ICSE Solutions for Selina Concise Chapter 9 Matrices Class 10 Maths
Exercise 9(A)
1. State, whether the following statements are true or false. If false, give a reason.
(i) If A and B are two matrices of orders 3×2 and 2×3 respectively; then their sum A + B is possible.
(ii) The matrices A2×3 and B2×3 are conformable for subtraction.
(iii) Transpose of a 2×1 matrix is a 2×1 matrix.
(iv) Transpose of a square matrix is a square matrix.
(v) A column matrix has many columns and one row.
Solution
(i) False.
The sum of matrices A + B is possible only when the order of both the matrices A and B are same.
(ii) True
(iii) False. Transpose of a 2 ×1 matrix is a 1×2 matrix.
(iv) True
(v) False. A column matrix has only one column and many rows.
2. Given:
, find x, y and z.
Solution
If two matrices are said to be equal, then their corresponding elements are also equal.
∴ x = 3,
y + 2 = 1 so, y = -1
z – 1 = 2 so, z = 3
3. Solve for a, b and c if
(i)
(ii)
Solution
If two matrices are said to be equal, then their corresponding elements are also equal.
Then,
(i) a + 5 = 2 ⇒ a = -3
-4 = b + 4 ⇒ b = -8
2 = c – 1 ⇒ c = 3
(ii) a = 3
a – b = -1
⇒ b = a + 1 = 4
b + c = 2
⇒ c = 2 – b = 2 – 4 = -2
4. If A = [8 -3] and B = [4 -5]; find:
(i) A + B (ii) B – A
Solution
(i) A + B = [8 -3] + [4 -5] = [8+4 -3-5] = [12 -8]
(ii) B – A = [4 -5] – [8 -3] = [4-8 -5-(-3)] = [-4 -2]
5. If A=
, B =
and C =
; find:
(i) B + C
(ii) A – C
(iii) A + B – C
(iv) A – B +C
Solution
(i) B + C = 
(ii) A – C = 
(iii) 
(iv) 
Exercise 9(B)
1. Evaluate:
(i) 3[5 -2]
Solution
3[5 -2] = [3×5 3x-2] = [15 -6]
(ii) 
Solution
(iii) 
Solution
(iv) 
Solution
2. Find x and y if:
(i) 3[4 x] + 2[y -3] = [10 0]
Solution
Taking the L.H.S, we have
3[4 x] + 2[y -3] = [12 3x] + [2y -6] = [(12+2y) (3x–6)]
Now, equating with R.H.S we get
[(12+2y) (3x–6)] = [10 0]
⇒ 12 + 2y = 10 and 3x – 6 = 0
⇒ 2y = -2 and 3x = 6
⇒ y = -1 and x = 2
(ii) 
Solution
We have,
So, equating the matrices we get
-x + 8 = 7 and 2x – 4y = -8
⇒ x = 1 and 2(1) – 4y = -8
⇒ 2 – 4y = -8
⇒ 4y = 10
⇒ y = 5/2
3. 
(i) 2A – 3B + C (ii) A + 2C – B
Solution
(i) 2A – 3B + C
(ii) A + 2C – B
4. 
Solution
Given,
5. 
(i) find the matrix 2A + B.
(ii) find a matrix C such that:
Solution
(i) 2A + B
(ii)
Exercise 9(C)
1. Evaluate: if possible:
If not possible, give reason.
Solution
= [6 + 0] = [6]
= [-2+2 3-8] = [0 -5]
The multiplication of these matrices is not possible as the rule for the number of columns in the first is not equal to the number of rows in the second matrix.
2. If 
and I is a unit matrix of order 2×2, find:
(i) AB
(ii) BA
(iii) AI
(iv) IB
(v) A2
(vi) B2A
Solution
(i) AB 
(ii) BA 
(iii) AI 
(iv) IB 
(v) A2 
(vi) B2 
B2A 
3. If 
find x and y when x and y when A2 = B.
Solution
A2 =
A2 = B
On comparing corresponding elements, we have
4x = 16
⇒ x = 4
And,
1 = -y
y = -1
4. Find x and y, if:
Solution
(i) 
On comparing the corresponding terms, we have
5x – 2 = 8
⇒ 5x = 10
⇒ x = 2
And,
20 + 3x = y
⇒ 20 + 3(2) = y
20 + 6 = y
y = 26
(ii) 
On comparing the corresponding terms, we have
x = 2
And,
-3 + y = -2
⇒ y = 3 – 2 = 1
5. 
(i) (AB) C (ii) A (BC)
Solution
(i) (AB)
(AB)C
(ii) BC
A(BC)
∴ its seen that (AB)C = A(BC)
6. 
is the following possible:
(i) AB (ii) BA (iii) A2
Solution
(i) AB
(ii) BA
(iii) A2 = A x A, is not possible since the number of columns of matrix A is not equal to its number of rows.
7. 
Find A2 + AC – 5B.
