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

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

Question 1.
(a) State whether each product is defined. If so, give its dimensions.
(i) A_4×5 and B_5×3 ; AB
(ii) A_2×1 and Q_2×3 ; PQ
(iii) E_6×2 and F_2×6 ; FE
(iv) C_7×7 and D_6×7 ; CD
(v) P_9×5 and Q_5×9 ; QP
(vi) A_3×5 and B_5×1 ; AB
(b) B is a 5 x 12 matrix. For AB to be defined, which characteristics must A have?
(i) 5 columns
(ii) 12 columns
(iii) 5 rows
(iv) 12 rows
(c) Tell whether matrix multiplication is commutative.
(d) A is a 4 x 2 matrix. Can you find A²? Why or why not ?
Solution:
(a) (i) since no. of columns in A = no. of rows in B
∴ AB is defined and it is of order 4 x 3.

(ii) Since no. of columns in P ≠ no. of rows in Q
∴ PQ is not defined is not.

(iii) Since no. of columns in F = 6 = No. of rows in E
∴ FE exists and is of order 2 x 2

(iv) Since the no. of columns in C ≠ No. of rows in ∀. Thus CD is not defined.

(v) Since the no. of columns in Q = 9 = No. of rows in P. Thus QP is defined and is of order 5 x 5.

(vi) Since the no. of columns in A = 5 = No. of rows in B. Thus AB is defined and is of order 3 x 1.

(b) Since B be a matrix of order 5 x 12 Now AB is to be defined if no. of columns in matrix A = no. of
rows in matrix B. Thus matrix A have 5 columns.

(c) Matrix multiplication is not commutative even if in some cases, AB is defined but BA is not defined.

(d) Given A be a matrix of order 4 x 2
∴ A² = A. A does not exists.
Since number of columns in A ≠ no. of rows in A.

Question 2.
Calculate :
(i) \(\left[\begin{array}{ll}
3 & -1
\end{array}\right]\left[\begin{array}{r}
-2 \\
-10
\end{array}\right]\)
(ii) \(\left[\begin{array}{llll}
5 & 2 & 3 & 4
\end{array}\right]\left[\begin{array}{r}
1 \\
0 \\
-1 \\
6
\end{array}\right]\)
(iii) \(\left[\begin{array}{rr}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\)
(iv) \(\left[\begin{array}{rrr}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{rrr}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]\)
(v) \(\left[\begin{array}{rrr}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{rr}
2 & -3 \\
1 & 0 \\
3 & 1
\end{array}\right]\)
Solution:

Question 3.
(a) Find the value of x in the following:
(i) \(\left[\begin{array}{ll}
x & 7
\end{array}\right]\left[\begin{array}{l}
4 \\
x
\end{array}\right]=[22]\)
(ii) \(\left[\begin{array}{lll}
-2 & x & 4
\end{array}\right]\left[\begin{array}{l}
x \\
3 \\
5
\end{array}\right]=[15]\)
(b) If \(\left[\begin{array}{ll}
2 x & 4
\end{array}\right]\left[\begin{array}{r}
x \\
-8
\end{array}\right]\), find the positive value of x.
(c) \(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\left[\begin{array}{rr}
1 & -3 \\
-2 & 4
\end{array}\right]=\left[\begin{array}{ll}
-4 & 6 \\
-9 & x
\end{array}\right]\)
Solution:
(a) (i) \(\left[\begin{array}{ll}
x & 7
\end{array}\right]\left[\begin{array}{l}
4 \\
x
\end{array}\right]\) = [22]
⇒ [11x] = [22] ⇒ 11x = 22 ⇒ x = 2

(ii) \(\left[\begin{array}{lll}
-2 & x & 4
\end{array}\right]\left[\begin{array}{l}
x \\
3 \\
5
\end{array}\right]\) = [15]
⇒ [- 2x + 3x + 20] – [15]
⇒ x + 20 = 15 ⇒ x = – 5

(b) Given \(\left[\begin{array}{ll}
2 x & 4
\end{array}\right]\left[\begin{array}{r}
x \\
-8
\end{array}\right]\) = 0
⇒ [2x² – 32] = [0] ∴ 2x² – 32 = 0
⇒ 2x² = 32 ⇒ x ± 4
∴ x = 4 [∵ x > 0]

