Frank Solutions for Class 10 Maths Chapter 11 Matrices

Frank Solutions for Class 10 Maths Chapter 11 Matrices are available here. When students feel stressed about searching for the most comprehensive and detailed Frank Solutions for Class 10 Maths, we at BYJU’S have prepared step by step solutions with detailed explanations. We advise students who want to score good marks in Maths, to go through these solutions and strengthen their knowledge

Chapter 11 – Matrices is defined as a set of numbers arranged in rows and columns so as to form a rectangular array. The numbers are called the elements, or entries, of the matrix. This chapter contains the topics related to classification of matrices, finding the unknown values in the matrices. So we suggest students to practice Frank Solutions for Class 10 Maths to get more marks in Maths.

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Access answers to Frank Solutions for Class 10 Maths Chapter 11 Matrices

1. Classify the following matrices:

(i)

Frank Solutions for Class 10 Maths Chapter 11 Image 1

Solution:-

The order of the given matrix is, 3 × 2.

Therefore, the given matrix is a rectangular matrix of 3 × 2.

(ii) Frank Solutions for Class 10 Maths Chapter 11 Image 2

Solution:-

The order of the given matrix is, 1 × 2.

Therefore, the given matrix is a rectangular matrix of 1 × 2.

(iii)

Frank Solutions for Class 10 Maths Chapter 11 Image 3

Solution:-

The order of the given matrix is, 2 × 3.

Therefore, the given matrix is a rectangular matrix of 2 × 3.

(iv)


Frank Solutions for Class 10 Maths Chapter 11 Image 4

Solution:-

The order of the given matrix is, 2 × 2.

Therefore, the given matrix is a square matrix of 2 × 2.

(v)


Frank Solutions for Class 10 Maths Chapter 11 Image 5

Solution:-

The order of the given matrix is, 3 × 4.

Therefore, the given matrix is a rectangular matrix of 3 × 4.

2. Find the values of a and b, if [2a + 3b a – b] = [19 2]

Solution:-

From the question,

2a + 3b = 19 … [equation (i)]

a – b = 2 … [equation (ii)]

a = 2 + b

Now, substitute the value of a in equation (i),

2(2 + b) + 3b = 19

4 + 2b + 3b = 19

5b = 19 – 4

5b = 15

b = 15/5

b = 3

Then, a = 2 + b

= 2 + 3

= 5

Therefore, the value of a is 5 and b is 3.

3. Find the values of x and y, if
Frank Solutions for Class 10 Maths Chapter 11 Image 6 =
Frank Solutions for Class 10 Maths Chapter 11 Image 7

Solution:-

Consider the given two matrices,

Frank Solutions for Class 10 Maths Chapter 11 Image 8=
Frank Solutions for Class 10 Maths Chapter 11 Image 9

The given two matrices are rectangular matrices of 2 × 1.

3x – y = 7 … [equation (i)]

x + y = 5 … [equation (ii)]

x = 5 – y

Now, substitute the value of x in equation (i),

3(5 – y) – y = 7

15 – 3y – y = 7

15 – 4y = 7

15 – 7 = 4y

8 = 4y

y = 8/4

y = 2

Then, x + y = 5

x = 5 – y

x = 5 – 2

x = 3

Therefore, the value of x is 3 and y is 2.

4. Find the values of a, b, c and d, if
Frank Solutions for Class 11 Maths Chapter 10 Image 10.

Solution:-

From the given matrices,

a + 3b = 5 … [equation (i)]

