ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices Ex 3.3

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices Ex 3.3

Question 1.
If A = [- 1 2 – 5] and B = \(\left[\begin{array}{r}
2 \\
-1 \\
7
\end{array}\right]\), write the orders of AB and BA.
Solution:
Given A = [- 1 2 – 5]
and B = \(\left[\begin{array}{r}
2 \\
-1 \\
7
\end{array}\right]\)
∴ AB = [- 1 2 – 5] \(\left[\begin{array}{r}
2 \\
-1 \\
7
\end{array}\right]\)
= [- 1 × 2 + 2 × (- 1) – 5 × 7]
= [- 39]_{1 × 1}
∴ order of AB be 1 × 1
and BA = \(\left[\begin{array}{r}
2 \\
-1 \\
7
\end{array}\right]\) [- 1 2 – 5]
= \(\left[\begin{array}{rrr}
-2 & 4 & -10 \\
1 & -2 & 5 \\
-7 & 14 & -35
\end{array}\right]\)
Thus, order of matrix is BA be 3 × 3.

Question 1 (old).
Write the order of the product matrix : \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\left[\begin{array}{lll}
2 & 3 & 4
\end{array}\right]\)
Solution:
\(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)_{3 × 1} [2 3 4]_{1 × 3} = \(\left[\begin{array}{rrr}
2 & 3 & 4 \\
4 & 6 & 18 \\
6 & 9 & 12
\end{array}\right]\)_{3 × 3}
Thus the product matrix be 3 × 3.

Question 2.
If A = \(\left[\begin{array}{lll}
2 & 1 & 4 \\
4 & 1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
3 & -1 \\
2 & 2 \\
1 & 3
\end{array}\right]\), then write the orders of AB and BA.
Solution:
Given A be a matrix of order 2 × 3
and B be a matrix of order 3 × 2
Sjnce number of columns in A = no. of rows in B = 3
Tiitis AB exists and it is of order 2 × 2.
Also, No. of columns in B = no. of rows in A = 2
∴ BA exists and is of order 3 × 3.

Question 3.
If A = \(\left[\begin{array}{lll}
2 & 1 & 4 \\
4 & 1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
3 & -1 \\
2 & 2 \\
1 & 3
\end{array}\right]\) A is a matrix of order 3 × 3 and B is a matrix of order 2 × 3, can you find AB?
Solution:
Given A be a rnatri of order 3 × 3
and B be a matrix of order 2 × 3
Here no. of column in A = 3 ≠ no. of rows in B = 2
Thus, AB does not exists.

Question 3 (old).
A = \(\left[\begin{array}{rr}
0 & -1 \\
0 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
3 & 5 \\
0 & 0
\end{array}\right]\), find AB.
Solution:
Given A = \(\left[\begin{array}{rr}
0 & -1 \\
0 & 2
\end{array}\right]\)
and B = \(\left[\begin{array}{ll}
3 & 5 \\
0 & 0
\end{array}\right]\)
∴ AB = \(\left[\begin{array}{rr}
0 & -1 \\
0 & 2
\end{array}\right]\left[\begin{array}{ll}
3 & 5 \\
0 & 0
\end{array}\right]\)
= \(\left[\begin{array}{rr}
0 \times 3-1 \times 0 & 0 \times 5-1 \times 0 \\
0 \times 3+2 \times 0 & 0 \times 5+2 \times 0
\end{array}\right]\)
= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{rr}
1 & 2 \\
-1 & 1 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
3 & -2 & 1 \\
1 & 4 & 2
\end{array}\right]\) and AB = [C_ij], find C₂₃ + C₃₁.
Solution:
AB = \(\left[\begin{array}{rr}
1 & 2 \\
-1 & 1 \\
2 & 3
\end{array}\right]\)_{3 × 2} \(\left[\begin{array}{rrr}
3 & -2 & 1 \\
1 & 4 & 2
\end{array}\right]\)_{2 × 3}
= \(\left[\begin{array}{rrr}
3+2 & -2+8 & 1+4 \\
-3+1 & 2+4 & -1+2 \\
6+3 & -4+12 & 2+6
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
5 & 6 & 5 \\
-2 & 6 & 1 \\
9 & 8 & 8
\end{array}\right]\)
= [C_ij]
∴ C₂₃ + C₃₁ = 1 + 9 = 10.

Question 5.
If A = \(\left[\begin{array}{rr}
2 & 4 \\
3 & 1 \\
-1 & 2
\end{array}\right]\), and B = \(\left[\begin{array}{rrr}
1 & -3 & 2 \\
2 & 1 & 4
\end{array}\right]\) and BA = [C_ij], find C₂₁ + C₂₂.
Solution:
BA = [C_ij]
= \(\left[\begin{array}{rrr}
1 & -3 & 2 \\
2 & 1 & 4
\end{array}\right]_{2 \times 3}\) \(\left[\begin{array}{rr}
2 & 4 \\
3 & 1 \\
-1 & 2
\end{array}\right]_{3 \times 2}\)
= \(\left[\begin{array}{ll}
2-9-2 & 4-3+4 \\
4+3-4 & 8+1+8
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-9 & -5 \\
3 & 17
\end{array}\right]_{2 \times 2}\)
∴ C₂₁ + C₂₂ = 3 + 17 = 20..

Question 5 (old).
Compute the indicated products :
(i) \(\left[\begin{array}{rr}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{rr}
a & -b \\
b & a
\end{array}\right]\) (NCERT)
(iii) \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\) [2 3 4] (NCERT)
Solution:
(i) \(\left[\begin{array}{rr}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{rr}
a & -b \\
b & a
\end{array}\right]\) = \(\left[\begin{array}{rr}
a^2+b^2 & -a b+b a \\
-b a+a b & b^2+a^2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
a^2+b^2 & 0 \\
0 & a^2+b^2
\end{array}\right]\)

(iii) \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\) [2 3 4] = \(\left[\begin{array}{ccc}
1 \times 2 & 1 \times 3 & 1 \times 4 \\
2 \times 2 & 2 \times 3 & 2 \times 4 \\
3 \times 2 & 3 \times 3 & 3 \times 4
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
2 & 3 & 4 \\
4 & 6 & 8 \\
6 & 9 & 12
\end{array}\right]\)

Question 6.
If [x 1] \(\left[\begin{array}{rr}
1 & 0 \\
-2 & 0
\end{array}\right]\) = O, find the value of x.
Solution:
Given [x 1] \(\left[\begin{array}{rr}
1 & 0 \\
-2 & 0
\end{array}\right]\) = 0

[x – 2 0] = [0, 0]
Thus their corresponding elements are equal.
∴ x – 2 = 0
⇒ x = 2.

