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

### Exercises of Concise Selina Solutions Class 10 Maths Chapter 9 Matrices

## 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 3Â xÂ 2 and 2Â xÂ 3 respectively; then their sum A + B is possible.**

**(ii) The matricesÂ A _{2 x 3}Â andÂ B_{2 x 3}Â are conformable for subtraction.**

**(iii) Transpose of a 2Â xÂ 1 matrix is a 2Â xÂ 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 2Â xÂ 1 matrix is a 1Â xÂ 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.

Therefore,

x = 3,

y + 2 = 1Â so, y = -1

z – 1 = 2Â so, z = 3

**3. Solve for a, b and c if**

**(i)Â **

**(ii)Â **

**Solution: **

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

Then,

(i) a + 5 = 2Â â‡’Â a = -3

-4 = b + 4Â â‡’ b = -8

2 = c – 1Â â‡’Â c = 3

(ii) a = 3

a – b = -1

â‡’Â b = a + 1 = 4

b + c = 2

â‡’ c = 2 – b = 2 – 4 = -2

**4. If A = [8 -3]Â and B =Â [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 â€“ A = [4 -5] – [8 -3] = [4-8 -5-(-3)] = [-4 -2]

**5. If A=, B =Â Â and C =Â ; find:**

**(i) B + C (ii) A – C**

**(iii) A + B – C (iv) A – B +C**

**Solution**:

(i)B + C =Â

(ii)A – C =Â

(iii)

(iv)

Exercise 9(B) Page No: 120

**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 â€“ 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 â€“ 4y = -8

x = 1 and 2(1) â€“ 4y = -8

2 â€“ 4y = -8

4y = 10

y = 5/2

**3.**

**(i) 2A â€“ 3B + C (ii) A + 2C – B**

**Solution: **

(i) 2A â€“ 3B + C

(ii) A + 2C â€“ 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) Page No: 129

**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) A ^{2} (iv) B^{2}A **

**Solution:**

(i) AB

(ii) BA

(iii) AI

(iv) IB

(v) A^{2}

(vi) B^{2}

B^{2}A

**3. If find x and y when x and y when A ^{2} = B.**

**Solution: **

A^{2}

A^{2} = 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 â€“ 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 â€“ 2 = 1

**5. **

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

**Solution: **

(i) (AB)

(AB) C

(ii) BC

A (BC)

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

**6. is the following possible: **

**(i) AB (ii) BA (iii) A ^{2}**

**Solution: **

(i) AB

(ii) BA

(iii) A^{2} = A x A, is not possible since the number of columns of matrix A is not equal to its number of rows.

**7. Find A ^{2} + AC â€“ 5B. **

**Solution:**

A^{2}

AC

5B

A^{2} + AC â€“ 5B =

**8. If and I is a unit matrix of the same order as that of M; show that: **

** M ^{2} = 2M + 3I**

**Solution: **

M^{2}

2M + 3I

Thus, M^{2} = 2M + 3I

**9. and BA = M ^{2}, find the values of a and b. **

**Solution:**

BA

M^{2}

So, BA =M^{2}

On comparing the corresponding elements, we have

-2b = -2

b = 1

And,

a = 2

**10. **

**(i) A â€“ B (ii) A ^{2 } (iii) AB (iv) A^{2} â€“ AB + 2B**

**Solution: **

(i) A â€“ B

(ii) A^{2}

(iii) AB

(iv) A^{2} â€“ AB + 2B =

**11. **

**(i) (A + B) ^{2} (ii) A^{2} + B^{2} **

**(iii) Is (A + B) ^{2} = A^{2} + B^{2} ?**

**Solution: **

(i) (A + B)

So, (A + B)^{2} = (A + B)(A + B) =

(ii) A^{2}

B^{2}

A^{2} + B^{2}

Thus, its seen that (A + B)^{2} â‰ A^{2} + B^{2+}

**12. Find the matrix A, if B = and B ^{2} = B + Â½A.**

**Solution: **

B^{2}

B^{2} = B + Â½A

Â½A = B^{2} â€“ B

A = 2(B^{2} â€“ B)

**13. If and A ^{2} = I, find a and b. **

**Solution:**

A^{2}

And, given A^{2 }= 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 â€“ AC.**

**Solution: **

(i) A(B + C)

AB + AC

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

(ii) (B – A)C

BC â€“ AC

Thus, (B – A)C = BC â€“ AC

**15. If simplify: A ^{2} + BC. **

**Solution: **

A^{2} + BC

Exercise 9(D) Page No: 131

**1. Find x and y, if:**

**Solution: **

On comparing the corresponding terms, we have

6x â€“ 10 = 8 and -2x + 14 = 4y

6x = 18 and y = (14 – 2x)/4

x = 3 and y = (14 â€“ 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, yÂ âˆˆÂ W (whole numbers)**

**(ii) x, yÂ âˆˆÂ Z (integers)**

**Solution: **

From the question, we have

x^{2} + y^{2} = 25 and -2x^{2} + y^{2} = -2

(i) x, yÂ âˆˆÂ W (whole numbers)

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

(ii) x, yÂ âˆˆ Z (integers)

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

**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

**5. Evaluate: **

**Solution:**

**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, itâ€™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 â€“ 9y = 12

y = 10/3 and 12x â€“ 9(10/3) = 12

12x â€“ 30 = 12

12x = 42

x = 42/12 = 7/2

Therefore,

Matrix M =

**7. **

**find the values of a, b and c. **

**Solution: **

On comparing the corresponding elements, we have

a + 1 = 5 â‡’ a = 4

b + 2 = 0 â‡’ b = -2

-1 â€“ c = 3 â‡’ 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 â€˜Xâ€™ and matrix â€˜Yâ€™.**

**Solution: **

Now,

On comparing with the corresponding terms, we have

-28 â€“ 3x = 10

3x = -38

x = -38/3

And,

20 â€“ 3y = -8

3y = 28

y = 28/3

Therefore,

**11. Given find the matrix X such that: **

**A + X = 2B + C**

**Solution**:

**12. Find the value of x, given that A ^{2} = B, **

**Solution: **

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.*