2a – b = 3 … [equation (ii)]

b = 2a – 3

Now, substitute the value of b in equation (i), we get

a + 3(2a – 3) = 5

a + 6a – 9 = 5

By transposing,

7a = 5 + 9

7a = 14

a = 14/7

a = 2

Again substitute the value of a in equation (i),

2 + 3b = 5

3b = 5 – 2

3b = 3

b = 3/3

b = 1

Then,

3c + d = 8 … [equation (iii)]

c – 2d = 5 … [equation (iv)]

c = 5 + 2d

Now substitute the value of c in equation (iii),

3(5 + 2d) + d = 8

15 + 6d + d = 8

7d = 8 – 15

7d = – 7

d = -7/7

d = -1

substitute the value of d in equation (iv),

c – 2d = 5

c = 5 + 2d

c = 5 + 2(-1)

c = 5 – 2

c = 3

Frank Solutions for Class 10 Maths Chapter 11 Image 11

Solution:-

Given two matrices are square matrices of 2 × 2

Frank Solutions for Class 10 Maths Chapter 11 Image 12

Frank Solutions for Class 10 Maths Chapter 11 Image 13

6. If A = [4 7] and B = [3 1], find: (i) A + 2B (ii) A –B (iii) 2A – 3B

Solution:-

From the question it is given that,

A = [4 7]1× 2

B = [3 1]1× 2

Then,

(i) A + 2B

2B = [3 × 2 1 × 2]

= [6 2]

So, A + 2B = [4 + 6 7 + 2]

= [10 9] 1× 2

(ii) A – B

A – B = [4 – 3 7 + 1]

= [1 6] 1× 2

(iii) 2A – 3B

2A= [4 × 2 7 × 2]

= [8 14]

3B = [3 × 3 1 × 3]

= [9 3]

So, 2A – 3B = [8 – 9 14 – 3]

= [-1 11] 1× 2


Frank Solutions for Class 10 Maths Chapter 11 Image 14

Solution:-

Given two matrices are square matrices of 2 × 2

Then,

(i) 2P + 3Q

Frank Solutions for Class 10 Maths Chapter 11 Image 15

Frank Solutions for Class 10 Maths Chapter 11 Image 16

Frank Solutions for Class 10 Maths Chapter 11 Image 17

Solution:-

Given two matrices are rectangular matrices of 2 × 3

Then,

Frank Solutions for Class 10 Maths Chapter 11 Image 18

Frank Solutions for Class 10 Maths Chapter 11 Image 19

Frank Solutions for Class 10 Maths Chapter 11 Image 20

Solution:-

Given two matrices are rectangular matrices of 3 × 2

Then,

Frank Solutions for Class 10 Maths Chapter 11 Image 21


Frank Solutions for Class 10 Maths Chapter 11 Image 22

Solution:-

Given matrix is a rectangular matrix of 2 × 3.

Then,

Transpose of a matrix by switching its rows with its columns

Frank Solutions for Class 10 Maths Chapter 11 Image 23

Frank Solutions for Class 10 Maths Chapter 11 Image 24

Frank Solutions for Class 10 Maths Chapter 11 Image 25

Frank Solutions for Class 10 Maths Chapter 11 Image 26

We cannot add them, because to add two matrices their corresponding number of rows and number of columns should be same. But in the above case B and Bt are not same.

Frank Solutions for Class 10 Maths Chapter 11 Image 27

Then,

5 + q = 9

q = 9 – 5

q = 4

r + 4 = 7

r = 7 – 4

r = 3

p + 3 = 5

p = 5 – 3

p = 2

7 + s = 8

s = 8 – 7

s = 1

Frank Solutions for Class 10 Maths Chapter 11 Image 28

Then,

4p = 12

p = 12/4

p = 3

6q = 6

q = 6/6

q = 1

8 + 2q = 2r

8 + 2(1) = 2r

8 + 2 = 2r

r = 10/2

r = 5

12 = 3s

s = 12/3

s = 4

Frank Solutions for Class 10 Maths Chapter 11 Image 29

Then,

2a + b = 4 … [equation (i)]

3a – b = 6 … [equation (ii)]

Now we have to add both equation (i) and equation (ii) we get,

5a = 10

a = 10/5

a = 2

substitute the value of a in equation (i)we get,

2(2) + b = 4

4 + b = 4

b = 0

Then,

c = 3a

c = 3(2)

c = 6

d = 7

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