Chapter 5 Algebra of Matrices Ex 5.1

1. If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?

Solution:

If a matrix is of order m × n elements, it has m n elements. So, if the matrix has 8 elements, we will find the ordered pairs m and n.

m n = 8

Then, ordered pairs m and n will be

m × n be (8 × 1),(1 × 8),(4 × 2),(2 × 4)

Now, if it has 5 elements

Possible orders are (5 × 1), (1 × 5).

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 1

Solution:

(i)

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 2

Now, Comparing with equation (1) and (2)

a22 = 4 and b21 = – 3

a22 + b21 = 4 + (– 3) = 1

(ii)

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 3

Now, Comparing with equation (1) and (2)

a11 = 2, a22 = 4, b11 = 2, b22 = 4

a11 b11 + a22 b22 = 2 × 2 + 4 × 4 = 4 + 16 = 20

3. Let A be a matrix of order 3 × 4. If R1 denotes the first row of A and C2 denotes its second column, then determine the orders of matrices R1 and C2.

Solution:

Given A be a matrix of order 3 × 4.

So, A = [ai j3×4

R1 = first row of A = [a11, a12, a13, a14]

So, order of matrix R1 = 1 × 4

C2 = second column of

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 4

Therefore order of C2 = 3 × 1

4. Construct a 2 ×3 matrix A = [aj j] whose elements aj j are given by:

(i) ai j = i × j

(ii) ai j = 2i – j

(iii) ai j = i + j

(iv) ai j = (i + j)2/2

Solution:

(i) Given ai j = i × j

Let A = [ai j]2 × 3

So, the elements in a 2 × 3 matrix are[a11, a12, a13, a21, a22, a23]

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 5

a11 = 1 × 1 = 1

a12 = 1 × 2 = 2

a13 = 1 × 3 = 3

a21 = 2 × 1 = 2

a22 = 2 × 2 = 4

a23 = 2 × 3 = 6

Substituting these values in matrix A we get,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 6

(ii) Given ai j = 2i – j

Let A = [ai j]2×3

So, the elements in a 2 × 3 matrix are

a11, a12, a13, a21, a22, a23

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 7

a11 = 2 × 1 – 1 = 2 – 1 = 1

a12 = 2 × 1 – 2 = 2 – 2 = 0

a13 = 2 × 1 – 3 = 2 – 3 = – 1

a21 = 2 × 2 – 1 = 4 – 1 = 3

a22 = 2 × 2 – 2 = 4 – 2 = 2

a23 = 2 × 2 – 3 = 4 – 3 = 1

Substituting these values in matrix A we get,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 8

(iii) Given ai j = i + j

Let A = [a i j2×3

So, the elements in a 2 × 3 matrix are

a11, a12, a13, a21, a22, a23

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 9

a11 = 1 + 1 = 2

a12 = 1 + 2 = 3

a13 = 1 + 3 = 4

a21 = 2 + 1 = 3

a22 = 2 + 2 = 4

a23 = 2 + 3 = 5

Substituting these values in matrix A we get,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 10

(iv) Given ai j = (i + j)2/2

Let A = [ai j]2×3

So, the elements in a 2 × 3 matrix are

a11, a12, a13, a21, a22, a23

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 11

Let A = [ai j]2×3

So, the elements in a 2 × 3 matrix are

a11, a12, a13, a21, a22, a23

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 12

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 13

a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 14

a13 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 15

a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 16

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 17

a23 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 18

Substituting these values in matrix A we get,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 19
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 20

5. Construct a 2 × 2 matrix A = [ai j] whose elements ai j are given by:

(i) (i + j)/2

(ii) ai j = (i – j)/2

(iii) ai j = (i – 2j)/2

(iv) ai j = (2i + j)/2

(v) ai j = |2i – 3j|/2

(vi) ai j = |-3i + j|/2

(vii) ai j = e2ix sin x j

Solution:

