Selina Solutions Concise Maths Class 10 Chapter 9 Matrices

 

Selina Solutions Concise Maths Class 10 Chapter 9 Matrices

Salina Maths class 10 solutions Class 10 Chapter 9 Matrices have been provided to you through an online medium, using which you can understand the difficult questions of ICSE Class 10 Mathematics, and can prepare for the upcoming Board exam.

Selina Class 10 ICSE solution will give strength to you and the board will also clear the exam, Mathematics Subject demands time and practice, then keep practicing and you will definitely get success.

Exercise 9(A) Solutions

Exercise 9(B) Solutions

Exercise 9(C) Solutions

Exercise 9(D) Solutions

Concise Maths Class 10 Chapter 9 Matrices Exercise 9(A) 

Question 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.

Question 2. Given: , 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

Question 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

Question 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]

Question 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) 

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                                     Exercise 9(B) 

Question 1. Evaluate:

(i) 3[5 -2]

Solution:

3[5 -2] = [3×5 3x-2] = [15 -6]

(ii) 

Solution:

(iii) 

Solution:

(iv) 

Solution:

Question 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

Question 3.

(i) 2A – 3B + C (ii) A + 2C – B

Solution:

(i) 2A – 3B + C

(ii) A + 2C – B

Question 4. 

Solution:

Given,

Question 5. 

(i) find the matrix 2A + B.

(ii) find a matrix C such that:

Solution:

(i) 2A + B

(ii)

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Exercise 9(C)

Question 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.

Question 2. If  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 

(ii) BA 

(iii) AI 

(iv) IB

(v) A2 

(vi) B2 

B2A 

Question 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

Question 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

Question 5. 

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

Solution:

(i) (AB)

(AB) C

(ii) BC

A (BC)

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

Question 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.

Question 7.  Find A2 + AC – 5B.

Solution:

A2

AC

5B

A2 + AC – 5B =

Question 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

Question 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

Question 10. 

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

Solution:

(i) A – B

(ii) A2

(iii) AB

(iv) A2 – AB + 2B =

Question 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+

Question 12. Find the matrix A, if B =  and B2 = B + ½A.

Solution:

B2

B2 = B + ½A

½A = B2 – B

A = 2(B2 – B)

Question 13. If  and A2 = I, find a and b.

Solution:

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

Question 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

Question 15. If  simplify: A2 + BC.

Solution:

A2 + BC

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Exercise 9(D) Page No: 131

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

Question 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

Question 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.

Question 4. 

(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

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 

Question 5. Evaluate:

Solution:

Question 6.  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)

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

Therefore,

Matrix M =

Question 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

Question 8. 

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

Solution:

(i) A (BA)

(ii) (AB) B

Question 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

Question 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

Therefore,

Question 11. Given  find the matrix X such that:

A + X = 2B + C

Solution:

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

Solution:

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

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