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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 9 - Matrices

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Selina Textbook Solutions Chapter 9 - Matrices

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 9 - Matrices.

All Selina textbook questions of Chapter 9 - Matrices solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam. 

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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 9 - Matrices Page/Excercise 9(A)

Solution 1

(i) False

The sum A + B is possible 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.

Solution 2

If two matrices are equal, then their corresponding elements are also equal. Therefore, we have:

x = 3,

y + 2 = 1 y = -1

z - 1 = 2 z = 3

Solution 3

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

(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

Solution 4

(i) A + B =

(ii) B - A

Solution 5

(i)B + C =

(ii)A - C =

(iii)A + B - C =

= =

(iv)A - B +C =

= =

Solution 6

(i)

 

(ii)

 

(iii) Addition is not possible, because both matrices are not of same order.

Solution 7

(i)

Equating the corresponding elements, we get,

3 - x = 7 and y + 2 = 2

Thus, we get, x = - 4 and y = 0.

(ii)

Equating the corresponding elements, we get,

-8 + y = -3 and x - 2 =2

Thus, we get, x = 4 and y = 5.

Solution 8

M =

Mt =

(i)

(i)

Solution 9

We know additive inverse of a matrix is its negative.

Additive inverse of A =

Additive inverse of B =

Additive inverse of C =

Solution 10

(i) X + B = C - A


(ii) A - X = B + C

Solution 11

(i) A + X = B

X = B - A


 

(ii) A - X = B

X = A - B

 


 

(iii) X - B = A

X = A + B

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 9 - Matrices Page/Excercise 9(B)

Solution 1

Solution 2

Comparing the corresponding elements, we get,

12 + 2y = 10 and 3x - 6 = 0

Simplifying, we get, y = -1 and x = 2.

Comparing corresponding the elements, we get,

-x + 8 = 7 and 2x - 4y = -8

Simplifying, we get,

x = 1 and y = = 2.5

Solution 3

(i) 2A - 3B + C

(ii) A + 2C - B

Solution 4

Solution 5

(i)

(ii) C + B =

C = - B =

Solution 6

Comparing the corresponding elements, we get,

2x + 9 = -7 2x = -16 x = -8

3y = 15 y = 5

z = 9

Solution 7

(i) 2A + 3At

(ii) 2At - 3A

(iii)

(iv)

Solution 8

(i) X + 2A = B

X = B - 2A

(ii) 3X + B + 2A = O

3X = -2A - B

(iii) 3A - 2X = X - 2B

3A + 2B = X + 2X

3X = 3A + 2B

Solution 9

3M + 5N

Solution 10

(i) M - 2I =

(ii) 5M + 3I =

Solution 11

2M =

M =

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 9 - Matrices Page/Excercise 9(C)

Solution 1

The number of columns in the first matrix is not equal to the number of rows in the second matrix. Thus, the product is not possible.

Solution 2

Solution 3

  

Solution 4

Comparing the corresponding elements, we get,

5x - 2 = 8 x = 2

20 + 3x = y y = 20 + 6 = 26

Comparing the corresponding elements, we get,

x = 2

-3 + y = -2 y = 1

Solution 5

Hence, A(BC) = (AB)C.

Solution 6

(iii) Product AA (=A2) is not possible as the number of columns of matrix A is not equal to its number of rows.

Solution 7

  

Solution 8

 

Hence, M2 = 2M + 3I.


Solution 9

Given, BA = M2

Comparing the corresponding elements, we get,

a = 2

-2b = -2 b = 1

Solution 10

Solution 11

(iii) Clearly, (A + B)2 A2 + B2

Solution 12

B2 = B + A

A = B2 - B

A = 2(B2 - B)

Solution 13

It is given that A2 = I.

 

Comparing the corresponding elements, we get,

1 + a = 1

Therefore, a = 0

 

 

 

-1 + b = 0 

Therfore, b = 1

Solution 14

Solution 15

Solution 16(iii)

  

Solution 16(ii)

Solution 16(i)

Solution 17

We know, the product of two matrices is defined only when the number of columns of first matrix is equal to the number of rows of the second matrix.

 

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

Clearly, the order of matrix M is 1 x 2.

 

Comparing the corresponding elements, we get,

a = 1 and a + 2b = 2 2b = 2 - 1 = 1 b =

 

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

Clearly, the order of matrix M is 2 x 1.

 

Comparing the corresponding elements, we get,

a + 4b = 13 ....(1)

2a + b = 5 ....(2)

 

Multiplying (2) by 4, we get,

8a + 4b = 20 ....(3)

 

Subtracting (1) from (3), we get,

7a = 7 a = 1

From (2), we get,

b = 5 - 2a = 5 - 2 = 3

Solution 18

Solution 19

Solution 20

AB = BA = B

We know that I × B = B × I = B, where I is the identity matrix.