(i) Given (i + j)/2

Let A = [ai j]2×2

So, the elements in a 2 × 2 matrix are

a11, a12, a21, a22

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 21

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 22

a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 23

a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 24

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 25

Substituting these values in matrix A we get,

(ii) Given ai j = (i – j)/2

Let A = [ai j]2×2

So, the elements in a 2 × 2 matrix are

a11, a12, a21, a22

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 29

a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 30

a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 31

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 32

Substituting these values in matrix A we get,

(iii) Given ai j = (i – 2j)/2

Let A = [ai j]2×2

So, the elements in a 2 × 2 matrix are

a11, a12, a21, a22

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 36

a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 37

a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 38

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 39

Substituting these values in matrix A we get,

(iv) Given ai j = (2i + j)/2

Let A = [ai j]2×2

So, the elements in a 2 × 2 matrix are

a11, a12, a21, a22

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 43

a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 44

a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 45

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 46

Substituting these values in matrix A we get,

(v) Given ai j = |2i – 3j|/2

Let A = [ai j]2×2

So, the elements in a 2×2 matrix are

a11, a12, a21, a22

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 50

a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 51

a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 52

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 53

Substituting these values in matrix A we get,

(vi) Given ai j = |-3i + j|/2

Let A = [ai j]2×2

So, the elements in a 2 × 2 matrix are

a11, a12, a21, a22

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 57

a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 58

a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 59

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 60

Substituting these values in matrix A we get,

(vii) Given ai j = e2ix sin x j

Let A = [ai j]2×2

So, the elements in a 2 × 2 matrix are

a11, a12, a21, a22,

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 64

a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 65

a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 66

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 67

Substituting these values in matrix A we get,

6. Construct a 3×4 matrix A = [ai j] whose elements ai j are given by:
(i) ai j = i + j

(ii) ai j = i – j

(iii) ai j = 2i

(iv) ai j = j

(v) ai j = ½ |-3i + j|

Solution:

(i) Given ai j = i + j

Let A = [ai j]2×3

So, the elements in a 3 × 4 matrix are

a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 69

a11 = 1 + 1 = 2

a12 = 1 + 2 = 3

a13 = 1 + 3 = 4

a14 = 1 + 4 = 5

a21 = 2 + 1 = 3

a22 = 2 + 2 = 4

a23 = 2 + 3 = 5

a24 = 2 + 4 = 6

a31 = 3 + 1 = 4

a32 = 3 + 2 = 5

a33 = 3 + 3 = 6

a34 = 3 + 4 = 7

Substituting these values in matrix A we get,

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 70

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 71

(ii) Given ai j = i – j

Let A = [ai j]2×3

So, the elements in a 3×4 matrix are

a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 72

a11 = 1 – 1 = 0

a12 = 1 – 2 = – 1

a13 = 1 – 3 = – 2

a14 = 1 – 4 = – 3

a21 = 2 – 1 = 1

a22 = 2 – 2 = 0

a23 = 2 – 3 = – 1

a24 = 2 – 4 = – 2

a31 = 3 – 1 = 2

a32 = 3 – 2 = 1

a33 = 3 – 3 = 0

a34 = 3 – 4 = – 1

Substituting these values in matrix A we get,

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 73

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 74

(iii) Given ai j = 2i

Let A = [ai j]2×3

So, the elements in a 3×4 matrix are

a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 75

a11 = 2×1 = 2

a12 = 2×1 = 2

a13 = 2×1 = 2

a14 = 2×1 = 2

a21 = 2×2 = 4

a22 = 2×2 = 4

a23 = 2×2 = 4

a24 = 2×2 = 4

a31 = 2×3 = 6

a32 = 2×3 = 6

a33 = 2×3 = 6

a34 = 2×3 = 6

Substituting these values in matrix A we get,

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 76

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 77

(iv) Given ai j = j

Let A = [ai j]2×3

So, the elements in a 3×4 matrix are

a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 78

a11 = 1

a12 = 2

a13 = 3

a14 = 4

a21 = 1

a22 = 2

a23 = 3

a24 = 4

a31 = 1

a32 = 2

a33 = 3

a34 = 4

Substituting these values in matrix A we get,

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 79

(vi) Given ai j = ½ |-3i + j|

Let A = [ai j]2×3

So, the elements in a 3×4 matrix are

a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 81

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 82a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 83a13 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 84a14 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 85a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 86