(c) Given
\(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\left[\begin{array}{rr}
1 & -3 \\
-2 & 4
\end{array}\right]=\left[\begin{array}{ll}
-4 & 6 \\
-9 & x
\end{array}\right]\)
⇒ \(\left[\begin{array}{rr}
2-6 & -6+12 \\
5-14 & -15+28
\end{array}\right]=\left[\begin{array}{ll}
-4 & 6 \\
-9 & x
\end{array}\right]\)
⇒ \(\left[\begin{array}{rr}
-4 & 6 \\
-9 & 13
\end{array}\right]=\left[\begin{array}{ll}
-4 & 6 \\
-9 & x
\end{array}\right]\)
∴ x = 13

Question 4.
If A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 1
\end{array}\right]\), B = \(\left[\begin{array}{rr}
-1 & 2 \\
1 & -1
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
2 & 1 \\
-1 & 2
\end{array}\right]\) in each of the problem through (i) to (xii), find a 2 x 2 matrix equal to the given product.
(i) AB
(ii) BA
(iii) AC
(iv) CA
(v) BC
(vi) CB
(vii) A²
(viii) B²
(ix) (A+B)C
(x) C(A+B)
(xi) (A+B)²
(xii) (C-A)²
Do you find that AB ≠ BA, AC ≠ CA, BC ≠ CB, (A + B) C ≠ C (A + B)?
Solution:
Given A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 1
\end{array}\right]\), B = \(\left[\begin{array}{rr}
-1 & 2 \\
1 & -1
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
2 & 1 \\
-1 & 2
\end{array}\right]\)

From (i) & (ii), we find that AB ≠ BA
From (iii) & (iv); AC ≠ CA
From (v) & (vi); BC ≠ CB
From (ix) & (x); (A + B)C ≠ C(A+B)

Question 5.
(a) If A = \(\left[\begin{array}{rr}
i & 0 \\
0 & -i
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
0 & i \\
i & 0
\end{array}\right]\) show that AB ≠ BA, where i² = – 1
(b) If A = \(\left[\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right]\), show that AB ≠ 0 but BA = 0.
Solution:

Question 6.
If A = \(\left[\begin{array}{rrr}
2 & 0 & 3 \\
-1 & 4 & 9
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
1 & 0 & -1 \\
0 & -1 & 1 \\
-1 & 0 & 11
\end{array}\right]\), C = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \\
1 & 0 & 1
\end{array}\right]\) and D = \(\left[\begin{array}{rr}
-1 & -1 \\
2 & 2 \\
-3 & -3
\end{array}\right]\), state the order of each of the following matrices :
(i) AB
(ii) DA
(iii) AD
(iv) CB
(v) BD
Solution:
(i) Since A be a matrix of order 2 x 3 & B be a matrix of order 3 x 3
Here no. of columns in A = no. of rows in B = 3.
Thus AB is defined and is of order 2 x 3

(ii) Here D be a matrix of order 3 x 2 and A be a matrix of order 2 x 3
Here no. of columns in D = 2 = no. of rows in A.
∴ AD is defined and is of order 3 x 3

(iii) Here no. of columns in A = 3 = no. of rows in D
∴DA exists and is of order 2 x 2

(iv) Here C be a matrix of order 4 x 3 and
B be a matrix of order 3 x 3.
Here, no. of columns in C = no. of rows in B = 3
∴ CB exists and is of order 4 x 3.

(v) B be a matrix of order 3 x 3 and D be a matrix of order 3 x 2.
∴ No. of columns in B = no. of rows in D = 3
∴ BD exists and is of order 3 x 2.

Question 7.
If A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
1 & -2 \\
-1 & 0 \\
2 & -1
\end{array}\right]\), obtain the product AB and explain why BA is not defined.
Solution:
Given A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right]_{3 \times 3}\)
& B = \(\left[\begin{array}{rr}
1 & -2 \\
-1 & 0 \\
2 & -1
\end{array}\right]_{3 \times 2}\)
Here no. of columns in A = No. of rows in B = 3
∴ AB exists.
Thus, AB = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
-1 & 0 \\
2 & -1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
0-1+4 & 0+0-2 \\
1-2+6 & -2+0-3 \\
2-3+8 & -4+0-4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3 & -2 \\
5 & -5 \\
7 & -8
\end{array}\right]\)
No. of columns in B ≠ no. of rows in A
∴ BA does not exists.

