ML Aggarwal Solutions for Class 10 Maths Chapter 8 Matrices consists of accurate solutions, which help the students to complete their homework and prepare well for the exams. These Solutions of ML Aggarwal provide students with an advantage in practical questions. Each step in the solution is explained to match students’ understanding. To score good marks in the Class 10 Mathematics examination, it is advised they solve questions provided at the end of each chapter in the ML Aggarwal Textbooks for Class 10. This chapter deals with matrices, types of matrices and product of matrices. A rectangular array of m x n numbers (real or complex) in the form of m horizontal lines (called rows) and n vertical lines (called columns), is called a matrix of order m by n, written as m x n matrix. Such an array is enclosed by [ ] or ( ). This chapter of ML Aggarwal Solutions for Class 10 contains three exercises with chapter test. These solutions provided by BYJU’S cover all these concepts, with detailed explanations.

ML Aggarwal Solutions for Class 10 Maths Matrices:

Download PDF

carouselExampleControls111

Previous



Next

Access answers to ML Aggarwal Solutions for Class 10 Maths Chapter 8 – Matrices

Exercise 8.1

1. Classify the following matrices:

Solution:

It is a square matrix of order 2

Solution:

It is a row matrix of order 1 × 3

Solution:

It is a column matrix of order 3 × 1

Solution:

It is a matrix of order 3 × 2

Solution:

It is a matrix of order 2 × 3

Solution:

It is a zero matrix of order 2 × 3

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

Solution:

It can have 1 × 4, 4 × 1 or 2 × 2 order.

(ii) If a matrix has 4 elements, what are the possible orders it can have?

Solution:

It can have 1 × 8, 8 × 1, 2 × 4 or 4 × 2 order.

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

(i) a_ij = 2i – j

(ii) a_ij =i.j

Solution:

(i) Given a_ij = 2i – j

Therefore matrix of order 2 × 2 is

(ii) Given a_ij =i.j

Therefore matrix of order 2 × 2 is

4. Find the values of x and y if:

Solution:

Given

Now by comparing the corresponding elements,

2x + y = 5 ….. i

3x – 2y = 4 ….ii

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

4x + 2y = 10 and 3x – 2y = 4

By adding we get

7x = 14

x = 14/7

x = 2

Substituting the value of x in (i)

4 + y = 5

y = 5 – 4

y = 1

Hence x = 2 and y = 1

5. Find the value of x if

Solution:

Given

Comparing the corresponding terms of the given matrix, we get

-y = 2

Therefore y = -2

Again we have

3x + y = 1

3x = 1 – y

Substituting the value of y we get

3x = 1 – (-2)

3x = 1 + 2

3x = 3

x = 3/3

x = 1

Hence x = 1 and y = -2

6. If

Find the values of x and y.

Solution:

Given

Comparing the corresponding terms, we get

x + 3 = 5

x = 5 – 3

x = 2

Again we have

y – 4 = 3

y = 3 + 4

y = 7

Hence x = 2 and y = 7

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

Solution:

Given

Comparing the corresponding elements of the given matrix, then we get

x + 2 = -5

x = -5 – 2

x = -7

Also we have 5z = -20

z = -20/5

z = – 4

Again from the given matrix, we have

y² + y – 6 = 0

The above equation can be written as

y² + 3y – 2y – 6 = 0

y (y + 3) – 2 (y + 3) = 0

y + 3 = 0 or y – 2 = 0

y = -3 or y = 2

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

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

Solution:

Given

Comparing the corresponding elements

x – 2 = 3 and y = 1

x = 2 + 3

x = 5

again we have

a + 2b = 5….. i

3a – b = 1 ……ii

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

a + 2b = 5

6a – 2b = 2

Now by adding the above equations we get

7a = 7

a = 7/7

a = 1

Substituting the value of a in (i), we get

1 + 2b = 5

2b = 5 -1

2b = 4

b = 4/2

b = 2

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

Solution:

Given

Comparing the corresponding terms, we get

3 = d

d = 3

Also we have

5 + c = -1

c = -1 – 5

c = -6

Also we have,

a + b = 6 and a b = 8

we know that,

(a – b)² = (a + b)² – 4 ab

(6)² – 32 = 36 – 32 = 4 = (± 2)²

a – b = ± 2

If a – b = 2

a + b = 6

Adding the above two equations we get

2a = 4

a = 4/2

a = 2

b = 6 – 4

b = 2

Again we have a – b = -2

And a + b = 6

Adding above equations we get

2a = 4

a = 4/2

a = 2

Also, b = 6 – 2 = 4

a = 2 and b = 4

Exercise 8.2

Solution:

Solution:

3. Simplify:

Solution:

Given,

4.

Solution:

5.

Find the matrix X if:

(i) 3A + X = B

(ii) X – 3B = 2A

Solution:

6. Solve the matrix equation

Solution:

7.

Solution:

8.

Solution:

9.

Solution:

10.

Solution:

On comparing the corresponding elements, we have

8 + y = 0

Then, y = -8

And, 2x + 1 = 5

2x = 5 – 1 = 4

x = 4/2 = 2

Therefore, x = 2 and y = -8

11.