Question 7.
If \(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\left[\begin{array}{rr}
1 & -3 \\
-2 & 4
\end{array}\right]\) = \(\left[\begin{array}{ll}
-4 & 6 \\
-9 & x
\end{array}\right]\), write the value of x.
Solution:
Given \(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\left[\begin{array}{rr}
1 & -3 \\
-2 & 4
\end{array}\right]=\left[\begin{array}{ll}
-4 & 6 \\
-9 & x
\end{array}\right]\)
\(\left[\begin{array}{ll}
2 \cdot 1+3(-2) & 2(-3)+3 \cdot 4 \\
5 \cdot 1+7(-2) & 5(-3)+7 \cdot 4
\end{array}\right]=\left[\begin{array}{ll}
-4 & 6 \\
-9 & x
\end{array}\right]\)
\(\left[\begin{array}{rr}
-4 & 6 \\
-9 & 13
\end{array}\right]=\left[\begin{array}{ll}
-4 & 6 \\
-9 & x
\end{array}\right]
\)
Thus, x = 13.

Question 8.
Write the value of x + y + z if \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
1 \\
-1 \\
0
\end{array}\right]\).
Solution:
Given \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
1 \\
-1 \\
0
\end{array}\right]\)
\(\left[\begin{array}{l}
1 \cdot x+0 \cdot y+0 \cdot z \\
0 \cdot x+1 \cdot y+0 \cdot z \\
0 \cdot x+0 \cdot y+1 \cdot z
\end{array}\right]=\left[\begin{array}{r}
1 \\
-1 \\
0
\end{array}\right]\)
\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
1 \\
-1 \\
0
\end{array}\right]\)
Thus x = 1 ; y = – 1 and z = 0
∴ x + y + z = 1 – 1 + 0 = 0.

Question (old).
Give an example of two non-zero 2 × 2 matrices A and B such that AB = O.
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
and B = \(\left[\begin{array}{ll}
– 1 & – 1 \\
1 & 1
\end{array}\right]\) ;
Here A ≠ O, B ≠ O
Here AB = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
= O

Question 9.
(i) \(\left[\begin{array}{rr}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\) (NCERT)
(ii) \(\left[\begin{array}{rrr}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{rr}
2 & -3 \\
1 & 0 \\
3 & 1
\end{array}\right]\) (NCERT)
Solution:
(i) \(\left[\begin{array}{rr}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
1 & -2 \\
2 & 3
\end{array}\right]_{2 \times 2}\) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]_{2 \times 3}\)
= \(\left[\begin{array}{lll}
1-4 & 2-6 & 3-2 \\
2+6 & 4+9 & 6+3
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-3 & -4 & 1 \\
8 & 13 & 9
\end{array}\right]_{2 \times 3}\)

(ii) \(\left[\begin{array}{rrr}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{rr}
2 & -3 \\
1 & 0 \\
3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
3 \times 2-1 \times 1+3 \times 3 & 3(-3)-1 \times 0+3 \times 1 \\
-1 \times 2+0 \times 1+2 \times 3 & -1(-3)+0 \times 0+2 \times 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
6-1+9 & -9-0+3 \\
-3+0+6 & 3+0+2
\end{array}\right]\)
= \(\left[\begin{array}{rr}
14 & -6 \\
3 & 5
\end{array}\right]\)

Question 9 (old).
Find the product [x y z] \(\left[\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\).
Solution:
[x y z] \(\left[\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
= [x y z] \(\left[\begin{array}{l}
a x+h y+g z \\
h x+b y+f z \\
g x+f y+c z
\end{array}\right]\)
= [x (ax + hy + gz) + y (hx + by + fz) + z (gx + fy + cz)]
= [ax² + 2hxy + 2gzx + 2hxy + 2fyz]_{1 × 1}

Question 10.
If A = \(\left[\begin{array}{rr}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\) and A² = I, find the value of α² + βγ.
Solution:
A² = I
⇒ \(\left[\begin{array}{rr}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\left[\begin{array}{rr}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\) = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \alpha^2+\beta \gamma
\end{array}\right]\) = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
α² + βγ = 1.

Question 11.
If A = \(\left[\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
0 & 1 \\
1 & 0
\end{array}\right]\), find AB and BA.
Solution:
Given A = \(\left[\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
0 & 1 \\
1 & 0
\end{array}\right]\)
Since A and B are matrices of same order.
∴ AB and BA exists.
Now AB = \(\left[\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)

= \(\left[\begin{array}{ll}
1 \cdot 0+0 \cdot 1 & 1 \cdot 1+0 \cdot 0 \\
0 \cdot 0-1 \cdot 1 & 0 \cdot 1-1 \cdot 0
\end{array}\right]\)

= \(\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\)

and BA = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right]\)

= \(\left[\begin{array}{ll}
0 \cdot 1+1 \cdot 0 & 0 \cdot 0-1(1) \\
1 \cdot 1+0 \cdot 0 & 1 \cdot 0+0 \cdot(-1)
\end{array}\right]\)

= \(\left[\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right]\)

Question 12.
(i) If A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right]\), find AB and BA.
(ii) If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
a & 0 \\
0 & b
\end{array}\right]\) and AB = BA, then show that B is a scalar matrix.
Solution:
(i) Given A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
and B = \(\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right]\)
Since A and B are matrices of same order.
Thus AB and BA exists.
∴ AB = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right]\)

= \(\left[\begin{array}{ll}
1 \times 1+1(-1) & 1(-1)+1 \times 1 \\
1 \times 1+1(-1) & 1(-1)+1 \times 1
\end{array}\right]\)

= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

and BA = \(\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)

= \(\left[\begin{array}{rr}
1 \times 1-1 \times 1 & 1 \times 1-1 \times 1 \\
-1 \times 1+1 \times 1 & -1 \times 1+1 \times 1
\end{array}\right]\)

= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

Thus AB = BA.

(ii) Given A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
and B = \(\left[\begin{array}{ll}
a & 0 \\
0 & b
\end{array}\right]\)
Also, AB = BA
⇒ \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
a & 0 \\
0 & b
\end{array}\right]\) = \(\left[\begin{array}{ll}
a & 0 \\
0 & b
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)

⇒ \(\left[\begin{array}{rr}
a+0 & 0+2 b \\
3 a+0 & 0+4 b
\end{array}\right]\) = \(\left[\begin{array}{rr}
a+0 & 2 a+0 \\
0+3 b & 0+4 b
\end{array}\right]\)

⇒ \(\left[\begin{array}{rr}
a & 2 b \\
3 a & 4 b
\end{array}\right]\) = \(\left[\begin{array}{rr}
a & 2 a \\
3 b & 4 b
\end{array}\right]\)

Thus their corresponding entries are equal.
∴ 2b = 2a
⇒ b = a
and 3a = 3b
⇒ a = b.
∴ B = \(\left[\begin{array}{ll}
a & 0 \\
0 & a
\end{array}\right]\)
which is clearly a scalar matrix since all diagonal elements are equal and all non diagonal entries are zero.

Question 13.
If A = \(\left[\begin{array}{cc}
\alpha & 0 \\
1 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]\) and A = B, then find the value (s) of α.
Solution:
Given A = \(\left[\begin{array}{cc}
\alpha & 0 \\
1 & 1
\end{array}\right]\)
and B = \(\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]\)
Since A² = B
∴ α = 1
⇒ α = ± 1
and α + 1 = 2
⇒ α = 1
Thus common value of α = 1.