Question 8.
Find AB and BA, if
(i) A = \(\left[\begin{array}{lll}
4 & -2 & 5
\end{array}\right] \text { and } B=\left[\begin{array}{l}
2 \\
0 \\
1
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{llll}
1 & 2 & 3 & 4
\end{array}\right] \text { and } B=\left[\begin{array}{l}
1 \\
2 \\
3 \\
4
\end{array}\right]\)
(iii) A = \(\left[\begin{array}{ll}
3 & 5 \\
0 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
0 & 11 \\
0 & 7
\end{array}\right]\)
(iv) A = \(\left[\begin{array}{ll}
1 & 2 \\
\cdot 2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
4 & 5 \\
5 & 6
\end{array}\right]\)
Solution:
(i) A = [4 – 2 5]_1×3 and B = \(\left[\begin{array}{l}
2 \\
0 \\
1
\end{array}\right]_{3 \times 1}\)
Here no. of columns in A = no. of rows in B = 3
∴ AB exists.
∴ AB = \(\left[\begin{array}{lll}
4 & -2 & 5
\end{array}\right]\left[\begin{array}{l}
2 \\
0 \\
1
\end{array}\right]\)
= [4 x 2 – 2 x 0 + 5 x 1] = [13]
Here, no. of columns in B = no. of rows in A = 1
∴ BA exists
Thus BA = \(\left[\begin{array}{l}
2 \\
0 \\
1
\end{array}\right]\left[\begin{array}{lll}
4 & -2 & 5
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
8 & -4 & 10 \\
0 & 0 & 0 \\
4 & -2 & 5
\end{array}\right]\)

(ii) Given A = \(\left[\begin{array}{llll}
1 & 2 & 3 & 4
\end{array}\right]_{1 \times 4} ; \mathrm{B}=\left[\begin{array}{l}
1 \\
2 \\
3 \\
4
\end{array}\right]_{4 \times 1}\)
Here, no. of columns in A = no. of rows in B = 4
∴ AB exists.
Thus, AB = \(\left[\begin{array}{llll}
1 & 2 & 3 & 4
\end{array}\right]\left[\begin{array}{l}
1 \\
2 \\
3 \\
4
\end{array}\right]\)
= [1 + 4 + 9 +16] = [30]
Number of columns in B = no. of rows in A = 1
∴ BA is defined

Question 9.
(i) If A = \(\left[\begin{array}{rr}
2 & -3 \\
-2 & 4
\end{array}\right]\), find – A² + 6A.
(ii) If A = \(\left[\begin{array}{rr}
2 & -1 \\
3 & 2
\end{array}\right]\) and \(\left[\begin{array}{rr}
1 & 4 \\
-1 & 7
\end{array}\right]\) find 3A² – 2B.
Solution:

Question 10.
(i) If A = \(\left[\begin{array}{rr}
0 & 3 \\
-7 & 5
\end{array}\right]\), find k so that kA² = 5A – 21 I.
(ii) If A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\), find k such that A² = kA – 2 I.
Solution:
(i) A² = \(\left[\begin{array}{rr}
0 & 3 \\
-7 & 5
\end{array}\right]\left[\begin{array}{rr}
0 & 3 \\
-7 & 5
\end{array}\right]\)

Thus, their corresponding elements are equal.
– 21k = – 21 ⇒ k = 1 other elements also gives k = 1.

(ii) Here A² = A.A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\)

Thus, their corresponding elements are equal.
∴ 1 = 3K – 2 ⇒ K = 1 ; – 2 = – 2K ⇒ K = 1
Also 4 = 4K ⇒ K = 1 & – 4 = – 2K – 2 ⇒ K = 1

Question 11.
(i) If A = \(\left[\begin{array}{rr}
2 & -1 \\
0 & 1
\end{array}\right], \mathbf{B}=\left[\begin{array}{rr}
1 & 0 \\
-1 & -1
\end{array}\right]\), verify whether (A + B) (A + B) = A² + 2AB + B².
Explain your result with proper reasoning.
(ii) If A = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\), B = \(\left[\begin{array}{rr}
0 & -i \\
i & 0
\end{array}\right]\) where i² = – 1, verify whether (A + B) (A + B) = A² + 2AB + B².
Solution:
(i) A + B = \(\left[\begin{array}{rr}
2 & -1 \\
0 & 1
\end{array}\right]+\left[\begin{array}{rr}
1 & 0 \\
-1 & -1
\end{array}\right]=\left[\begin{array}{rr}
3 & -1 \\
-1 & 0
\end{array}\right]\)

