# ML Aggarwal Solutions for Chapter 8 Matrices Class 10 Maths ICSE

Here, we are providing the solutions for Chapter 8 Matrices from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of the eighth chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 8 Matrices ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on classifying the matrices, addition, subtraction and multiplication of matrices, finding the value of variables of equation using matrix and also construction of matrices.

### Exercise 8.1

1. Classify the following matrices:

(i) It is square matrix of order 2

(ii) It is column matrix of order 1 × 3

(iii) It is column matrix of order 3 × 1

(iv) It is matrix of order 3 × 2

(v) It is matrix of order 2 × 3

(vi) It is zero matrix or order 2 × 3

2. (i) If a matrix has 4 elements, what are the possible order it can have ?

(ii) If a matrix has 8 elements, what are the possible order it can have ?

(i) It can have 1 × 4, 4 × 1, or 2 × 2 order

(ii) It can have 1 × 8, 8 × 1, 2 × 4 or 4 × 2 order

3. Construct a 2×2 matrix whose elements aij are given by

(a) aij = 2i – j

(b) aij = i, j

Comparing corresponding elements,

2x + y = 5 ...(i)

3x – 2y = 4 …(ii)

Multiply (i) by 2 and (ii) by ‘1’ we get

4x + 2y = 10, 3x – 2y = 4

Adding we get, 7x = 14

⇒ x = 2

Substituting the value of x in (i)

2 × 2 + y = 5

⇒ 4 + y = 5

y = 5 – 4 = 1

Hence x = 2, y = 1

Comparing the corresponding terms, we get,

- y = 2

⇒ y = - 2

3x + y = 1 ⇒ 3x ≠ 1 – y

⇒ 3x = 1 – (-2)

= 1 + 2

= 3

⇒ x = 3/3 = 1

Hence x = 1, y = - 2

Comparing the corresponding terms, we get,

x + 3 = 5

⇒ x = 5 – 3 = 2

⇒ y – 4 = 3

⇒ y = 3 + 4 = 7

x = 2, y = 7

7. Find the values of x, y and z if

Comparing the corresponding elements of equal determinants,

x + 2 = - 5

⇒ x = - 5 – 2 = - 7

∴ x = - 7, 5z = - 20

⇒ z = - 20/5 = - 4

⇒ y2 + y = 6

⇒ y2 + y – 6 = 0

⇒ y2 + 3y - 2y – 6 = 0

⇒ y(y + 3) – 2(y+ 3) = 0

⇒ (y + 3)(y – 2) = 0

Either y + 3 = 0, then y = - 3 or y – 2 = 0, then y = 2

Hence x = - 7, y = - 3, 2, z = - 4

8. Find the values of x, y, a and b if

Comparing corresponding elements

x – 2 = 3, y = 1

x = 3 + 2 = 5

a + 2b = 5 ….(i)

3a – b = 1 ....(ii)

Multiplying (i) by 1 and (ii) by 2

a + 2b = 5, 6a – 2b = 2

Adding, we get, 7a = 7

⇒ a = 1

Substituting the value of a in (i)

1 + 2b = 5

⇒ 2b = 5 – 1 = 4

⇒ b = 2

Hence x = 5, y = 1, a = 1, b = 2

9. Find the values of a, b, c and d if  =

Comparing the corresponding terms, we get

3 = d

⇒ d = 3

⇒ 5 + c = - 1

⇒ c = - 1 – 5

⇒ c = - 6

a + b = 6 and ab = 8

∴ (a – b)2 = (a + b)2 – 4ab

= (6)2 – 4 × 8 = 36 – 32

= 4

= (± 2)2

∴ a - b = ± 2

(i) If a – b = 2

a + b = 6

Adding, we get 2a = 8 ⇒ a = 4

a + b = 6

⇒ 4 + b = 6

⇒ b = 6 – 4 = 2

(ii) If a – b = - 2

a + b = 6

Adding, we get, 2a = 4

⇒ a = 4/2 = 2

a + b = 6

⇒ 2 + b = 6

⇒ b = 6 – 2 = 4

∴ a = 2, b = 4

Hence, a = 4, b = 2, or a = 2, b = 4

c = - 6 and d = 3

10. Find the values of x, y and b, if

Comparing the corresponding terms, we get,

3x + 4y = 2 …..(i)

x – 2y = 4 …(ii)

Multiplying (i) by 1 and (ii) by 2

3x + 4y = 2, 2x – 4y = 8

Adding we get, 5x = 10

⇒ x = 2

Substituting the value of x in (i)

3 × 2 + 4y = 2,

6 + 4y = 2,

4y = 2 – 6 = - 4

y = - 1

∴ x = 2, y = - 1

a + b = 5 …(iii)

