Selina Solutions Concise Maths Class 10 Chapter 9 Matrices

A rectangular arrangement of numbers, arranged in rows and columns is called a matrix. Order, elements, types, transpose and operations of a matrix are the major topics covered in this chapter. The exercise problems of Chapter 9 are solved by a very well experienced faculty having in-depth knowledge about these concepts. In addition, the solutions are prepared as per the weightage allotted in the board exam. To secure good marks in the ICSE Class 10 examinations, students can use the solutions PDF for a quick reference and guidance while solving problems of Concise Mathematics Selina. Also, students can find Selina Solutions for Class 10 Mathematics Chapter 9 Matrices PDF, from the links available below.

Selina Solutions Concise Maths Class 10 Chapter 9 Matrices Download PDF

 

selina solutions concise maths class 10 chapter 9a
selina solutions concise maths class 10 chapter 9a
selina solutions concise maths class 10 chapter 9b
selina solutions concise maths class 10 chapter 9b
selina solutions concise maths class 10 chapter 9b
selina solutions concise maths class 10 chapter 9b
selina solutions concise maths class 10 chapter 9c
selina solutions concise maths class 10 chapter 9c
selina solutions concise maths class 10 chapter 9c
selina solutions concise maths class 10 chapter 9c
selina solutions concise maths class 10 chapter 9c
selina solutions concise maths class 10 chapter 9c
selina solutions concise maths class 10 chapter 9c
selina solutions concise maths class 10 chapter 9c
selina solutions concise maths class 10 chapter 9c
selina solutions concise maths class 10 chapter 9d
selina solutions concise maths class 10 chapter 9d
selina solutions concise maths class 10 chapter 9d
selina solutions concise maths class 10 chapter 9d
selina solutions concise maths class 10 chapter 9d
selina solutions concise maths class 10 chapter 9d

 

Exercises of Concise Selina Solutions Class 10 Maths Chapter 9 Matrices

Exercise 9(A) Solutions

Exercise 9(B) Solutions

Exercise 9(C) Solutions

Exercise 9(D) Solutions

Access Selina Solutions Concise Maths Class 10 Chapter 9 Matrices

Exercise 9(A) Page No: 120

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 x 2 and 2 x 3 respectively; then their sum A + B is possible.

(ii) The matrices A2 x 3 and B2 x 3 are conformable for subtraction.

(iii) Transpose of a 2 x 1 matrix is a 2 x 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 x 1 matrix is a 1 x 2 matrix.

(iv) True

(v) False

A column matrix has only one column and many rows.

2. Given: Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(A) - 1, find x, y and z.

Solution:

If two matrices are said to be equal, then their corresponding elements are also equal.

Therefore,

x = 3,

y + 2 = 1 so, y = -1

z – 1 = 2 so, z = 3

3. Solve for a, b and c if

(i) Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(A) - 2

(ii) Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(A) - 3

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=Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(A) - 4, B = Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(A) - 5 and C = Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(A) - 6; find:

(i) B + C (ii) A – C

(iii) A + B – C (iv) A – B +C

Solution:

(i)B + C = Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(A) - 7

(ii)A – C = Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(A) - 8

(iii) Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(A) - 9

(iv) Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(A) - 10


Exercise 9(B) Page No: 120

1. Evaluate:

(i) 3[5 -2]

Solution:

3[5 -2] = [3×5 3x-2] = [15 -6]

(ii) Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 1

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 2

(iii) Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 3

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 4

(iv) Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 5

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 6

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) Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 7

Solution:

We have,

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 8

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.
Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 9

(i) 2A – 3B + C (ii) A + 2C – B

Solution:

(i) 2A – 3B + C

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 10

(ii) A + 2C – B

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 11

4. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 12

Solution:

Given,

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 13

5. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 14

(i) find the matrix 2A + B.

(ii) find a matrix C such that:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 15

Solution:

(i) 2A + B

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 16

(ii)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(B) - 17


Exercise 9(C) Page No: 129

1. Evaluate: if possible:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 1

If not possible, give reason.