From (1) & (2), we have
(A + B)(A + B) ≠ A² + 2AB + B²
Now, (A + B)(A + B) = A(A + B) + B(A + B)
= A.A + AB + BA + BB
= A² + AB + BA + B²
[since distribution law under multiplication over addition holds]
≠ A2 + 2 AB + B²
[∵ AB ≠ BA as matrix multiplication is not commutative]

(ii) Given A = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) & B = \(\left[\begin{array}{rr}
0 & -i \\
i & 0
\end{array}\right]\)

From (1) & (2); we have
L.H.S. = R.H.S.

Question 12.
If A = \(\left[\begin{array}{ll}
4 & 2 \\
1 & 1
\end{array}\right]\) find (A – 2I) (A – 3I) where I is the unit matrix, and express the above product in a matrix form.
Solution:

Question 13.
(i) If f(x) = x² – 5x + 7, find f(A) when A = \(\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\).
(ii) Given f(x) = x² – 5x + 6, find f(A) if A = \(\left[\begin{array}{rrr}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\).
Solution:
(i) Given f(x) = x² – 5x + 7
∴ f(A) = A² – 5A + 7I

Question 14.
(i) Without using the concept of inverse of a matrix, find the matrix \(\left[\begin{array}{ll}
x & y \\
z & u
\end{array}\right]\) such that \(\left[\begin{array}{rr}
5 & -7 \\
-2 & 3
\end{array}\right]\left[\begin{array}{ll}
x & y \\
z & u
\end{array}\right]=\left[\begin{array}{rr}
-16 & -6 \\
7 & 2
\end{array}\right]\)
(ii) Without using the concept of inverse of a matrix, find the matrix X so that
\(\left[\begin{array}{ll}
5 & 4 \\
1 & 1
\end{array}\right] X=\left[\begin{array}{rr}
1 & -2 \\
1 & 3
\end{array}\right]\), where X is a 2 x 2 matrix.
Solution:
(i) Given \(\left[\begin{array}{rr}
5 & -7 \\
-2 & 3
\end{array}\right]\left[\begin{array}{ll}
x & y \\
z & u
\end{array}\right]=\left[\begin{array}{rr}
-16 & -6 \\
7 & 2
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
5 x-7 z & 5 y-7 u \\
-2 x+3 z & -2 y+3 u
\end{array}\right]=\left[\begin{array}{cc}
-16 & -6 \\
7 & 2
\end{array}\right]\)
Thus, their corresponding entries are equal.
5x – 7z = – 16 …(1)
– 2x + 3z = – 7 …(2)
5y – 7u = – 6 …(3)
-2y + 3z = 2 …(4)
on solving (1) & (2); we have
x = 1; z = 3
on solving (3) & eqn. (4); we have
y = – 4 & u = – 2
Thus, required matrix be \(\left[\begin{array}{ll}
x & y \\
z & u
\end{array}\right]=\left[\begin{array}{ll}
1 & -4 \\
3 & -2
\end{array}\right]\)

(ii) Given \(\left[\begin{array}{ll}
5 & 4 \\
1 & 1
\end{array}\right] \mathbf{X}=\left[\begin{array}{rr}
1 & -2 \\
1 & 3
\end{array}\right]\)
⇒ AX = B
where \(A=\left[\begin{array}{ll}
5 & 4 \\
1 & 1
\end{array}\right]_{2 \times 2} \& B=\left[\begin{array}{rr}
1 & -2 \\
1 & 3
\end{array}\right]_{2 \times 2}\)
AX is defined if No. of columns in A = No. of rows in X = 2
Also matrix B which is on R.H.S. of eqn. (1) is of order 2 x 2. Thus, clearly matrix X is of order 2 x 2.
Let X = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Thus, eqn. (1) becomes; \(\left[\begin{array}{ll}
5 & 4 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{cc}
1 & -2 \\
1 & 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
5 a+4 c & 5 b+4 d \\
a+c & b+d
\end{array}\right]=\left[\begin{array}{cc}
1 & -2 \\
1 & 3
\end{array}\right]\)
5a + 4c = 1 … (1)
a + c = 1 … (2)
5b + 4d = – 2 … (3)
b + d = 3 … (4)
Mulitple eqn (2) by (4) and then subtracting from (1); we have
a = -3; c = 4
on solving (3) & eqn. (4); we have
b = – 14; d = 3 – b = 17
Thus, required matrix be \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{cc}
-3 & -14 \\
4 & 17
\end{array}\right]\)