Question 14.
If A is a square matrix such that A² = I, then find the simplified value of (A – I)³ + (A + I)³ – 7A.
Solution:
(A – I)³ = (A – I) (A – I) (A – I)
= (A – I) (A² – A . I – I . A + I²)
= (A – I) (A² – A – A + 1)
= (A – 1) (A² – 2A + I) [∵ AI = IA = A and I² = I]
= A (A² – 2A + 1) – I (A² – 2A + I)
= A³ – 2A² + A.I – I.A² + 2I . A – I²
= A³ – 2A² + A – A² + 2A – I = A³ – 3A² + 3A – I ……………(1)
and (A + I)³ = (A + I) (A + I) (A + I)
= (A + I) (A² + A + A + I)
= (A + I) (A² + 2A + I)
= A³ + 2A² + A + A² + 2A + I
= A³ + 3A² + 3A + 1 ……………(2)
∴ (A – I)³ + (A + I)³ – 7A = A³ – 3A² + 3A – I + A³ + 3A² + 3A + I – 7A
[using (1) and (2)]
= 2A³ + 6A – 7A
= 2 A³ – A
= 2A . A² – A
= 2A . I – A [∵ A² = I]
= 2 A – A = A.

Question 14 (old).
Find AB, if A = \(\left[\begin{array}{ll}
6 & 9 \\
2 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
2 & 6 & 0 \\
7 & 9 & 8
\end{array}\right]\). Can you find BA?
Solution:
Given, A = \(\left[\begin{array}{ll}
6 & 9 \\
2 & 3
\end{array}\right]\)_{2 × 2}
and B = \(\left[\begin{array}{lll}
2 & 6 & 0 \\
7 & 9 & 8
\end{array}\right]\)_{2 × 3}
Since number of columns in A = no. of rows in B = 2
Thus AB exists.
∴ AB = \(\left[\begin{array}{ll}
6 & 9 \\
2 & 3
\end{array}\right] \cdot\left[\begin{array}{lll}
2 & 6 & 0 \\
7 & 9 & 8
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
12+63 & 36+81 & 0+72 \\
4+21 & 12+27 & 0+24
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
75 & 117 & 72 \\
25 & 39 & 24
\end{array}\right]\)
Since, no. of columns in B = 3 ≠ no. of rows in A = 2 BA does not exists.
∴ BA does not exists.

Question 15.
Write the following as a single matrix :
(i) [1 – 2 3] \(\left[\begin{array}{rrr}
2 & -1 & 5 \\
0 & 2 & 4 \\
7 & 5 & 0
\end{array}\right]\) – [2 – 5 7]
(ii) \(\left[\begin{array}{rrr}
3 & 2 & 5 \\
7 & -4 & 0
\end{array}\right]\left[\begin{array}{rr}
2 & 2 \\
2 & -1 \\
3 & 5
\end{array}\right]\) – \(\left[\begin{array}{rr}
7 & -8 \\
5 & 9
\end{array}\right]\)
Solution:
(i) [1 – 2 3] \(\left[\begin{array}{rrr}
2 & -1 & 5 \\
0 & 2 & 4 \\
7 & 5 & 0
\end{array}\right]\) – [2 – 5 7]
= [1 × 2 – 2 × 0 + 3 × 7 1 (- 1) – 2 × 2 + 3 × 5 1 × 5 – 2 × 4 + 3 × 0] – [2 – 5 7]
= [23 10 – 3] – [2 – 5 7]
= [23 – 2 10 + 5 – 3 – 7]
= [21 15 – 10]

(ii) \(\left[\begin{array}{rrr}
3 & 2 & 5 \\
7 & -4 & 0
\end{array}\right]\left[\begin{array}{rr}
2 & 2 \\
2 & -1 \\
3 & 5
\end{array}\right]\) – \(\left[\begin{array}{rr}
7 & -8 \\
5 & 9
\end{array}\right]\)

= \(\left[\begin{array}{ll}
3 \times 2+2 \times 2+5 \times 3 & 3 \times 2+2(-1)+5 \times 5 \\
7 \times 2-4 \times 2+0 \times 3 & 7 \times 2-4(-1)+0 \times 5
\end{array}\right]\) – \(\left[\begin{array}{rr}
7 & -8 \\
5 & 9
\end{array}\right]\)

= \(\left[\begin{array}{rr}
25 & 29 \\
6 & 18
\end{array}\right]-\left[\begin{array}{rr}
7 & -8 \\
5 & 9
\end{array}\right]\)

= \(\left[\begin{array}{rr}
18 & 37 \\
1 & 9
\end{array}\right]\)

Question 15 (old).
If A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\), then find AB and BA. Is AB = BA?
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\)_{2 × 3}
and B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\)_{3 × 2}
Since no. of columns in A = no. of rows in B = 3
Thus AB exists.
AB = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
1(2)-2(4)+3(2) & 1(3)-2(5)+3(1) \\
-4(2)+2(4)+5(2) & -4(3)+2(5)+5(1)
\end{array}\right]\)
= \(\left[\begin{array}{rr}
2-8+6 & 3-10+3 \\
-8+8+10 & -12+10+5
\end{array}\right]\)
= \(\left[\begin{array}{rr}
0 & -4 \\
10 & 3
\end{array}\right]\)

and BA = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]_{3 \times 2}\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]_{2 \times 3}\)

[Since no. of columns in B = no. of rows in A = 2]
∴ BA exists.

= \(\left[\begin{array}{lll}
2(1)+3(-4) & 2(-2)+3(2) & 2(3)+3(5) \\
4(1)+5(-4) & 4(-2)+5(2) & 4(3)+5(5) \\
2(1)+1(-4) & 2(-2)+1(2) & 2(3)+1(5)
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-10 & 2 & 21 \\
-16 & +2 & 37 \\
-2 & -2 & 11
\end{array}\right]_{3 \times 3}\)
Thus AB ≠ BA.

Question 16.
(i) A = \(\left[\begin{array}{rr}
\cos 2 \alpha & \sin 2 \alpha \\
-\sin 2 \alpha & \cos 2 \alpha
\end{array}\right]\), find A².
(ii) If M (θ) = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), show that M (x) M (y) = M (x + y).
Solution:
(i) Given A = \(\left[\begin{array}{rr}
\cos 2 \alpha & \sin 2 \alpha \\
-\sin 2 \alpha & \cos 2 \alpha
\end{array}\right]\)

∴ A² = \(\left[\begin{array}{rr}
\cos 2 \alpha & \sin 2 \alpha \\
-\sin 2 \alpha & \cos 2 \alpha
\end{array}\right]\left[\begin{array}{rr}
\cos 2 \alpha & \sin 2 \alpha \\
-\sin 2 \alpha & \cos 2 \alpha
\end{array}\right]\)

= \(\left[\begin{array}{rr}
\cos ^2 2 \alpha-\sin ^2 2 \alpha & 2 \cos 2 \alpha \sin 2 \alpha \\
-2 \sin 2 \alpha \cos 2 \alpha & \cos ^2 2 \alpha-\sin ^2 2 \alpha
\end{array}\right]\)
= \(\left[\begin{array}{rr}
\cos 4 \alpha & \sin 4 \alpha \\
-\sin 4 \alpha & \cos 4 \alpha
\end{array}\right]\)
[∵ cos 2θ = cos² θ – sin² θ
and sin 2θ = 2 sin θ cos θ]