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 2

= [6 + 0] = [6]

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 3

= [-2+2 3-8] = [0 -5]

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 4=
Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 5

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 6

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 Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 7 and I is a unit matrix of order 2×2, find:

(i) AB (ii) BA (iii) AI

(iv) IB (v) A2 (iv) B2A

Solution:

(i) AB Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 8

(ii) BA Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 9

(iii) AI Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 10

(iv) IBSelina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 11

(v) A2 Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 12

(vi) B2 Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 13

B2A Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 14

3. If Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 15 find x and y when x and y when A2 = B.

Solution:

A2 Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 16

A2 = B

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 17

On comparing corresponding elements, we have

4x = 16

x = 4

And,

1 = -y

y = -1

4. Find x and y, if:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 18

Solution:

(i)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 19

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

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 20(ii)

On comparing the corresponding terms, we have

x = 2

And,

-3 + y = -2

y = 3 – 2 = 1

5. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 21

(i) (AB) C (ii) A (BC)

Solution:

(i) (AB)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 22

(AB) C

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 23

(ii) BC

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 24

A (BC)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 25

Therefore, its seen that (AB) C = A (BC)

6. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 26 is the following possible:

(i) AB (ii) BA (iii) A2

Solution:

(i) AB

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 27

(ii) BA

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 28

(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. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 29 Find A2 + AC – 5B.

Solution:

A2

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 30

AC

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 31

5B

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 32

A2 + AC – 5B =
Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 33

8. If Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 34 and I is a unit matrix of the same order as that of M; show that:

M2 = 2M + 3I

Solution:

M2

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 35

2M + 3I

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 36

Thus, M2 = 2M + 3I

9. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 37 and BA = M2, find the values of a and b.

Solution:

BA

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 38

M2

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 39

So, BA =M2

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 40

On comparing the corresponding elements, we have

-2b = -2

b = 1

And,

a = 2

10. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 41

(i) A – B (ii) A2 (iii) AB (iv) A2 – AB + 2B

Solution:

(i) A – B

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 42

(ii) A2

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 43

(iii) AB

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 44

(iv) A2 – AB + 2B =

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 45

11. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 46

(i) (A + B)2 (ii) A2 + B2

(iii) Is (A + B)2 = A2 + B2 ?

Solution:

(i) (A + B)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 47

So, (A + B)2 = (A + B)(A + B) =

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 48

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 49

(ii) A2

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 50

B2

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 51

A2 + B2

Thus, its seen that (A + B)2 ≠ A2 + B2+

12. Find the matrix A, if B = Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 52 and B2 = B + ½A.

Solution:

B2

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 53

B2 = B + ½A

½A = B2 – B

A = 2(B2 – B)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 54

13. If Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 55 and A2 = I, find a and b.

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 56

A2

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 Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 57 then show that:

(i) A(B + C) = AB + AC

(ii) (B – A)C = BC – AC.

Solution:

(i) A(B + C)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 58

AB + AC

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 59

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

(ii) (B – A)C

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 60

BC – AC

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 61

Thus, (B – A)C = BC – AC

15. If Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 62 simplify: A2 + BC.

Solution:

A2 + BC

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(C) - 63


Exercise 9(D) Page No: 131

1. Find x and y, if:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 1

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 2

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:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 3

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 4

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; Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 5 ; 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. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 6

(i) The order of the matrix X.

(ii) The matrix X.

Solution:

(i) Let the order of the matrix be a x b.

Then, we know that

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 7

Thus, for multiplication of matrices to be possible

a = 2

And, form noticing the order of the resultant matrix

b = 1

(ii)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 8

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 Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 9

5. Evaluate:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 10

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 11

6. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 12 3A x M = 2B; find matrix M.

Solution:

Given,

3A x M = 2B

And let the order of the matric of M be (a x b)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 13

Now, it’s clearly seen that

a = 2 and b = 1

So, the order of the matrix M is (2 x 1)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 14

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

Therefore,

Matrix M =Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 15

7. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 16

find the values of a, b and c.

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 17

On comparing the corresponding elements, we have

a + 1 = 5 ⇒ a = 4

b + 2 = 0 ⇒ b = -2

-1 – c = 3 ⇒ c = -4

8. Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 18

(i) A (BA) (ii) (AB) B.

Solution:

(i) A (BA)

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 19

(ii) (AB) B

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 20

9. Find x and y, if: Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 21

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 22

Thus, on comparing the corresponding terms, we have

2x + 3x = 5 and 2y + 4y = 12

5x = 5 and 6y = 12

x = 5 and y = 2

10. If matrix Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 23 find the matrix ‘X’ and matrix ‘Y’.

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 24

Now,

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 25

On comparing with the corresponding terms, we have

-28 – 3x = 10

3x = -38

x = -38/3

And,

20 – 3y = -8

3y = 38

y = 38/3

Therefore,

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 26

11. Given Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 27 find the matrix X such that:

A + X = 2B + C

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 28

12. Find the value of x, given that A2 = B,

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 29

Solution:

Selina Solutions Concise Class 10 Maths Chapter 9 ex. 9(D) - 30

Thus, on comparing the terms we get x = 36.

The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2020-21 will be updated soon.


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