Question 15.
If A = \(\left[\begin{array}{rrr}
-1 & 1 & 0 \\
3 & -3 & 3 \\
5 & -5 & 5
\end{array}\right]\) B = \(\left[\begin{array}{rrr}
0 & 4 & 3 \\
1 & -3 & -3 \\
-1 & 4 & 4
\end{array}\right]\) show that A²B² = A².
Solution:

Question 16.
If A = \(\left[\begin{array}{rr}
2 & 3 \\
-1 & 2
\end{array}\right]\), then show that A² – 4A + 7I. Using this result calculate A³.
Solution:

Question 17.
Given A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
3 & -2 & 1 \\
-2 & 1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
1 & 2 \\
2 & 4 \\
1 & -2
\end{array}\right]\)
Is it possible to compute AB? If it is possible to compute AB, then write
(i) the order of the matrix AB, and
(ii) the value of each of the elements a₁₁, a₁₂ and a₃₁ of the matrix AB.
[Note. Element a_jk means in the jth row and kth column of the matrix.]
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
3 & -2 & 1 \\
-2 & 1 & 0
\end{array}\right]_{3 \times 3}\)
& B = \(\left[\begin{array}{rr}
1 & 2 \\
2 & 4 \\
1 & -2
\end{array}\right]_{3 \times 2}\)
Here number of columns in A = Number of rows in B = 3
∴ AB is defined and it is a matrix of order 3 x 2
Thus, AB = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
3 & -2 & 1 \\
-2 & 1 & 0
\end{array}\right]\left[\begin{array}{rr}
1 & 2 \\
2 & 4 \\
1 & -2
\end{array}\right]\)
= \(\left[\begin{array}{rr}
1-2+1 & 2-4-2 \\
3-4+1 & 6-8-2 \\
-2+2+0 & -4+4+0
\end{array}\right]\)
= \(\left[\begin{array}{rr}
0 & -4 \\
0 & -4 \\
0 & 0
\end{array}\right]\)
∴ a₁₁ = 0, a₁₂ = – 4, and a₃₁ = 0.

Question 18.
Given A = \(\left[\begin{array}{rr}
3 & -1 \\
1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{l}
3 \\
1
\end{array}\right]\), C = \(\left[\begin{array}{r}
1 \\
-2
\end{array}\right]\), find the matrix X, such AX = 3B + 2C.
Solution:
Given
A = \(\left[\begin{array}{rr}
3 & -1 \\
1 & 2
\end{array}\right]_{2 \times 2}\), B = \(\left[\begin{array}{l}
3 \\
1
\end{array}\right]_{2 \times 1}\), C = \(\left[\begin{array}{r}
1 \\
-2
\end{array}\right]_{2 \times 1}\)
∴ 3B + 2C be a matrix of order 2 x 1
Now AX is defined if no. of columns in A = no. of rows in X = 2
∴ AX = 3B +2C is only defined if X be a matrix of order 2 x 1
Let X = \(\left[\begin{array}{l}
a \\
b
\end{array}\right]\)
Thus, \(\left[\begin{array}{cc}
3 & -1 \\
1 & 2
\end{array}\right]\left[\begin{array}{l}
a \\
b
\end{array}\right]=3\left[\begin{array}{l}
3 \\
1
\end{array}\right]+2\left[\begin{array}{c}
1 \\
-2
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
3 a-b \\
a+2 b
\end{array}\right]=\left[\begin{array}{l}
9 \\
3
\end{array}\right]+\left[\begin{array}{c}
2 \\
-4
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
3 a-b \\
a+2 b
\end{array}\right]=\left[\begin{array}{l}
9+2 \\
3-4
\end{array}\right]=\left[\begin{array}{l}
11 \\
-1
\end{array}\right]\)
∴ 3a – 2b = 11 … (1)
& a + 2b = – 1 … (2)
eqn. (1) x 2 + eqn. (2); we have
6a – 2b + a + 2b = 22 – 1 = 21
⇒ 7a = 21 ⇒ a = 3
from (2) ; 2b = – 4 ⇒ b = – 2
Thus, required matrix X = \(\left[\begin{array}{l}
a \\
b
\end{array}\right]=\left[\begin{array}{c}
3 \\
-2
\end{array}\right]\)

Question 19.
A is a 5 x p matrix. B is a 2xq matrix. A is a conformable to B, for pre-multiplication (i.e., AB can be worked out). AB works out to be a 5 x 4 matrix. Write the values of p and q.
Solution:
Given A be a matrix of order 5 x q and
B be a matrix of order 2 x q.
Now AB is defined,
Thus, No. of columns in A = No. of rows in B
i.e. p = 2
Also order of matrix AB is given to be 5×4. Thus, q = 4.