(ii) Given, M (θ) = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\)
L.H.S. = M (x) M (y)
= \(\left[\begin{array}{rr}
\cos x & \sin x \\
-\sin x & \cos x
\end{array}\right]\left[\begin{array}{rr}
\cos y & \sin y \\
-\sin y & \cos y
\end{array}\right]\)

= \(\left[\begin{array}{rr}
\cos x \cos y-\sin x \sin y & \cos x \sin y+\sin x \cos y \\
-\sin x \cos y-\cos x \sin y & \cos x \cos y-\sin x \sin y
\end{array}\right]\)

= \(\left[\begin{array}{rr}
\cos (x+y) & \sin (x+y) \\
-\sin (x+y) & \cos (x+y)
\end{array}\right]\)

= M (x + y) = R.H.S.
[∵ sin (A + B) = sin A cos B + cos A sin B
and cos (A + B) = Cos A cos B – sin A sin B]

Question 17.
If A = \(\left[\begin{array}{rr}
2 & -1 \\
3 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
0 & 4 \\
-1 & 7
\end{array}\right]\), find 3A² – 2B + I.
Solution:
Given, A = \(\left[\begin{array}{rr}
2 & -1 \\
3 & 2
\end{array}\right]\) ; B = \(\left[\begin{array}{rr}
0 & 4 \\
-1 & 7
\end{array}\right]\)
Now A² = A . A
= \(\left[\begin{array}{rr}
2 & -1 \\
3 & 2
\end{array}\right]\left[\begin{array}{rr}
2 & -1 \\
3 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ll}
4-3 & -2-2 \\
6+6 & -3+4
\end{array}\right]\)
= \(\left[\begin{array}{rr}
1 & -4 \\
12 & 1
\end{array}\right]\)

Now, 3A² – 2B + I = \(3\left[\begin{array}{rr}
1 & -4 \\
12 & 1
\end{array}\right]-2\left[\begin{array}{rr}
0 & 4 \\
-1 & 7
\end{array}\right]+\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

= \(\left[\begin{array}{cc}
3 & -12 \\
36 & 3
\end{array}\right]-\left[\begin{array}{cc}
0 & 8 \\
-2 & 14
\end{array}\right]+\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

= \(\left[\begin{array}{cc}
3-0+1 & -12-8+0 \\
36+2+0 & 3-14+1
\end{array}\right]\)

= \(\left[\begin{array}{rr}
4 & -20 \\
38 & -10
\end{array}\right]\)

Question 17 (old).
If A = \(\left[\begin{array}{rrr}
-1 & 3 & 5 \\
1 & -3 & -5 \\
-1 & 3 & 5
\end{array}\right]\), show that A² = A.
Solution:
Given A = \(\left[\begin{array}{rrr}
-1 & 3 & 5 \\
1 & -3 & -5 \\
-1 & 3 & 5
\end{array}\right]\)
∴ A² = A . A
= \(\left[\begin{array}{rrr}
-1 & 3 & 5 \\
1 & -3 & -5 \\
-1 & 3 & 5
\end{array}\right]\left[\begin{array}{rrr}
-1 & 3 & 5 \\
1 & -3 & -5 \\
-1 & 3 & 5
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
-1(-1)+3 \times 1+5(-1) & -1(3)+3(-3)+5 \times 3 & -1 \times 5+3 \times(-5)+5 \times 5 \\
1 \times(-1)-3 \times 1-5 \times(-1) & 1 \times 3-3(-3)-5 \times 3 & 1 \times 5-3 \times(-5)-5 \times 5 \\
-1(-1)+3 \times 1+5 \times(-1) & -1 \times 3+3(-3)+5 \times 3 & -1 \times 5+3 \times(-5)+5 \times 5
\end{array}\right]\)

= \(\left[\begin{array}{rrr}
-1 & 3 & 5 \\
1 & -3 & -5 \\
-1 & 3 & 5
\end{array}\right]\) = A.

Question 18.
If A = [2 1], B = \(\left[\begin{array}{lll}
5 & 3 & 4 \\
8 & 7 & 6
\end{array}\right]\) and C = \(\left[\begin{array}{rrr}
-1 & 2 & 1 \\
0 & 1 & 2
\end{array}\right]\), verify that A (B + C) = AB + AC. (NCERT Exampler)
Solution:
Given A = [2 1], B = \(\left[\begin{array}{lll}
5 & 3 & 4 \\
8 & 7 & 6
\end{array}\right]\) and C = \(\left[\begin{array}{rrr}
-1 & 2 & 1 \\
0 & 1 & 2
\end{array}\right]\)
L.H.S. = A (B + C)
= [2 1] \(\left(\left[\begin{array}{lll}
5 & 3 & 4 \\
8 & 7 & 6
\end{array}\right]+\left[\begin{array}{rrr}
-1 & 2 & 1 \\
0 & 1 & 2
\end{array}\right]\right)\)
= [2 1] \(\left[\begin{array}{lll}
4 & 5 & 5 \\
8 & 8 & 8
\end{array}\right]\)
= [2 × 4 + 1 × 8 2 × 5 + 1 × 8 × 2 × 5 + 1 × 8]_{1 × 3}
= [16 18 18]
R.H.S. = AB + AC
= [2 1] \(\left[\begin{array}{lll}
5 & 3 & 4 \\
8 & 7 & 6
\end{array}\right]\) + [2 1] \(\left[\begin{array}{rrr}
-1 & 2 & 1 \\
0 & 1 & 2
\end{array}\right]\)
= [10 + 8 6 + 7 8 + 6] + [- 2 + 0 4 + 1 2 + 2]
= [18 13 14] + [- 2 5 4]
= [16 18 18]
∴ L.H.S. = R.H.S.

Thus, A (B + C) = AB + AC.

Question 19.
(i) Solve for x and y, given that \(\left[\begin{array}{rr}
2 & -5 \\
1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
1 \\
3
\end{array}\right]\).
(ii) Solve for x and y, given that \(\left[\begin{array}{rr}
x & y \\
3 y & x
\end{array}\right]\left[\begin{array}{l}
1 \\
2
\end{array}\right]=\left[\begin{array}{l}
3 \\
5
\end{array}\right]\).
Solution:
(i) Given, \(\left[\begin{array}{rr}
2 & -5 \\
1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
1 \\
3
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
2 x-3 y \\
x+y
\end{array}\right]=\left[\begin{array}{l}
1 \\
3
\end{array}\right]\)
∴ 2x – 3y = 1 …………..(1)
x + y = 3 …………..(2)
Multiply eqn. (2) by 3 + eqn. (1) ; we get
3x + 2x = 9 + 1
⇒ 5x = 10
⇒ x = 2.
∴ from (2) ;
y = 3 – 2 = 1.