Solution:

On comparing the corresponding terms, we have

2x + 1 = 5

2x = 5 -1 = 4

x = 4/2 = 2

And,

8 + y = 0

y = -8

And, z = 7

Therefore, x = 2, y = -8 and z = 7.

12.

Solution:

Now, comparing the corresponding terms, we get

4 – 4x = -8

4 + 8 = 4x

12 = 4x

x = 12/4

x = 3

And, y + 5 = 2

y = 2 – 5 =

y = -3

Therefore, x = 3 and y = -3

13.

Find the value of a, b and c.

Solution:

Next, on comparing the corresponding terms, we have

a + 1 = 5 ⇒ a = 4

b + 2 = 0 ⇒ b = -2

-c = 3 ⇒ c = -3

Therefore, the value of a, b and c are 4, -2 and -3 respectively.

14. and 5A + 2B = C, find the values of a, b and c.

Solution:

On comparing the corresponding terms, we get

5a + 6 = 9

5a = 9 – 6

5a = 3

a = 3/5

And,

25 + 2b = -11

2b = -11 – 25

2b = -36

b = -36/2

b = -18

And, c = 6

Therefore, the value of a, b and c are 3/5, -18 and 6 respectively.

Exercise 8.3

Solution:

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

That is order of matrix is 2 × 1

Solution:

3.

Solution:

4.

Solution:

5. Given matrices:

Find the products of

(i) ABC

(ii) ACB and state whether they are equal.

Solution:

Now consider,

6.

Solution:

Given

7.

Solution:

Solution:

(i) A(B + C) (ii) (B + C)A

Solution:

10.

Find the matrix C(B – A).

Solution:

Given,

Find A² + AB + B².

Solution:

Given,

12. , find A² + AC – 5B.

Solution:

Given,

13. If , find AC + B² – 10C.

Solution:

Given,

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

Solution:

Given,

From above, its clearly seen that A³ = A.

15. If show that 6X – X² = 9I where I is the unit matrix.

Solution:

Given,

– Hence proved

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

the unit matrix of order 2.

Solution:

Given,

17. Find the matrix 2 × 2 which satisfies the equation

Solution:

Given,

18. If find the value of x, so that A² – 0

Solution:

Given,

On comparing,

1 + x = 0

∴ x = -1

19.

Solution:

Comparing the corresponding elements,

– 3x + 4 = -5

-3x = -5 – 4 = -9

x = -9/-3 = 3

Therefore, x = 3 and y = -10.

Comparing, we get

8x = 16

⇒ x = 16/8 = 2

And, 9y = 9

y = 9/9 = 1

20. Find the values of x and y if

Solution:

Given,

On comparing the corresponding elements, we have

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

Therefore, x = -1 and y = 5.

Chapter test

Solution:

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 = 2

Hence a = 2, b = 2 or 1

Solution:

Now comparing the corresponding elements

3a = 4 + a

a – a = 4

2a = 4

Therefore, a = 2

3b = a + b + 6

3b – b = 2 + 6

2b = 8

Therefore, b = 4

3d = 3 + 2d

3d – 2d = 3

Therefore, d = 3

3c = c + d – 1

3c – c = 3 – 1

2c = 2

Therefore, c = 1

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

3. Determine the matrices A and B when

Solution:

Given,

4.

Solution:

Comparing the corresponding elements, we have

4 + 2a = 18

2a = 18 – 4 = 14

a = 14/2

⇒ a = 7

1 + 2b = 7

2b = 7 – 1 = 6

b = 6/2

⇒ b = 3

2 + 2c = 14

2c = 14 – 2 = 12

2c = 12

c = 12/2

⇒ c = 6

3 + 2d = 11

2d = 11 – 3

d = 8/2

⇒ d = 4

Therefore, a = 7, b = 3, c = 6 and d = 4.

5. If , find the each of the following and state it they

are equal:

(i) (A + B) (A – B)

(ii) A² – B²

Solution:

Given,

Hence, its clearly seen that (A + B) (A – B) ≠ A² – B².

6. If , find A² – 5A – 14I, where I is unit matrix of order 2 × 2.

Solution:

Given,

7. If and A² = 0, find p and q.

Solution:

On comparing the corresponding elements, we have

9 + 3p = 0

3p = -9

p = -9/3

p = -3

And,

9 + 3q = 0

3q = -9

q = -9/3

q = -3

Therefore, p = -3 and q = -3.

8. If find a, b, c and d.

Solution:

Given,

On comparing the corresponding elements, we have

-a = 1 ⇒ a = -1

-b = 0 ⇒ b = 0

c = 0 and d = -1

Therefore, a = -1, b = 0, c = 0 and = -1.

9. Find a and b if

Solution:

On comparing the corresponding terms, we have

2a – 4 = 0

2a = 4

a = 4/2

a = 2

And, 2a – 2b = -2

2(2) – 2b = -2

4 – 2b = -2

2b = 4 + 2

b = 6/2

b = 3

Therefore, a = 2 and b = 3.

10. If

Find (i) 2A – 3B (ii) A² (iii) BA

Solution: