# Class 12-science RD SHARMA Solutions Maths Chapter 5 - Algebra of Matrices

## Algebra of Matrices Exercise MCQ

### Solution 46

As A and B are symmetric matrices, we have

A^{T} = A
and B^{T} = B … (i)

Consider,

Hence, AB^{T}
- BA^{T} is a skew-symmetric matrix.

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

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### Solution 8

### Solution 9

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### Solution 19

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### Solution 31

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### Solution 40

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### Solution 44

### Solution 45

Correct option: (d)

A matrix is called Diagonal matrix if all the elements, except those in the leading diagonal, are zero.

## Algebra of Matrices Exercise Ex. 5.1

### Solution 1

We know that if a matrix is of the order , it has mn elements. Thus, to find all the possible orders of a matrix having 8 elements, we have to find all the ordered pairs of natural numbers whose products is 8.

The ordered pairs are:

are the ordered pairs of natural numbers whose product is 5.

Hence, the possible orders of a matrix having 5 elements are

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 5(vii)

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### Solution 7

### Solution 8

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### Solution 19

### Solution 20

### Solution 21

As A = B,

## Algebra of Matrices Exercise Ex. 5.2

### Solution 1

### Solution 2(i)

### Solution 2(ii)

### Solution 2(iii)

### Solution 2(iv)

### Solution 3

(ii)

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15(i)

### Solution 15(ii)

### Solution 15(iii)

### Solution 16

### Solution 17

### Solution 18

(i)

(ii)

### Solution 19(i)

### Solution 19(ii)

### Solution 20

### Solution 21

### Solution 22

Let 3x and 4x be the monthly incomes of Aryan and Babban respectively.

Let 5y and 7y be their monthly expenditures respectively.

As each individual saves Rs. 15000 per month, we have

3x - 5y = 15000 … (i)

4x - 7y = 15000 … (ii)

The above equations can be written in matrix form as follows

Let AX = B, where

Let's find A^{-1}

Therefore, x = Rs. 30000 and y = Rs. 15000

So, monthly income of Aryan = Rs. 90,000 and monthly income of Babban is Rs. 120,000.

## Algebra of Matrices Exercise Ex. 5.3

### Solution 1

### Solution 2(i)

### Solution 2(ii)

### Solution 2(iii)

### Solution 3(i)

### Solution 3(ii)

### Solution 3(iii)

### Solution 3(iv)

### Solution 4(i)

### Solution 4(ii)

### Solution 5(i)

### Solution 5(ii)

### Solution 5(iii)

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

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### Solution 14

### Solution 15

### Solution 16(i)

### Solution 16(ii)

### Solution 17(i)

### Solution 17(ii)

### Solution 18

### Solution 19

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### Solution 21

### Solution 22

### Solution 23

### Solution 24(i)

### Solution 24(ii)

### Solution 25

### Solution 26

### Solution 27

### Solution 28

### Solution 29

### Solution 31

### Solution 32

### Solution 33

### Solution 34

### Solution 35

### Solution 36

### Solution 37

### Solution 38

### Solution 39

### Solution 40(i)

### Solution 40(ii)

### Solution 40(iii)

### Solution 41

### Solution 42

### Solution 43

### Solution 44

### Solution 45

### Solution 46

### Solution 47

### Solution 48(i)

### Solution 48(ii)

### Solution 48(iii)

### Solution 48(iv)

### Solution 49

### Solution 51

### Solution 52

### Solution 53

### Solution 54(i)

### Solution 54(ii)

### Solution 55

### Solution 56

### Solution 57

### Solution 58

### Solution 59

### Solution 60

### Solution 61

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### Solution 64

### Solution 65(i)

### Solution 65(ii)

### Solution 65(iii)

### Solution 65(iv)

### Solution 66

### Solution 67

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### Solution 77

### Solution 30

Given: A^{2}
= A

(2 + A)^{3}
- 19A

= 8I + A^{3}
+ 12A + 6A^{2} - 19A

= A^{3} +
6A^{2} - 7A + 8I

= A^{2} +
6A - 7A + 8I

= A - A + 8I

= 8I

Hence, (2 + A)^{3} - 19A = 8I.