(ii) Given, \(\left[\begin{array}{rr}
x & y \\
3 y & x
\end{array}\right]\left[\begin{array}{l}
1 \\
2
\end{array}\right]=\left[\begin{array}{l}
3 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
x+2 y \\
3 y+2 x
\end{array}\right]=\left[\begin{array}{l}
3 \\
5
\end{array}\right]\)
∴ x + 2y = 3 ………….(1)
2x + 3y = 5 ………….(2)
Multiply eqn. (1) by 2 – eqn. (2) ; we get
y = 1
∴ from (1) ;
x = 3 – 2 = 1
Thus, x = 1 and y = 1.

Question 20.
(i) If A = \(\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]\), show that A² – 5A – 14 I = O.
(ii) If A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 1
\end{array}\right]\), find the value of A² + 2A + 7 I.
Solution:
(i) Here, A² – 5A – 14 I = \(\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]-5\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]-14\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

= \(\left[\begin{array}{rr}
29 & -25 \\
-20 & 24
\end{array}\right]-\left[\begin{array}{rr}
15 & -25 \\
-20 & 10
\end{array}\right]-\left[\begin{array}{rr}
14 & 0 \\
0 & 14
\end{array}\right]\)

= \(\left[\begin{array}{cc}
29-15-14 & -25+25-0 \\
-20+20+0 & 24-10-14
\end{array}\right]\)

= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\) = O

(ii) Given A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 1
\end{array}\right]\)
∴ A² + 2A + 7I = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 1
\end{array}\right]+2\left[\begin{array}{ll}
1 & 2 \\
4 & 1
\end{array}\right]+7\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

= \(\left[\begin{array}{rr}
1+8 & 2+2 \\
4+4 & 8+1
\end{array}\right]+\left[\begin{array}{ll}
2 & 4 \\
8 & 2
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right]\)

= \(\left[\begin{array}{ll}
9 & 4 \\
8 & 9
\end{array}\right]+\left[\begin{array}{ll}
2 & 4 \\
8 & 2
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right]\)

= \(\left[\begin{array}{rr}
18 & 8 \\
16 & 18
\end{array}\right]\)

Question 20 (old).
If A = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right]\), B = \(\left[\begin{array}{lll}
4 & 5 & 6 \\
0 & 1 & 2
\end{array}\right]\) and C = \(=\left[\begin{array}{rrr}
1 & -4 & 1 \\
-2 & 5 & -3 \\
3 & 6 & 5
\end{array}\right]\), verify that (AB) C = A (BC).
Solution:
Given A = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right]\) ;
B = \(\left[\begin{array}{lll}
4 & 5 & 6 \\
0 & 1 & 2
\end{array}\right]\)
and C = \(=\left[\begin{array}{rrr}
1 & -4 & 1 \\
-2 & 5 & -3 \\
3 & 6 & 5
\end{array}\right]\)

From (1) and (2) ; we have
(AB) C = A (BC)

Question 21.
If A = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\), show that (aA + bB) (aA – bB) = (a² + b²) A.
Solution:
Given A = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\)

aA + bB = \(a\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]+b\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\)
= \(\left[\begin{array}{ll}
a & 0 \\
0 & a
\end{array}\right]+\left[\begin{array}{rr}
0 & b \\
-b & 0
\end{array}\right]\)
= \(\left[\begin{array}{rr}
a & b \\
-b & a
\end{array}\right]\)

and aA – bB = \(a\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-b\left[\begin{array}{rr}
0 & 1 \\
-1 & 0
\end{array}\right]\)
= \([\left[\begin{array}{ll}
a & 0 \\
0 & a
\end{array}\right]-\left[\begin{array}{rr}
0 & b \\
-b & 0
\end{array}\right]\)
= \(\left[\begin{array}{rr}
a & -b \\
b & a
\end{array}\right]\)

∴ (aA + bB) (aA – bB) = \(\left[\begin{array}{rr}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{rr}
a & -b \\
b & a
\end{array}\right]\)
= \(\left[\begin{array}{rr}
a^2+b^2 & -a b+a b \\
-a b+a b & a^2+b^2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
a^2+b^2 & 0 \\
0 & a^2+b^2
\end{array}\right]\)
= (a² + b²) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= (a² + b²) A

Question 22.
If A = \(\left[\begin{array}{rr}
4 & 2 \\
-1 & 1
\end{array}\right]\), prove that (A – 2I) (A – 3I) = O.
Solution:
Given A = \(\left[\begin{array}{rr}
4 & 2 \\
-1 & 1
\end{array}\right]\)
Thus, A – 2I = \(\left[\begin{array}{rr}
4 & 2 \\
-1 & 1
\end{array}\right]\) – 2 \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
4 & 2 \\
-1 & 1
\end{array}\right]-\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]\)
= \(\left[\begin{array}{rr}
2 & 2 \\
-1 & -1
\end{array}\right]\)

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

∴ (A – 2I) (A – 3I) = \(\left[\begin{array}{rr}
2 & 2 \\
-1 & -1
\end{array}\right]\left[\begin{array}{rr}
1 & 2 \\
-1 & -2
\end{array}\right]\)
= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\) = O

Question 23.
If A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\), find k so that A² = kA = 2I.
Solution:
Given, A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\)
⇒ A² = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\)
= \(\left[\begin{array}{rr}
9-8 & -6+4 \\
12-8 & -8+4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]\)

Now A² = kA = 2I
⇒ \(\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]=k\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]-\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]\)

⇒ k \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) = \(\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]+\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\)
⇒ k = 1.

Question 26 (old).
If A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
3 & 3 & 3
\end{array}\right]\), B = \(\left[\begin{array}{rr}
3 & 1 \\
5 & 2 \\
-2 & 4
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
4 & 2 \\
-3 & 5 \\
5 & 0
\end{array}\right]\), show that AB = AC though B ≠ C, A ≠ O.
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
3 & 3 & 3
\end{array}\right]\) ;
B = \(\left[\begin{array}{rr}
3 & 1 \\
5 & 2 \\
-2 & 4
\end{array}\right]\)
and C = \(\left[\begin{array}{rr}
4 & 2 \\
-3 & 5 \\
5 & 0
\end{array}\right]\)
Here Clearly B ≠ C, A ≠ O
AB = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
3 & 3 & 3
\end{array}\right]_{2 \times 3}\left[\begin{array}{rr}
3 & 1 \\
5 & 2 \\
-2 & 4
\end{array}\right]_{3 \times 2}\)
= \(\left[\begin{array}{rr}
3+5-2 & 1+2+4 \\
9+15-6 & 3+6+12
\end{array}\right]\)
= \(\left[\begin{array}{rr}
6 & 7 \\
18 & 21
\end{array}\right]\)

AC = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
3 & 3 & 3
\end{array}\right]\left[\begin{array}{rr}
4 & 2 \\
-3 & 5 \\
5 & 0
\end{array}\right]\)
= \(\left[\begin{array}{rr}
4-3+5 & 2+5+0 \\
12-9+15 & 6+15+0
\end{array}\right]\)
= \(\left[\begin{array}{rr}
6 & 7 \\
18 & 21
\end{array}\right]\)
Thus AB = AC ;
But B ≠ C, A ≠ 0.