2a – b = - 5 …(iv)

### Exercise 8.2

find M + 2N

Find 2A – 3B

3. If

Compute 3A + 4B

4. Given

(i) Find the matrix 2A + B

(ii) Find a matrix 2A + B

Find A + 2B – 3C

6. If

Find the matrix X if:

(i) 3A + X = B

(ii) X – 3B = 2A

(i) 3A + X = B

⇒ X = B – 3A

(ii) X – 3B = 2A

⇒ X = 2A + 3B

7. Solve the matrix equation

find the matrix M

9. Given

Find the matrix X such that A + 2X = 2B + C

A + 2X = 2B + C

2X = 2B + C – A

Find the values of x and y

Find the values of x and y.

Comparing the corresponding terms, we get

4 – 4x = - 8

⇒ - 4x = - 8 – 4

⇒ - 4x = - 12

⇒ x = (- 12/-4) = 3

And y + 5 = 2

⇒ y = 2 – 5 = - 3

∴ x = 3, y = - 3

Find the value of a, b and c.

Comparing the corresponding elements:

a + 1 = 5 ⇒ a = 4

b + 2 = 0 ⇒ b = - 2

- c = 3 ⇒ c = - 3

15. If and 5A + 2B = C, Find the values of a, b, c.

Comparing each term

5a + 6 = 9

⇒ 5a = 9 – 6 = 3

⇒ a = 3/5

⇒ 25 + 2b = - 11

⇒ 2b = - 11 – 25 = - 36

⇒ b = - (36/2) = - 18

c = 6

Hence a = 3/5, b = - 18 and c = 6

### Exercise 8.3

1. If is the product AB possible ? Give a reason. If yes, find AB.

Yes, the product is possible because of number of column in A = number of row in B

i.e., (2 × 2) . (2 × 1) = (2 × 1) is the order of the matrix.

Find 2PQ.

7. Given the matrices:

Find the products of (i) ABC (ii) ACB and state whether they are equal.

find the matrix AB + BA

10. If Find each of the following and state if they are equal.

(i) CA + B

(ii) A + CB

(i) CA + B

(ii)

Compute:

(i) A(B + C)

(ii) (B + C)A

Given that

find A2 and A3. Also state that which of these is equal to A.

From above, it is clear that A3 = A

Show that 6X – X2 = 9I where I is the unit matrix.

Given that

= 9I = R.H.S

Hence, proved.

18. Show that is a solution of the matrix equation X2 – 2X – 3I = 0, where I is the unit matrix of order 2

Given, X2 – 2X – 3I = 0

∴ X2 – 2X – 3I = 0

Hence proved.

20. If A = find the value of x, so that A2 – 0

Comparing 1+ x = 0 ⇒ x = - 1

x= -1

Comparing the corresponding elements

- 3x + 4 = - 5

⇒ - 3x = - 5 – 4 = - 9

- 10 = y

⇒ y = - 10

Hence x = 3, y = - 10

(ii)

Comparing, we get

8x = 16

⇒ x = 16/8 = 2

And 9y = 9

⇒ y = 9/9 = 1

Here x = 2, y = 1

Comparing the corresponding elements

2x + y = 3 …(i)

3x + y = 2 ....(ii)

Subtracting, we get

- x = 1

⇒ x = - 1

Substituting the value of x in (i)

2(-1) + y = 3

⇒ - 2 + y = 3

⇒ y = 3 + 2 = 5

Hence, x = - 1, y = 5

Comparing the corresponding elements

2y = 0 ⇒ y = 0

3x = 9 ⇒ x = 3

Hence x = 3, y = 0

Write down the values of a, b, c and d

Comparing the corresponding elements

a = 3, b = 4, c = 2, d = 5

26. Find the value of x given that A2 = B

A2 = B

⇒ A × A = B

Comparing the corresponding elements of two equal matrices, x = 36.

find the value of x, given that A2 – B.