Question 20.
(i) Verily that A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\) satisfies the equation A³ – 4A² + A = 0.
(ii) Show that the matrix A = \(\left[\begin{array}{rrr}
5 & 3 & 1 \\
2 & -1 & 2 \\
4 & 1 & 3
\end{array}\right]\) satisfies the equation A³ – 7A² – 5A + 13I = 0.
(iii) If A = \(\left[\begin{array}{lll}
0 & 1 & 0 \\
0 & 0 & 1 \\
p & q & r
\end{array}\right]\) and I is the identity matrix of order 3, show that A³ = pI + qA + rA².
Solution:

Question 21.
(i) If A = \(\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
a & b & -1
\end{array}\right]\), show that A² = I.
(ii) If A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\), show that A² = 3A.
Solution:
(i) Here,
A² = A.A = \(\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
a & b & -1
\end{array}\right]\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
a & b & -1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
1+0+0 & 0+0+0 & 0+0+0 \\
0+0+0 & 0+1+0 & 0+0+0 \\
a+0-a & 0+b-b & 0+0+1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = 1

(ii) Here,
A² = A.A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
1+1+1 & 1+1+1 & 1+1+1 \\
1+1+1 & 1+1+1 & 1+1+1 \\
1+1+1 & 1+1+1 & 1+1+1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
3 & 3 & 3 \\
3 & 3 & 3 \\
3 & 3 & 3
\end{array}\right]=3\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\) = 3A

Question 22.
Solve the following matrix equation for x :
(i) \(\left[\begin{array}{lll}
1 & 1 & x
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 1 & 0
\end{array}\right]\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]\) = 0
(ii) \(\left[\begin{array}{ll}
x-5 & -1
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
4 \\
1
\end{array}\right]\) = 0
Solution:

Question 23.
(i) If A = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), show that A² = \(\left[\begin{array}{rr}
\cos 2 \theta & \sin 2 \theta \\
-\sin 2 \theta & \cos 2 \theta
\end{array}\right]\)
(ii) If A = \(\left[\begin{array}{rr}
\cos 2 \theta & \sin 2 \theta \\
-\sin 2 \theta & \cos 2 \theta
\end{array}\right]\), show that A² = \(\left[\begin{array}{rr}
\cos 4 \theta & \sin 4 \theta \\
-\sin 4 \theta & \cos 4 \theta
\end{array}\right]\).
Solution:

Question 24.
Matrix R(t) is given by R(t) = \(\left[\begin{array}{rr}
\cos t & \sin t \\
-\sin t & \cos t
\end{array}\right]\), show that R(s) R(t) = R(s + t).
Solution:
R(s) R(t) = \(\left[\begin{array}{cc}
\cos s & \sin s \\
-\sin s & \cos s
\end{array}\right]\left[\begin{array}{cc}
\cos t & \sin t \\
-\sin t & \cot t
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos s \cos t-\sin t \sin s & \sin t \cos t+\cos t \sin s \\
-\sin s \cos t-\cos s \sin t & \cos s \cos t-\sin s \sin t
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos (s+t) & \sin (s+t) \\
-\sin (s+t) & \cos (s+t)
\end{array}\right]\) = R(s + t) = R.H.S

Question 25.
If A is a square matrix such that A² = A, then write the value of 7A – (I + A)³, where I is an identity matrix.
Solution:
Now, 7A – (I + A)³ = 7A – (I + A)(I + A)²
= 7A – (I + A)(I + AI + IA + A²)
= 7A – (I + A)(I + 2A + A) (∵ A² = A& IA = AI = A)
= 7A – (I + A)(I + 3A)
= 7A – I² – 3AI – AI – 3A²
= 7A – I – 3A – A – 3A (∵ A² = A)
= 7A – I – 7A
= 7A – 7A – I (∵ A + B = B + A)
= O – I = – I