Solution
A2
AC
5B
A2 + AC – 5B 
8. If
and I is a unit matrix of the same order as that of M; show that:
M2 = 2M + 3I
Solution
M2
2M + 3I
Thus, M2 = 2M + 3I
9. 
and BA = M2, find the values of a and b.
Solution
BA
M2
So, BA =M2
On comparing the corresponding elements, we have
-2b = -2
⇒ b = 1
And, a = 2
10. 
(i) A – B
(ii) A2
(iii) AB
(iv) A2 – AB + 2B
Solution
(i) A – B
(ii) A2
(iii) AB
(iv) A2 – AB + 2B
11. 
(i) (A + B)2
(ii) A2 + B2
(iii) Is (A + B)2 = A2 + B2 ?
Solution
(i) (A + B)
So, (A + B)2 = (A + B)(A + B)
(ii) A2
B2
A2 + B2
Thus, its seen that (A + B)2 ≠ A2 + B2+
12. Find the matrix A, if B =
and B2 = B + ½A.
Solution
B2
B2 = B + ½A
½A = B2 – B
A = 2(B2 – B)
13. If 
and A2 = I, find a and b.
Solution
And, given A2 = I
So on comparing the corresponding terms, we have
1 + a = 1
Thus, a = 0
And, -1 + b = 0
Thus, b = 1
14. If
then show that:
(i) A(B + C) = AB + AC
(ii) (B – A)C = BC – AC.
Solution
(i) A(B + C)
AB + AC
Thus, A(B + C) = AB + AC
(ii) (B – A)C
BC – AC
Thus, (B – A)C = BC – AC
15. If
Simplify: A2 + BC.
Solution
A2 + BC
Exercise 9(D)
1. Find x and y, if:
Solution
On comparing the corresponding terms, we have
6x – 10 = 8 and -2x + 14 = 4y
⇒ 6x = 18 and y = (14 – 2x)/4
⇒ x = 3 and y = (14 – 2(3))/4
⇒ y = (14 – 6)/ 4
⇒ y = 8/4 = 2
Thus, x = 3 and y = 2
2. Find x and y, if:
Solution
On comparing the corresponding terms, we have
3x + 18 = 15 and 12x + 77 = 10y
⇒ 3x = -3 and y = (12x + 77)/10
⇒ x = -1 and y = (12(-1) + 77)/10
⇒ y = 65/10 = 6.5
Thus, x = -1 and y = 6.5
3. If
; find x and y, if:
(i) x, y ∈ W (whole numbers)
(ii) x, y ∈ Z (integers)
Solution
From the question, we have
x2 + y2 = 25 and -2x2 + y2 = -2
(i) x, y ∈ W (whole numbers)
It can be observed that the above two equations are satisfied when x = 3 and y = 4.
(ii) x, y ∈ Z (integers)
It can be observed that the above two equations are satisfied when x = ± 3 and y = ± 4.
4. 
(i) The order of the matrix X.
(ii) The matrix X.
Solution
(i) Let the order of the matrix be a×b.
Then, we know that
Thus, for multiplication of matrices to be possible
a = 2
And, form noticing the order of the resultant matrix
b = 1
(ii)
On comparing the corresponding terms, we have
2x + y = 7 and
⇒ -3x + 4y = 6
Solving the above two equations, we have
x = 2 and y = 3
Thus, the matrix X is
5. Evaluate:
Solution
6.
3A×M = 2B; find matrix M.
Solution
Given,
3A×M = 2B
And let the order of the matric of M be (a×b)
Now, it’s clearly seen that
a = 2 and b = 1
So, the order of the matrix M is (2 x 1)
Now, on comparing with corresponding elements we have
-3y = -10 and 12x – 9y = 12
⇒ y = 10/3 and 12x – 9(10/3) = 12
⇒ 12x – 30 = 12
⇒ 12x = 42
⇒ x = 42/12 = 7/2
∴ Matrix M = 
7. 
find the values of a, b and c.
Solution
On comparing the corresponding elements, we have
a + 1 = 5 ⇒ a = 4
b + 2 = 0 ⇒ b = -2
-1 – c = 3 ⇒ c = -4
8. 
(i) A(BA)
(ii) (AB)B.
Solution
(i) A(BA)
(ii) (AB)B
9. Find x and y, if:
Solution
Thus, on comparing the corresponding terms, we have
2x + 3x = 5 and 2y + 4y = 12
⇒ 5x = 5 and 6y = 12
⇒ x = 1 and y = 2
10. If matrix
find the matrix ‘X’ and matrix ‘Y’.
Solution
Now,
On comparing with the corresponding terms, we have
-28 – 3x = 10
⇒ 3x = -38
⇒ x = -38/3
And, 20 – 3y = -8
⇒ 3y = 28
⇒ y = 28/3
∴
11. Given 
find the matrix X such that:
A + X = 2B + C
Solution
12. Find the value of x, given that A2 = B,
Solution
Thus, on comparing the terms we get x = 36.