Corresponding the corresponding elements 3x = 36

⇒ x = 12

Hence x = 12

find x and y when A2 = B

⇒ 4x = 16 and 1 = - y

⇒ x = 4 and y = - 1

⇒ 2x = 6 and 2y = - 4

⇒ x = 6/2 and y = - 4/2

⇒ x = 3 and ⇒ y = - 2

Comparing the corresponding elements

3a + 2 = 11

⇒ 3a = 11 – 2 = 9

∴ a = 9/3 = 3

4a – 3 = b

⇒ b = 4 × 3 – 3

= 12 - 3 = 9

⇒ 3 = c

Hence a= 3, b = 9, c = 3

Comparing the corresponding elements

x – 2 = 8 – x

⇒ x + x = 8 + 2

⇒ 2x = 10

∴ x = 10/2

= 5

32. If A = find x and y so that A2 – xA + yI

Comparing the corresponding elements

3x = 12

⇒ x = 4

2x + y = 7

⇒ 2 ×4 + y = 7

⇒ 8 + y = 7

⇒ y = 7 – 8 = - 1

Hence x = 4, y = - 1

Find x and y such that PQ = 0

Comparing the corresponding elements

6 + 6y = 0

⇒ 6y = - 6

⇒ y = - 1

2x + 12 = 0

⇒ 2x = - 12

⇒ x = - 6

Hence x = - 6, y = - 1

(i) State the order of matrix M

(ii) Find the matrix M

Given (i) M is the order of 1 × 2

Comparing the corresponding elements

x = 1 and x + 2y = 2

⇒ 1 + 2y = 2

⇒ 2y = 2 – 1 = 1

⇒ y = 1/2

Hence x = 1, y = 1/2

(i) the order of matrix X

(ii) the matrix X

2x + y = 7 …(i)

- 3x + 4y = 6 ...(ii)

Multiplying (i) by 3 and (ii) by 2, and adding we get:

6x + 3y = 21

- 6x + 8y = 12

11y = 33 ⇒ y = 3

From (i), 2x = 7 – 3 = 4

⇒ x = 2

Comparing the corresponding elements.

4x = - 4 ⇒ x = - 1

4y = 8 ⇒ y = 2

37. (i) If find the matrix C such that AC = B.

(ii) If find the matrix C such that CA = B.

(i)

Comparing the corresponding elements,

2x – y = - 3 …(i)

-4x + 5y = 2 ….(ii)

Multiplying (i) by 5 and (ii) by 1

10x – 5y = - 15

- 4x + 5y = 2

Adding, we get 6x = - 13

⇒ x = - 13/6

Substituting the value of x in (i)

2(-13/6) – y = - 3

⇒ - 13/3 – y = - 3

⇒ - y = 3 + 13/3

= (-9 + 13)/3

= 4/3

∴ y = - (4/3)

(ii)

Comparing,

2x – 4y = 0

⇒ x – 2y = 0

∴ x = 2y

And –x + 5y = - 3

⇒ - 2y + 5y = - 3

⇒ 3y = - 3

⇒ y = - 1

∴ x = 2y = 2 × (-1) = - 2

38. If A = find matrix B such that BA = I, where I is unity matrix of order 2.

Comparing the corresponding terms, we get

3a – b = 1, - 4a + 2b = 0

⇒ 2b = 4a

⇒ b = 2a

∴ 3a – b = 1

⇒ 3a – 2a = 1

⇒ a = 1

and b = 2a

⇒ b = 2 × 1 = 2

∴ a = 1, b = 2

and 3c – d = 0 ⇒ d = 3c

- 4c + 2d = 1

⇒ - 4c + 2 × 3c = 1

⇒ - 4c + 6c = 1

⇒ 2c = 1

⇒ c = 1/2

And d = 3c = 3 × 1/2 = 3/2

Hence a = 1, b = 2, c = 1/2, d = 3/2

Comparing corresponding elements, we get

∵ - 4a + 5b = 17 ...(i)

2a – b = - 1 …(ii)

- 4c + 5d = 47 ...(iii)

2c – d = - 13 …(iv)

Multiplying (i) by 1 and (ii) by 2

⇒ - 4a + 5b = 17

4a – 2b = - 2

⇒ b = 15/3 = 5

2a – b = - 1

⇒ 2a – 5 = - 1

⇒ 2a = - 1 + 5 = 4

⇒ a = 4/2 = 2

∴ a = 2, b = 5

Again multiplying(iii) by 1 and (iv) by 2,

- 4c + 5d = 47

4c – 2d = - 26

⇒ d = 21/3 = 7

And 2c – d = - 13

⇒ 2c – 7 = - 13

⇒ 2c = - 13 + 7 = - 6

⇒ c = - 6/2 = - 3

∴ c = - 3, d = 7

### Multiple Choice Questions

Choose the correct answer from the given four options (1 to 14):

1. If where aij = i + j, then A is equal to

A = 2×2 where aij = i + j, then A is equal to

(a) x = 2, y = 7

(b) x = 7, y = 2

(c) x = 3, y = 6

(d) x = - 2, y = 7

(d) x = - 2, y = 7

Comparing we get

x + 3 = 5

⇒ x = 5 – 3 = 2

And y – 4 = 3

⇒ y = 3 + 4 = 7

x = 2, y = 7

(a) x = 2, y = 3

(b) x = 2, y = - 3

(c) x = - 2,

(d) x = 3, y = 2

(b) x = 2, y = - 3

Comparing, we get

3x = 6

⇒ x = 6/3 = 2

⇒ - y = 3

⇒ x = 2, y = - 3

(a) - 2

(b) 0

(c) 2

(d) 2

(d) 2

Comparing, we get

y = - 2

And x – 2y = 6

⇒ x – 2 × (-2) = 6

⇒ x + 4 = 6

⇒ x = 6 – 4 = 2

(a) – 3

(b) 1

(c) 3

(d) 5

(c) 3

Comparing, we get

3y = - 3

⇒ y = -3/3 = - 1

4x = 8

⇒ x = 8/4 = 2

x – y = 2 – (-1)