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 87

a23 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 88

a24 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 89a31 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 90a32 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 91a33 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 92a34 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 93

Substituting these values in matrix A we get,

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 94

Multiplying by negative sign we get,

7. Construct a 4 × 3 matrix A = [ai j] whose elements ai j are given by:

(i) ai j = 2i + i/j

(ii) ai j = (i – j)/ (i + j)

(iii) ai j = i

Solution:

(i) Given ai j = 2i + i/j

Let A = [ai j]4×3

So, the elements in a 4 × 3 matrix are

a11, a12, a13, a21, a22, a23, a31, a32, a33, a41, a42, a43

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 95

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 96

a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 97

a13 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 98

a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 99

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image100

a23 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 101

a31 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 102

a32 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 103

a33 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 104

a41 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 105

a42 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 106

a43 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 107

Substituting these values in matrix A we get,

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 108

(ii) Given ai j = (i – j)/ (i + j)

Let A = [ai j]4×3

So, the elements in a 4 × 3 matrix are

a11, a12, a13, a21, a22, a23, a31, a32, a33, a41, a42, a43

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 110

a11 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image111

a12 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 112

a13 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 113

a21 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 114

a22 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 115

a23 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 116

a31 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 117

a32 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 118

a33 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 119

a41 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 120

a42 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 121

a43 =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 122

Substituting these values in matrix A we get,

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 123

(iii) Given ai j = i

Let A = [ai j]4×3

So, the elements in a 4 × 3 matrix are

a11, a12, a13, a21, a22, a23, a31, a32, a33, a41, a42, a43

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 125

a11 = 1

a12 = 1

a13 = 1

a21 = 2

a22 = 2

a23 = 2

a31 = 3

a32 = 3

a33 = 3

a41 = 4

a42 = 4

a43 = 4

Substituting these values in matrix A we get,

A =
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 126

8. Find x, y, a and b if

Solution:

Given

Given that two matrices are equal.

We know that if two matrices are equal then the elements of each matrices are also equal.

Therefore by equating them we get,

3x + 4y = 2 …… (1)

x – 2y = 4 …… (2)

a + b = 5 …… (3)

2a – b = – 5 …… (4)

Multiplying equation (2) by 2 and adding to equation (1), we get

3x + 4y + 2x – 4y = 2 + 8

⇒ 5x = 10

⇒ x = 2

Now, substituting the value of x in equation (1)

3 × 2 + 4y = 2

⇒ 6 + 4y = 2

⇒ 4y = 2 – 6

⇒ 4y = – 4

⇒ y = – 1

Now by adding equation (3) and (4)

a + b + 2a – b = 5 + (– 5)

⇒ 3a = 5 – 5 = 0

⇒ a = 0

Now, again by substituting the value of a in equation (3), we get

0 + b = 5

⇒ b = 5

∴ a = 0, b = 5, x = 2 and y = – 1

9. Find x, y, a and b if

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 130

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 131

We know that if two matrices are equal then the elements of each matrices are also equal.

Given that two matrices are equal.

Therefore by equating them we get,

2a + b = 4 …… (1)

And a – 2b = – 3 …… (2)

And 5c – d = 11 …… (3)

4c + 3d = 24 …… (4)

Multiplying equation (1) by 2 and adding to equation (2)

4a + 2b + a – 2b = 8 – 3

⇒ 5a = 5

⇒ a = 1

Now, substituting the value of a in equation (1)

2 × 1 + b = 4

⇒ 2 + b = 4

⇒ b = 4 – 2

⇒ b = 2

Multiplying equation (3) by 3 and adding to equation (4)

15c – 3d + 4c + 3d = 33 + 24

⇒ 19c = 57

⇒ c = 3

Now, substituting the value of c in equation (4)

4 × 3 + 3d = 24

⇒ 12 + 3d = 24

⇒ 3d = 24 – 12

⇒ 3d = 12

⇒ d = 4

∴ a = 1, b = 2, c = 3 and d = 4

10. Find the values of a, b, c and d from the following equations:

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 132

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 133

We know that if two matrices are equal then the elements of each matrices are also equal.