Question 24.
(i) If A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\), show that f(A) = O, where f(x) = x² – 2x – 3.
(ii) If A = \(\left[\begin{array}{rr}
-1 & 2 \\
3 & 1
\end{array}\right]\), f(A), where f(x) = x² – 2x + 3.
Solution:
(i) Given A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\)
and f(x) = x² – 2x – 3
∴ f(A) = A² – 2A – 3I
= \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]-2\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]-3\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
5 & 4 \\
4 & 5
\end{array}\right]-\left[\begin{array}{ll}
2 & 4 \\
4 & 2
\end{array}\right]-\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)
= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
= O

(ii) Given A = \(\left[\begin{array}{rr}
-1 & 2 \\
3 & 1
\end{array}\right]\)
and f(x) = x² – 2x – 3
∴ f(A) = A² – 2A – 3I
∴ f(A) = \(\left[\begin{array}{rr}
-1 & 2 \\
3 & 1
\end{array}\right]\left[\begin{array}{rr}
-1 & 2 \\
3 & 1
\end{array}\right]-2\left[\begin{array}{rr}
-1 & 2 \\
3 & 1
\end{array}\right]+3\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
1+6 & -2+2 \\
-3+3 & 6+1
\end{array}\right]-\left[\begin{array}{rr}
-2 & 4 \\
6 & 2
\end{array}\right]+\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)
= \(\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right]-\left[\begin{array}{rr}
-2 & 4 \\
6 & 2
\end{array}\right]+\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)
= \(\left[\begin{array}{rr}
12 & -4 \\
-6 & 8
\end{array}\right]\)

Question 25.
If A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), then
(i) find λ, µ so that A² = λA + µI.
(ii) prove that A³ – 4A² + A = O.
Solution:
Given that A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\)
and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

(i) A² = λA + µI
⇒ \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\) = \(\lambda\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]+\mu\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

⇒ \(\left[\begin{array}{rr}
7 & 12 \\
4 & 7
\end{array}\right]=\left[\begin{array}{rr}
2 \lambda & 3 \lambda \\
\lambda & 2 \lambda
\end{array}\right]+\left[\begin{array}{rr}
\mu & 0 \\
0 & \mu
\end{array}\right]\)

⇒ \(\left[\begin{array}{rr}
7 & 12 \\
4 & 7
\end{array}\right]=\left[\begin{array}{cc}
2 \lambda+\mu & 3 \lambda \\
\lambda & 2 \lambda+\mu
\end{array}\right]\)
Thus their corresponding elements are equal.
2λ + µ = 7 ;
3λ = 12
⇒ λ = 4
∴ µ = 7 – 8 = – 1

(ii) A² = A . A
= \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\)

= \(\left[\begin{array}{rr}
7 & 12 \\
4 & 7
\end{array}\right]\)

Question 26.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\), verify that A² – 4A – 5I = O.
Solution:
Given A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\)
L.H.S. = A² – 4A – 5I
= \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-4\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-5\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right]-\left[\begin{array}{lll}
4 & 8 & 8 \\
8 & 4 & 8 \\
8 & 8 & 4
\end{array}\right]-\left[\begin{array}{lll}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5
\end{array}\right]\)
= \(\left[\begin{array}{lll}
9-4-5 & 8-8-0 & 8-8-0 \\
8-8-0 & 9-4-5 & 8-8-0 \\
8-8-0 & 8-8-0 & 9-4-5
\end{array}\right]\)
= \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
= O

Question 27.
(i) LetA and B be square matrices of same order. Does (A + B)² = A² + 2AB + B² hold ? If not, why?
(ii) let A and B be square matrices of order 3 X 3. Is (AB)² = A²B² ? Give reason. (NCERT Exemplar)
Solution:
(i) (A + B)² = (A + B) (A + B)
= A (A + B) + B (A + B)
(Matrix multiplication is distributive over matrix addition)
= A . A + AB + BA + B . B
= A² + AB + BA + B² ≠ A² + 2AB + B²
since AB ≠ BA as matrix multiplication in general not commutative.

(ii) (AB)² = (AB) (AB) [∵ A² = A . A]
= A (AB) B if AB = BA
= (AA) (BB) [associative law holds]
= A²B² , if AB = BA
Thus, (AB)² = A² B² if AB = BA.

Question 28.
If A and B are square matrices of the same order such that AB = BA, prove that
(i) (A + B) (A – B) = A² – B²
(ii) (A + B)² = A² + 2AB + B²
(iii) (A – B)² = A² – 2AB + B²
(iv) (A + B)² = A³ + 3A²B + 3AB² + B³
(v) (A – B)³ = A³ – 3A²B + 3AB² – B³
Solution:
Given AB = BA
(i) L.H.S. = (A + B) (A – B)
= A . A – A . B + B . A + B . B
= A² – B² [∵ AB = BA]
= R.H.S.

(ii) (A + B)² = (A + B) (A + B)
= A² + AB + BA + B²
= A² + 2AB + B² [∵ AB = BA]

(iii) L.H.S. = (A – B)²
= (A – B) (A – B)
= A . A – A . B – A . B + B . B
= A² – 2AB + B² [∵ AB = BA]

(iv) L.H.S. = (A + B)³
= (A + B)² . (A + B)
= (A² + 2AB + B²) . (A + B)
[∵ (A + B)² = (A + B) . (A + B)
= A² + 2AB + B², AS BA = AB]
L.H.S. = A³ + A²B + 2A (BA) + 2A (B . B) + B (BA) + B³
= A³ + 3A²B + 2AB² + (AB²) + B³
[∵AB = BA]
= A³ + 3A²B + 3AB² + B³

(v) L.H.S. = (A – B)³
= (A – B) . (A – B)²
= (A – B) (A – B) (A – B)
= (A – B) (A² – AB – BA + B²)
= (A – B) (A² – 2AB + B²)
[∵ AB = BA given]
= A³ – 2A (AB) + A . B² – B . A² + 2B (AB) – B . B²
= A³ – 2A²B + AB² – (BA) A + 2B (BA) – B³
= A³ – 2A²B + AB² – A (AB) + 2B (AB) – B³
= A³ – 2A²B + AB² – A²B + 2 (BA) B – B³
= A³ – 3A²B + AB² + 2AB² – B³
[∵AB = BA]
= A³ – 3A²B + 3AB² – B³

Question 29.
Find x, if
(i) [1 x 1] \(\left[\begin{array}{rrr}
1 & 3 & 2 \\
2 & 5 & 1 \\
15 & 3 & 2
\end{array}\right]\left[\begin{array}{l}
1 \\
2 \\
x
\end{array}\right]\) = O
(ii) [1 2 1] \(\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 0 & 1 \\
1 & 0 & 2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]\) = O.
Solution:
(i) [1 x 1] = \(\left[\begin{array}{rrr}
1 & 3 & 2 \\
2 & 5 & 1 \\
15 & 3 & 2
\end{array}\right]\left[\begin{array}{l}
1 \\
2 \\
x
\end{array}\right]\) = O

⇒ [1 x 1] \(\left[\begin{array}{c}
1+6+2 x \\
2+10+x \\
15+6+2 x
\end{array}\right]\) = O
⇒ [1 x 1] \(\left[\begin{array}{c}
7+2 x \\
12+x \\
21+2 x
\end{array}\right]\) = O
⇒ [1 (7 + 2x) + x (12 + x) + 1 (21 + 2x)] = O
⇒ [x² + 16x + 28] = O = [0]
⇒ x² + 16x + 28 = 0
⇒ x² + 2x + 14x + 28 = 0
⇒ x (x + 2) + 14 (x + 2) = 0
⇒ (x + 2) (x + 14) = 0
⇒ x = – 2 or x = – 14.

