\left[\begin{array}{rrr}
1 & 2 & 3 \\
-1 & -3 & 2
\end{array}\right]+k\left[\begin{array}{lll}
1 & 0 & 2 \\
3 & 4 & 5
\end{array}\right]=\left[\begin{array}{lll}
4 & 4 & 10 \\
4 & 2 & 14
\end{array}\right]\)
Solution:

Thus, their correspounding elements are equal.
∴ 2 + K = 4 ⇒ K = 2,
6 + 2K = 10 ⇒ K = 2
Also, – 2 + 3K = 4
⇒ K = 2; – 6 + 4K = 2
⇒ K = 2 & 4 + 5K = 14
⇒ 5K = 10
⇒ K = 2.
Hence the required value of K be 2.

Question 11.
If A = diag (1-4 8),
B = diag (- 2 3 5),
C = diag (-3 7 10), find
(i) 2A + 3B
(ii) B + 2C – A
(iii) 3A – B + 4C.
Solution:

Question 12.
Solve the matrix equation :
\(2\left[\begin{array}{rr}
x & y \\
z & t
\end{array}\right]+3\left[\begin{array}{rr}
1 & -1 \\
0 & 2
\end{array}\right]=3\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
Solution:
Given \(2\left[\begin{array}{rr}
x & y \\
z & t
\end{array}\right]+3\left[\begin{array}{rr}
1 & -1 \\
0 & 2
\end{array}\right]=3\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
⇒ \(\left[\begin{array}{rr}
2 x & 2 y \\
2 z & 2 t
\end{array}\right]+\left[\begin{array}{rr}
3 & -3 \\
0 & 6
\end{array}\right]=\left[\begin{array}{rr}
9 & 15 \\
12 & 18
\end{array}\right]\)
⇒ \(\left[\begin{array}{rr}
2 x+3 & 2 y-3 \\
2 z & 2 t+6
\end{array}\right]=\left[\begin{array}{rr}
9 & 15 \\
12 & 18
\end{array}\right]\)
Thus, their corresponding elements are equal.
2x + 3 = 9 ⇒ x = 3
& 2t + 6 = 18 ⇒ t = 6;
2y – 3 = 15 ⇒ y = 9
& 2z = 12 ⇒ z = 6

Question 13.
Find x, y, z and w if
\(3\left[\begin{array}{cc}
x & y \\
z & w
\end{array}\right]=\left[\begin{array}{rr}
x & 6 \\
-1 & 2 w
\end{array}\right]+\left[\begin{array}{cc}
4 & x+y \\
z+w & 3
\end{array}\right]\)
Solution:
Given
\(3\left[\begin{array}{ll}
x & y \\
z & w
\end{array}\right]=\left[\begin{array}{rr}
x & 6 \\
-1 & 2 w
\end{array}\right]+\left[\begin{array}{cc}
4 & x+y \\
z+w & 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
3 x & 3 y \\
3 z & 3 w
\end{array}\right]=\left[\begin{array}{cc}
x+4 & 6+x+y \\
-1+z+w & 2 w+3
\end{array}\right]\)
Thus, their corresponding entries are equal.
∴ 3x = x + 4 ⇒ x = 2
3y = 6 + x + y ⇒ 2y = 6 + 2 = 8
⇒ y = 4
3z = – 1 + z + w
⇒ 2z = w – 1 … (1)
Also, 3w = 2w + 3 ⇒ w = 3
∴ from (1); 2z = 3 – 1 = 2 ⇒ z = 1
Thus, x = 2 ; y = 4; z = 1; w = 3

Question 14.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4 \\
5 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
-3 & -2 \\
1 & -5 \\
4 & 3
\end{array}\right]\), then find a matrix C = \(\left[\begin{array}{ll}
p & q \\
r & s \\
t & u
\end{array}\right]\) such that A + B – C = 0.
Solution:
Since A + B – C = 0 ⇒ C = A + B
⇒ \(\left[\begin{array}{ll}
p & q \\
r & s \\
t & u
\end{array}\right]=\left[\begin{array}{ll}
1 & 2 \\
3 & 4 \\
5 & 6
\end{array}\right]+\left[\begin{array}{rr}
-3 & -2 \\
1 & -5 \\
4 & 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
p & q \\
r & s \\
t & u
\end{array}\right]=\left[\begin{array}{cc}
1-3 & 2-2 \\
3+1 & 4-5 \\
5+4 & 6+3
\end{array}\right]=\left[\begin{array}{rr}
-2 & 0 \\
4 & -1 \\
9 & 9
\end{array}\right]\)
⇒ p = – 2; q = 0′ r = 4 ; s = – 1; t = 9; u = 9

Question 15.
Let A = \(\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 0 & 2 \\
3 & 4 & 5
\end{array}\right]\), find a matrix B such that A + B – 4I = 0.
Solution:
Given A + B – 4I = 0
⇒ B = 4I – A
= 4\(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]-\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 0 & 2 \\
3 & 4 & 5
\end{array}\right]\)
⇒ B = \(\left[\begin{array}{lll}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{array}\right]-\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 0 & 2 \\
3 & 4 & 5
\end{array}\right]\)
\(\left[\begin{array}{ccc}
4-2 & 0-3 & 0-5 \\
0-1 & 4-0 & 0-2 \\
0-3 & 0-4 & 4-5
\end{array}\right]\)
⇒ B = \(\left[\begin{array}{rrr}
2 & -3 & -5 \\
-1 & 4 & -2 \\
-3 & -4 & -1
\end{array}\right]\)