Given that two matrices are equal.

Therefore by equating them we get,

2a + b = 4 …… (1)

And a – 2b = – 3 …… (2)

And 5c – d = 11 …… (3)

4c + 3d = 24 …… (4)

Multiplying equation (1) by 2 and adding to equation (2)

4a + 2b + a – 2b = 8 – 3

⇒ 5a = 5

⇒ a = 1

Now, substituting the value of a in equation (1)

2 × 1 + b = 4

⇒ 2 + b = 4

⇒ b = 4 – 2

⇒ b = 2

Multiplying equation (3) by 3 and adding to equation (4)

15c – 3d + 4c + 3d = 33 + 24

⇒ 19c = 57

⇒ c = 3

Now, substituting the value of c in equation (4)

4 × 3 + 3d = 24

⇒ 12 + 3d = 24

⇒ 3d = 24 – 12

⇒ 3d = 12

⇒ d = 4

∴ a = 1, b = 2, c = 3 and d = 4


Chapter 5 Algebra of Matrices Ex 5.2

1. Compute the following sums:

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 134
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 135

Solution:

(i) Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 136

Corresponding elements of two matrices should be added

Therefore, we get

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 137

Therefore,
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 138

(ii) Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 139

Therefore,

Find each of the following:

(i) 2A – 3B

(ii) B – 4C

(iii) 3A – C

(iv) 3A – 2B + 3C

Solution:

(i) Given

First we have to compute 2A

Now by computing 3B we get,

Now by we have to compute 2A – 3B we get

Therefore

(ii) Given
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 149

First we have to compute 4C,

Now,

Therefore we get,

(iii) Given

First we have to compute 3A,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 154

Now,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 155

Therefore,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 156

(iv) Given

First we have to compute 3A

Now we have to compute 2B

By computing 3C we get,

Therefore,

(i) A + B and B + C

(ii) 2B + 3A and 3C – 4B

Solution:

(i) Consider A + B,

A + B is not possible because matrix A is an order of 2 x 2 and Matrix B is an order of 2 x 3, so the Sum of the matrix is only possible when their order is same.

Now consider B + C

(ii) Consider 2B + 3A

2B + 3A also does not exist because the order of matrix B and matrix A is different, so we cannot find the sum of these matrix.

Now consider 3C – 4B,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 165
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 166

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 167

Now we have to compute 2A – 3B + 4C

5. If A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4), find

(i) A – 2B

(ii) B + C – 2A

(iii) 2A + 3B – 5C

Solution:

(i) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

(ii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

We have to find B + C – 2A

Here,

Now we have to compute B + C – 2A

(iii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

Now we have to find 2A + 3B – 5C

Here,

Now consider 2A + 3B – 5C

6. Given the matrices

Verify that (A + B) + C = A + (B + C)

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 176

Now we have to verify (A + B) + C = A + (B + C)

First consider LHS, (A + B) + C,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image177
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 178

Now consider RHS, that is A + (B + C)

Therefore LHS = RHS

Hence (A + B) + C = A + (B + C)

7. Find the matrices X and Y,

Solution:

Consider,

Now by simplifying we get,

Therefore,

Again consider,

Now by simplifying we get,

Therefore,

Solution:

Given

Now by transposing, we get

Therefore,

Solution:

Given

Now by multiplying equation (1) and (2) we get,

Now by adding equation (2) and (3) we get,

Now by substituting X in equation (2) we get,

Solution:

Consider

Now, again consider

Therefore,

And


Chapter 5 Algebra of Matrices Ex 5.3

1. Compute the indicated products:

Solution:

(i) Consider

On simplification we get,

(ii) Consider

On simplification we get,

(iii) Consider

On simplification we get,

2. Show that AB ≠ BA in each of the following cases:

Solution:

(i) Consider,

Again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 218

Now again consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 219

From equation (1) and (2), it is clear that

AB ≠ BA

(iii) Consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 220

Now again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

3. Compute the products AB and BA whichever exists in each of the following cases:

Solution:

(i) Consider,

BA does not exist

Because the number of columns in B is greater than the rows in A

(ii) Consider,

Again consider,

(iii) Consider,

AB = 0+(−1)+6+6

AB = 11

Again consider,

(iv) Consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 231
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 232

4. Show that AB ≠ BA in each of the following cases:

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 233
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 234

Solution:

(i) Consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 235

Again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

Again consider,

From equation (1) and (2) it is clear that,

AB ≠ BA

5. Evaluate the following:

Solution:

(i) Given

First we have to add first two matrix,

On simplifying, we get

(ii) Given,

First we have to multiply first two given matrix,

= 82

(iii) Given

First we have subtract the matrix which is inside the bracket,

Solution:

Given

We know that,

Again we know that,

Now, consider,

We have,

Now, from equation (1), (2), (3) and (4), it is clear that A= B2= C2= I2

Solution:

Given

Consider,

Now we have to find,

Solution:

Given

Consider,

Hence the proof.

Solution:

Given,

Consider,

Again consider,

Hence the proof.

Solution:

Given,

Consider,

Hence the proof.

Solution:

Given,

Consider,

We know that,

Again we have,

Solution:

Given,

Consider,

Again consider,

From equation (1) and (2) AB = BA = 03×3

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 285

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 286

Again consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 287

From equation (1) and (2) AB = BA = 03×3

Solution:

Given

Now consider,

Therefore AB = A

Again consider, BA we get,

Hence BA = B

Hence the proof.

Solution:

Given,

Consider,

Now again consider, B2

Now by subtracting equation (2) from equation (1) we get,

16. For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC)

Solution:

(i) Given

Consider,

Now consider RHS,

From equation (1) and (2), it is clear that (AB) C = A (BC)

(ii) Given,

Consider the LHS,

Now consider RHS,

From equation (1) and (2), it is clear that (AB) C = A (BC)

17. For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC.

Solution:

(i) Given

Consider LHS,

Now consider RHS,

From equation (1) and (2), it is clear that A (B + C) = AB + AC

(ii) Given,

Consider the LHS

Now consider RHS,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 317
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 318
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 319

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 320

Consider the LHS,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 321

Now consider RHS

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 322

From the above equations LHS = RHS

Therefore, A (B – C) = AB – AC.

19. Compute the elements a43 and a22 of the matrix:

Solution:

Given

From the above matrix, a43 = 8and a22 = 0

Solution:

Given

Consider,

Again consider,

Now, consider the RHS

Therefore, A3 = p I + q A + rA2

Hence the proof.

21. If ω is a complex cube root of unity, show that

Solution:

Given

It is also given that ω is a complex cube root of unity,

Consider the LHS,

We know that 1 + ω + ω2 = 0 and ω3 = 1

Now by simplifying we get,

Again by substituting 1 + ω + ω2 = 0 and ω3 = 1 in above matrix we get,

Therefore LHS = RHS

Hence the proof.

Solution:

Given,

Consider A2

Therefore A2 = A

Solution:

Given

Consider A2,

Hence A2 = I3

Solution:

(i) Given

= 2x+1+2+x+3 = 0

= 3x+6 = 0

= 3x = -6

x = -6/3

x = -2

(ii) Given,

On comparing the above matrix we get,

x = 13

Solution:

Given

⇒ (2x+4)x+4(x+2)–1(2x+4) = 0

⇒ 2x2 + 4x + 4x + 8 – 2x – 4 = 0

⇒ 2x2 + 6x + 4 = 0

⇒ 2x2 + 2x + 4x + 4 = 0

⇒ 2x (x + 1) + 4 (x + 1) = 0

⇒ (x + 1) (2x + 4) = 0

⇒ x = -1 or x = -2

Hence, x = -1 or x = -2

Solution:

Given

By multiplying we get,

Solution:

Given

Now we have to prove A2 – A + 2 I = 0

Solution:

Given

Solution:

Given

Hence the proof.