(ii) Given [1 2 1] \(\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 0 & 1 \\
1 & 0 & 2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]\) = 0
⇒ [1 2 1] \(\left[\begin{array}{c}
4 \\
x \\
2 x
\end{array}\right]\) = 0
⇒ [4 + 2x + 2] = 0
⇒ [4x + 4] = 0
⇒ 4x + 4 = 0
⇒ x = – 1.

Question 30.
If A = \(\left[\begin{array}{rr}
2 & 3 \\
-1 & 2
\end{array}\right]\) and find f(x) = x² – 4x + 7, show that f (A) = O. Use this to find A³ and A⁵.
Solution:
Given A = \(\left[\begin{array}{rr}
2 & 3 \\
-1 & 2
\end{array}\right]\)
and f(x) = x² – 4x + 7
f(A) = A² – 4A + 7I
= \(\left[\begin{array}{rr}
2 & 3 \\
-1 & 2
\end{array}\right]\left[\begin{array}{rr}
2 & 3 \\
-1 & 2
\end{array}\right]-4\left[\begin{array}{rr}
2 & 3 \\
-1 & 2
\end{array}\right]+7\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

= \(\left[\begin{array}{rr}
4-3 & 6+6 \\
-2-2 & -3+4
\end{array}\right]-\left[\begin{array}{rr}
8 & 12 \\
-4 & 8
\end{array}\right]+\left[\begin{array}{rr}
7 & 0 \\
0 & 7
\end{array}\right]\)

= \(\left[\begin{array}{rr}
1 & 12 \\
-4 & 1
\end{array}\right]-\left[\begin{array}{rr}
8 & 12 \\
-4 & 8
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right]\)

= \(\left[\begin{array}{rr}
1-8+7 & 12-12+0 \\
-4+4+0 & 1-8+7
\end{array}\right]\)

= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
= O
Thus, A² – 4A + 7I = O …………..(1)
Pre multiplying eqn. (1) by A, we have
⇒ A³ – 4A² + 7 A . I = A . O = O
⇒ A³ – 4 (4A – 7I) + 7A = 0 [using (1)]
⇒ A³ – 16A + 28I + 7A = 0
⇒ A³ – 9A + 28I = 0
∴ A³ = 9A – 28I
= \(9\left[\begin{array}{rr}
2 & 3 \\
-1 & 2
\end{array}\right]-28\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
18 & 27 \\
-9 & 18
\end{array}\right]-\left[\begin{array}{rr}
28 & 0 \\
0 & 28
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-10 & 27 \\
-9 & -10
\end{array}\right]\)
Since A³ = 9A – 28I …………..(2)
pre-multiplying eqn. (2) by A² ; we have
A⁵ = 9A³ – 28A² I3
= 9A³ – 28 A²
= 9 (9A – 28I) – 28 (4A – 7I)
[using (1) and (2)]
= 81A – 252I – 112A + 196I
= – 31A – 56I
= \(-31\left[\begin{array}{rr}
2 & 3 \\
-1 & 2
\end{array}\right]-56\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-62 & -93 \\
31 & -62
\end{array}\right]-\left[\begin{array}{rr}
56 & 0 \\
0 & 56
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-118 & -93 \\
31 & -118
\end{array}\right]\)

Question 31.
If A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{ll}
2^{n-1} & 2^{n-1} \\
2^{n-1} & 2^{n-1}
\end{array}\right]\), for all positive integers n.
Solution:
Given A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
We want to prove that
Aⁿ = \(\left[\begin{array}{ll}
2^{n-1} & 2^{n-1} \\
2^{n-1} & 2^{n-1}
\end{array}\right]\) ∀ n ∈ N
We shall prove it by using mathematical induction on n ∈ N.
For n = 1 ;
A’ = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2^{1-1} & 2^{1-1} \\
2^{1-1} & 2^{1-1}
\end{array}\right]\)
∴ result is true for n = 1
Let us assume that result is true for n = m ∈ N
i.e., A^m = \(\left[\begin{array}{ll}
2^{m-1} & 2^{m-1} \\
2^{m-1} & 2^{m-1}
\end{array}\right]\)
Now we shall prove the result for n = m + 1
i.e., A^m+1 = A^m . A
= \(\left[\begin{array}{ll}
2^{m-1} & 2^{m-1} \\
2^{m-1} & 2^{m-1}
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)

= \(\left[\begin{array}{ll}
2^{m-1}+2^{m-1} & 2^{m-1}+2^{m-1} \\
2^{m-1}+2^{m-1} & 2^{m-1}+2^{m-1}
\end{array}\right]\)

= \(\left[\begin{array}{ll}
2 \cdot 2^{m-1} & 2 \cdot 2^{m-1} \\
2 \cdot 2^{m-1} & 2 \cdot 2^{m-1}
\end{array}\right]\)

= \(\left[\begin{array}{cc}
2^{n m} & 2^m \\
2^m & 2^m
\end{array}\right]\)

= \(\left[\begin{array}{ll}
2^{m+1-1} & 2^{m+1-1} \\
2^{m+1-1} & 2^{m+1-1}
\end{array}\right]\)
Thus result is true for n = m + 1
Hence by mathematical induction, result is true for all n ∈ N.

Question 32.
(i) Find a 2 × 2 matrix B such that B \(\left[\begin{array}{rr}
1 & -2 \\
1 & 4
\end{array}\right]\) = 6 I₂
(ii) If A = \(\left[\begin{array}{rr}
3 & -4 \\
-1 & 2
\end{array}\right]\), find matrix B such that BA = I.
Solution:
(i) Given B \(\left[\begin{array}{rr}
1 & -2 \\
1 & 4
\end{array}\right]\) = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\) ………………(1)
Since product matrix is :f order 2 × 2. The
post multiplier of matrix B s of order 2 × 2.
∴ B must be a matrix of order 2 × 2.
Let B = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
∴ eqn. (1) becomes ;
\(\left[\begin{array}{cc}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
1 & 4
\end{array}\right]=\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\) ;
Thus, \(\left[\begin{array}{ll}
a+b & -2 a+4 b \\
c+d & -2 c+4 d
\end{array}\right]=\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)
∴ a + b = 6 …………….(1)
– 2a + 4b = 0
∴ a – 2b = 0 …………….(2)
c + d = 0 ……………..(3)
and – 2c + 4d = 6
⇒ c – 2d = – 3 ………….(4)
on solving eqn. (1) and eqn. (2) we have
b = 2 ; a = 4
and on solving eqn. (3) and eqn. (4) ; we have
d = 1, c = – 1
∴ required matrix = A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\)