= 2 + 1 = 3

(i) x = 2, y = 6

(b) x = 2, y = -6

(c) x = 3, y = - 4

(d) x = 3, y = - 6

(b) x = 2, y = - 6
Comparing, we get

3x = 6

⇒ x = 6/3 = 2

And 3x – y = 10

2 ×2 – y = 10

⇒ 4 – y = 10

⇒ - y = 10 – 4 = 6

⇒ y = - 6

∴ x = 2, y = -6

Given:

Comparing the corresponding elements

a + 3 = 2a + 1

⇒ 2a – a = 3 – 1

⇒ a = 2

b2 + 2 = 3b

⇒ b2 – 3b + 2 = 0

⇒ b2 – b – 2b + 2 = 0

⇒ b(b – 1) – 2(b – 1) = 0

(b – 1)(b – 2) = 0

Either b – 1 = 0, then b = 1

Or b -2 = 0, then b = 1

Or b – 2 = 0, then b = 2

Hence, a = 2, b = 2 or 1

Comparing the corresponding elements:

3a = 4 + a

⇒ 3a – a = 4

⇒ 2a = 4

∴ a = 2

3b = a + b + 6

⇒ 3b – b = 2 + 6

⇒ 2b = 8

∴ b = 4

3d = 3 + 2d ⇒ 3d – 2d = 3

∴ d = 3

3c = c + d – 1

⇒ 3c – c = 3 – 1

2c = 2

⇒ c = 1

Hence a = 2, b = 4, c = 1, d = 3

Given

4. Determine the matrices A and B when

5. (i) Find the matrix B if and A2 = A + 2B

(i)

Comparing the corresponding elements

4 + 2a = 18

⇒ 2a = 18 – 4 = 14

∴ a = 7

1 + 2b = 7

⇒ 2b = 7 – 1 = 6

∴ b = 3

2 + 2c = 14

⇒ 2c = 14 – 2 = 12

∴ c = 6

3 + 2d = 11

⇒ 2d = 11 – 3 = 8

∴ d = 4

Hence a = 7, b = 3, c = 6, d = 4

(ii)

6. If Compute (AB)C = (CB) A ?

Given

It is clear from above that (AB)C ≠ (CB)A.

7. If find each of the following and state if they are equal:

(i) (A + B)(A – B)

(ii) A2 – B2

Given

(ii)

We see that (A + B)(A – B) ≠ A2 – B2

8. If A = find A2 – 5A – 14I

Where I is unit of order 2 × 2

Comparing the corresponding elements

9 + 3p = 0

⇒3p = - 9

⇒ p = - 3

9 + 3q = 0

⇒ 3q = - 9

⇒ q = - 3

Hence p = -3, q = - 3

Given

Comparing the corresponding elements,

9/25 + 2/5.x = 1

⇒ 2/5.x = 1 = 9/25 = 16/25

x = 16/25 × 5/2 = 8/5

6/25 + 2/5.y = 0

⇒ 2/5y = -6/25

y = - 6/25 × 5/2 = -3/5

Hence x = 8/5, y = -3/5

Comparing the corresponding elements

- a = 1

⇒ a = - 1

-b = 0

⇒ b = 0

c = 0 and d = - 1

Hence a = - 1, b = 0, c = 0, d = - 1

Comparing the corresponding elements

2a – 4 = 0

⇒ 2a = 4

⇒ a = 2

2a – 2b = - 2

⇒ 2 × 2 – 2b = - 2

⇒ 4 – 2b = - 2

⇒ - 2b = - 2 – 4

= - 6

⇒ b = 3

Hence a = 2, b = 3

Find (i) 2A – 3B

(ii) A2

(iii) BA

Given

(∵ cot 45° = 1)

(i) 2A – 3B

(ii) A2 = A × A

(iii)

The solutions provided for Chapter 8 Matrices of ML Aggarwal Textbook. This solutions of ML Aggarwal Textbook of Chapter 8 Matrices contains answers to all the exercises given in the chapter. These solutions are very important if you are a student of ICSE boards studying in Class 10. While preparing the solutions, we kept this in our mind that these should based on the latest syllabus given by ICSE Board.

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