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 372
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 373

Hence the proof.

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 374

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

I is identity matrix, so

Also given,
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 403

Now, we have to find A2, we get

Now, we will find the matrix for 8A, we get

So,

Substitute corresponding values from eqn (i) and (ii), we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 413

Therefore, the value of k is 7

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 414

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 415

To show that f (A) = 0

Substitute x = A in f(x), we get

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 416

I is identity matrix, so

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 417

Now, we will find the matrix for A2, we get

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 418

Now, we will find the matrix for 2A, we get

Substitute corresponding values from eqn (ii) and (iii) in eqn (i), we get

So,

Hence Proved

Solution:

Given

So

Now, we will find the matrix for A2, we get

Now, we will find the matrix for λ A, we get

But given, A2 = λ A + μ I

Substitute corresponding values from equation (i) and (ii), we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence, λ + 0 = 4 ⇒ λ = 4

And also, 2λ + μ = 7

Substituting the obtained value of λ in the above equation, we get

2(4) + μ = 7 ⇒ 8 + μ = 7 ⇒ μ = – 1

Therefore, the value of λ and μ are 4 and – 1 respectively

39. Find the value of x for which the matrix product

Solution:

We know,

is identity matrix of size 3.

So according to the given criteria

Now we will multiply the two matrices on LHS using the formula cij = ai1b1j + ai2b2j + … + ain bnj, we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

So we get

So the value of x is
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 447


Chapter 5 Algebra of Matrices Ex 5.4

(i) (2A)T = 2 AT

(ii) (A + B)T = AT + BT

(iii) (A – B)T = AT – BT

(iv) (AB)T = BT AT

Solution:

(i) Given

Consider,

Put the value of A

L.H.S = R.H.S

(ii) Given

Consider,

L.H.S = R.H.S

Hence proved.

(iii) Given

Consider,

L.H.S = R.H.S

(iv) Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 471

So,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 472
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 473

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 474
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 475
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 476
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 477

L.H.S = R.H.S

So,
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 479

(i) A + B)T = AT + BT

(ii) (AB)T = BT AT

(iii) (2A)T = 2 AT

Solution:

(i) Given

Consider,

L.H.S = R.H.S

So,
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 487

(ii) Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 488

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 489
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 490
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 491
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 492
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 493

L.H.S = R.H.S

So,
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 494

(iii) Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 495

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 496
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 497
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 498
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 499

L.H.S = R.H.S

So,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 500
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 501

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 502

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 503
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 504
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 505
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 506

L.H.S = R.H.S

So,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 507
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 508

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 509

Now we have to find (AB)T

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 510
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 511
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 512
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 513

So,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 514

Chapter 5 Algebra of Matrices Ex 5.5

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 515

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 516

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 517
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 518
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 519

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 520 … (i)

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 521
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 522

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 523 … (ii)

From (i) and (ii) we can see that

A skew-symmetric matrix is a square matrix whose transpose equal to its negative, that is,

X = – XT

So, A – AT is a skew-symmetric.

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 524

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 525

Consider,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 526 … (i)

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 527
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 528

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 529 … (ii)

From (i) and (ii) we can see that

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that is,

X = – XT

So, A – AT is a skew-symmetric matrix.

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 530

Solution:

Given,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 531 is a symmetric matrix.

We know that A = [aij]m × n is a symmetric matrix if aij = aji

So,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 532
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 533
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 534
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 535

Hence, x = 4, y = 2, t = -3 and z can have any value.

4. Let. Find matrices X and Y such that X + Y = A, where X is a symmetric and y is a skew-symmetric matrix.

Solution:

Given,
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 537 Then
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 538

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 539
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 540
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 541
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 542
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 543

Now,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 544
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 545
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 546
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 547
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 548

Now,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 549

X is a symmetric matrix.

Now,

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 550
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 551

-Y T = Y

Y is a skew symmetric matrix.

RD Sharma Solutions for Class 12 Maths Chapter 5 Image 552
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 553
RD Sharma Solutions for Class 12 Maths Chapter 5 Image 554

Hence, X + Y = A


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