(ii) Given A = \(\left[\begin{array}{rr}
3 & -4 \\
-1 & 2
\end{array}\right]\)
and BA = I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) …………….(1)
Since B and I be of order 2 × 2
∴ B must be a matrix of order 2 × 2.
Let B = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]_{2 \times 2}\)
∴ from (1) ; we have
\(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{rr}
3 & -4 \\
-1 & 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

⇒ \(\left[\begin{array}{ll}
3 a-b & -4 a+2 b \\
3 c-d & -4 c+2 d
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]\)
∴ 3a – b = 1 …………..(1)
– 4a + 2b = 0
⇒ – 2a + b = 0 ……………..(2)
3c – d = 0 ………………(3)
and – 4c + 2d = 0
⇒ – 2a + b = 0 ……………(2)
3c – d = 0 ……………(3)
and – 4c + 2d = 1
⇒ – 2c + d = \(\frac{1}{2}\) …………….(4)
On solving (1) and (2) ;
a = 1 ; b = 2
On solving (3) and (4) ;
c = \(\frac{1}{2}\) ; d = \(\frac{3}{2}\)
Thus required matrix B = \(\left[\begin{array}{cc}
1 & 2 \\
\frac{1}{2} & \frac{3}{2}
\end{array}\right]\).

Question 33.
Find the matrix A such that \(\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] A=\left[\begin{array}{lll}
3 & 3 & 5 \\
1 & 0 &
\end{array}\right]\).
Solution:
\(\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] A=\left[\begin{array}{lll}
3 & 3 & 5 \\
1 & 0 & 1
\end{array}\right]_{2 \times 3}\) ………(1)
⇒ BA = C
No. of columns in B = 2 = No. of rows in A
also No. of columns in A = 3 = No. of columns in C
Thus A be a matrix of order 2 × 3
Let A = \(\left[\begin{array}{lll}
a & b & c \\
d & e & f
\end{array}\right]\)
∴ from (1) ;
\(\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\left[\begin{array}{lll}
a & b & c \\
d & e & f
\end{array}\right]\)
= \(\left[\begin{array}{lll}
3 & 3 & 5 \\
1 & 0 & 1
\end{array}\right]\)
\(\left[\begin{array}{ccc}
a+d & b+e & c+f \\
d & e & f
\end{array}\right]\) = \(\left[\begin{array}{lll}
3 & 3 & 5 \\
1 & 0 & 1
\end{array}\right]\)
∴ d = 1;
e = 0 and f = 1
Also, a + d = 3
⇒ a = 3 – 1 = 2
b + e = 3
⇒ b = 3
and c + f = 5
⇒ c = 5 – 1 = 4
Thus A = \(\left[\begin{array}{lll}
2 & 3 & 4 \\
1 & 0 & 1
\end{array}\right]\).

Question 37 (old).
Find the matrix A such that \(\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right] \quad A=\left[\begin{array}{rr}
-1 & -8 \\
1 & -2 \\
9 & 22
\end{array}\right]\).
Solution:
Since \(\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right]_{3 \times 2} \quad A=\left[\begin{array}{rr}
-1 & -8 \\
1 & -2 \\
9 & 22
\end{array}\right]_{3 \times 2}\) …………….(1)
Since, product of matrices on left hand side is possible
if no. of rows in A = 2 No. of columns of prematrix.
Further product matrix on R.H.S is of order 3 × 2.
∴ A must be a matrix of order 2 × 2.
Let A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]_{2 \times 2}\)
∴ from (1) ; we have
\(\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{rr}
-1 & -8 \\
1 & -2 \\
9 & 22
\end{array}\right]\)
∴ \(\left[\begin{array}{rr}
2 a-c & 2 b-d \\
a+0 & b+0 \\
-3 a+4 c & -3 b+4 d
\end{array}\right]=\left[\begin{array}{rr}
-1 & -8 \\
1 & -2 \\
9 & 22
\end{array}\right]\)
Thus their corresponding entries are equal.
∴ 2a- c = – 1;
a = 1
∴ 2 – c = – 1
⇒ c = 3
Also a = 1, c = 3 satisfies – 3a + 4c = 9
Also, 2b – d = – 8 ;
b = – 2
∴ – 4 – d = – 8
⇒ d = 4
Also b = – 2
and d = 4 satisfies – 3b + 4d = 22
Hence a = 1 ; c = 3 ; b = – 2 ; d = 4
∴ required matrix A = \(\left[\begin{array}{rr}
1 & -2 \\
3 & 4
\end{array}\right]_{2 \times 2}\).

Question 38 (old).
Find the matrix A such that \(\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right] \quad A=\left[\begin{array}{rrr}
-1 & -8 & -10 \\
1 & -2 & -5 \\
9 & 22 & 15
\end{array}\right]\).
Solution:
Since,
\(\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right]_{3 \times 2} \mathrm{~A}=\left[\begin{array}{rrr}
-1 & -8 & -10 \\
1 & -2 & -5 \\
9 & 22 & 15
\end{array}\right]_{3 \times 3}\)
Since product of matrices on L.H.S. is possible only if
no. of rows in A = 2 = No. of columns in prematrix.
Further, product matrix on R.H.S is of order 3 × 3.
∴ A must be a matrix of order 2 × 3
Let A = \(\left[\begin{array}{lll}
a & b & c \\
d & e & f
\end{array}\right]_{2 \times 3}\)
∴ from (1) ; we have
\(\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right]\left[\begin{array}{lll}
a & b & c \\
d & e & f
\end{array}\right]\) = \(\left[\begin{array}{rrr}
-1 & -8 & -10 \\
1 & -2 & -5 \\
9 & 22 & 15
\end{array}\right]\)

⇒ \(\left[\begin{array}{rrr}
2 a-d & 2 b-e & 2 c-f \\
a+0 & b+0 & c+0 \\
-3 a+4 d & -3 b+4 c & 3 c+4 f
\end{array}\right]\) = \(\left.\begin{array}{rr}
– 1 & -8 & -10 \\
1 & -2 & -5 \\
9 & 22 & 15
\end{array}\right]\)
Thus their corresponding entries are equal.
i.e. 2a – d = – 1 …………..(1) :
a = 1
∴ from (1) ;
d = 2 + 1 = 3
Clearly a = 1, d = 3 satisfies – 3a + 4d = 9
Also, 2b – e = – 8 …………….(2)
and b = – 2
∴ from (2); e = 4
Thus, b = – 2 and e = 4 satisfies – 3b + 4e = 22
Also. 2c – f = – 10 ………………(3)
and c = – 5
∴ from (3) ; f = 0
Also, c = – 5, f = 0 satisties – 3c + 4f = 15
Thus required matrix A = \(\left[\begin{array}{rrr}
1 & -2 & -5 \\
3 & 4 & 0
\end{array}\right]\).