### Solution 40(iv)

### Solution 48(v)

Given:

Let

Hence,

### Solution 48(vi)

Given:

Let

Solving equations
(i) and (ii), we get, a_{1} = 1, b_{1} = -2

Solving (iii) and
(iv), we get, a_{2} = 2, b_{2} = 0

From equations
(v) and (vi), we get, a_{3} = -5, b_{3} = 4

Hence,

### Solution 50

Given: and

As A^{2}
+ I = kA

Hence, k = -4.

### Solution 63

To prove

For n = 1,

Therefore, it is true for n = 1.

Suppose the result is true for n = k

Take n = k + 1

Thus, is true for all n ∈ N.

### Solution 78

Let 3x and 4x be the monthly incomes of Aryan and Babbar respectively.

Let 5y and 7y be their monthly expenditures respectively.

As each individual saves Rs. 15000 per month, we have

3x - 5y = 15000 … (i)

4x - 7y = 15000 … (ii)

The above equations can be written in matrix form as follows

Let AX = B, where

Let's find A^{-1}

Therefore, x = Rs. 30000 and y = Rs. 15000

So, monthly income of Aryan = Rs. 90,000 and monthly income of Babbar is Rs. 120,000.

This encourages us to understand the power of savings and we should save certain money for future.

### Solution 79

Let Rs. x and Rs. y is being invested in the first and second bonds respectively.

Let A be the investment matrix and B be the interest matrix.

Therefore,

The annual interest = AB =

If the interest had been interchanged, the total interest would be Rs. 100 less.

Equations (i) and (ii) can be expressed as

PX = Q, where

Now, |P| = 100 - 144 = -44

So, inverse of P exist.

Thus, x = 10000 and y = 15000

Hence, the total amount invested is Rs. 25,000.

## Algebra of Matrices Exercise Ex. 5.4

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 2

### Solution 3(i)

### Solution 3(ii)

### Solution 3(iii)

### Solution 4

### Solution 5

### Solution 6(i)

### Solution 6(ii)

### Solution 7

### Solution 8

### Solution 9

### Solution 10

## Algebra of Matrices Exercise Ex. 5.5

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

Given:

Consider,

Thus, A + A^{T}
is a symmetric matrix.

### Solution 10

As is a symmetric matrix, its transpose will be equal to itself.

Hence, the value of x is 5.

## Algebra of Matrices Exercise Ex. 5VSAQ

### Solution 64

As A is a symmetric matrix, we have

A^{T} = A

### Solution 65

Number of elements in a 2 × 2 matrix = 4

The first element can be 1, 2 or 3.

The second element can be 1, 2 or 3.

Similarly, the remaining two elements can take either of the 3 numbers.

So, for every element we have 3 choices.

Therefore, number
of ways of writing 1, 2 or 3 in a 2 × 2 matrix is 3^{4} which is 81.

Thus, the number of all possible matrices of order 2 × 2 with each entry 1, 2 or 3 is 81.

### Solution 66

Let

Order of matrix P is 1 × 3

Order of matrix Q is 3 × 3

Order of matrix R is 3 × 1

After multiplying P and Q, we'll get an output matrix B of order 1 × 3.

After multiplying B with R, we'll get an output matrix of order 1 × 1.

Hence, the order of matrix A is 1 × 1.

### Solution 67

As P is symmetric, we have

As Q is skew-symmetric, we have

### Solution 68

Matrix A is order 3 × 2

Matrix B is of order 2 × 4

Then the product matrix AB will have the order 3 × 4.

### Solution 69

As matrix A is skew-symmetric

Therefore, -A = A^{T}

Hence, the values of a and b are -2 and 3 respectively.

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### Solution 49

If a matrix is of order , then the number of elements in the matrix is the product .

Given that the required matrix is having 5 elements and 5 is a prime number.

Hence the prime factorization of 5 is either .

Thus, the order of the matrix is either .

### Solution 50

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### Solution 56

A^{2} = A

A^{3} = A^{2}
= A

7A - (I + A)^{3}

= 7A - (I^{3}
+ A^{3} + 3A^{2}I + 3AI^{2})

= 7A - (I + A + 3A + 3A)

= 7A - (I + 7A)

= -I

### Solution 57

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### Solution 63