Selina Solutions Concise Maths Class 10 Chapter 9 Matrices Exercise 9(C)

This exercise primarily deals with the multiplication of matrices. Students who wish to secure good marks in their exams can make use of the Selina Solutions for Class 10 Maths. The solutions are created by experts in the subject having vast academic experience. The solutions of the Concise Selina for Class 10 Maths Chapter 9 Matrices Exercise 9(C) are available in PDF format enclosed in the links given below.

Selina Solutions Concise Maths Class 10 Chapter 9 Matrices Exercise 9(C) Download PDF

 

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

 

Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 9 Matrices

Exercise 9(A) Solutions

Exercise 9(B) Solutions

Exercise 9(D) Solutions

Access Selina Solutions Concise Maths Class 10 Chapter 9 Matrices Exercise 9(C)

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


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