Hence, A is an identity matrix.

Solution 21

Comparing the corresponding elements, we get,

3a = 3 + a

2a = 3 

a =            

 

 

3b = b b = 0

4c = 4 + c 3c = 4 c =

Solution 22

Clearly, it can be said that:

(P + Q) (P - Q) = P2 - Q2 not true for matrix algebra.

Solution 23

 

 

Hence, ABC ≠ ACB.

Solution 24

 

Thus, CA + B   A + CB


Solution 25

Clearly, the order of matrix X is 2 x 1.

 

Comparing the two matrices, we get,

2x + y = 3 … (1)

x + 3y = -11 … (2)

Multiplying (1) with 3, we get,

6x + 3y = 9 … (3)

Subtracting (2) from (3), we get,

5x = 20

x = 4

From (1), we have:

y = 3 - 2x = 3 - 8 = -5

 

Solution 26

Solution 27

Solution 28

Hence, proved.


Solution 29

 

Comparing the corresponding elements, we get,

2x + 12 = 0

thus, x = -6

6 + 6y = 0

thus, y = -1

Solution 30

Solution 31

(i) True.

Addition of matrices is commutative.

(ii) False.

Subtraction of matrices is not commutative.

(iii) True.

Multiplication of matrices is associative.

(iv) True.

Multiplication of matrices is distributive over addition.

(v) True.

Multiplication of matrices is distributive over subtraction.

(vi) True.

Multiplication of matrices is distributive over subtraction.

(vii) False.

Laws of algebra for factorization and expansion are not applicable to matrices.

(viii) False.

Laws of algebra for factorization and expansion are not applicable to matrices.

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 9 - Matrices Page/Excercise 9(D)

Solution 1

Comparing the corresponding elements, we get,

6x - 10 = 8

6x = 18

x = 3

-2x + 14 = 4y

4y = -6+ 14 = 8

y = 2

Solution 2

Comparing the corresponding elements, we get,

3x + 18 = 15

3x = -3

x = -1

12x + 77 = 10y

10y = -12 + 77 = 65

y = 6.5

Solution 3

(i) x, y Î W (whole numbers)

It can be observed that the above two equations are satisfied when x = 3 and y = 4.

 

(ii) x, y Î Z (integers)

It can be observed that the above two equations are satisfied when x = 3 and y = 4.

Solution 4

(i)

 

 

 

(ii)

 

 

 

 

Solution 5

Solution 6

Let the order of matrix M be a x b.

 

3A x M = 2B

Clearly, the order of matrix M is 2 x 1.


Comparing the corresponding elements, we get,

-3y = -10


 y =           

12x - 9y = 12

 

 

Solution 7

 

Comparing the corresponding elements, we get,

                                 

a + 1 = 5 a = 4

2 + b = 0 b = -2

-1 - c = 3 c = -4

Solution 8

(i)

(ii)

Solution 9

Comparing the corresponding elements, we get,

5x = 5x = 1

6y = 12 y = 2

Solution 10

Solution 11

Given, A + X = 2B + C


Solution 12

 

Given, A2 = B


Comparing the corresponding elements, we get,

x = 36

Solution 13

Solution 14

Solution 15

Solution 16

A =

A2 = A A =

=

AB = A B =

=

=

B2 = B x B =

=

=

A2 + AB + B2 =

=

Solution 17

 

Comparing the corresponding elements, we get,

3a - 8 = 24 3a = 32 a =

24 - 2b = 0 2b = 24 b = 12

11 = 6c c =

Solution 18

A =

BA =

C2 =

BA = C2 =

By comparing,

-2q = -8 q = 4

And p = 8

Solution 19

AB =

Solution 20

=

=

Solution 21

 

 

 

 

 

 

 

 

Solution 22

A2 = 9A + MI

A2 - 9A = mI ….(1)

Now, A2 = AA

Substituting A2 in (1), we have

A2 - 9A = mI

  

Solution 23

 

(i) Let the order of matrix X = m × n

Order of matrix A = 2 × 2

Order of matrix B = 2 × 1

Now, AX = B

   

m = 2 and n = 1

 

Thus, order of matrix X = m × n = 2 × 1

 

Multiplying (1) by 2, we get

4x + 2y = 8 ….(3)

Subtracting (2) from (3), we get

3x = 3

x = 1

Substituting the value of x in (1), we get

2(1) + y = 4

2 + y = 4

y = 2

TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 9 - Matrices  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well. 

Text Book Solutions

ICSE X - Mathematics

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