# RD Sharma Solutions for Class 12 Maths Chapter 7 Adjoint and Inverse of a Matrix

## RD Sharma Solutions for Class 12 Maths Chapter 7 – Free PDF Download Updated for (2022-23)

RD Sharma Solutions for Class 12 Maths Chapter 7 – Adjoint and Inverse of a Matrix is provided here. The RD Sharma textbook contains a huge number of solved examples and illustrations. It also provides quality content, easy stepwise explanations of various difficult concepts, and a wide variety of questions for practice. RD Sharma Solutions for Class 12 Chapter 7 are completely based on the exam-oriented approach to help the students in board exams. The PDF of RD Sharma Solutions for Class 12 Maths Chapter 7 Adjoint and Inverse of a Matrix is provided here.

Practising these RD Sharma Solutions for Class 12 will ensure that the students can easily excel in their final examination of Mathematics. Students can refer to and download Chapter 7 Adjoint and Inverse of a Matrix from the given links. This chapter is based on the adjoint of a square matrix and its properties. RD Sharma Solutions cover all the topics related to it.

Some of the essential topics in RD Sharma Solutions of this chapter are listed below.

• Definition and meaning of adjoint of a square matrix
• The inverse of a matrix
• Some useful results on invertible matrices
• Determining the adjoint and inverse of a matrix
• Determining the inverse of a matrix when it satisfies the matrix equation
• Finding the inverse of a matrix by using the definition of inverse
• Finding a non – singular matrix when adjoint is given
• Elementary transformation or elementary operations of a matrix
• Method of finding the inverse of a matrix by elementary transformation

## RD Sharma Solutions For Class 12 Maths Chapter 7 Adjoint and Inverse of a Matrix:-

### Also, access RD Sharma Solutions for Class 12 Maths Chapter 7 Adjoint and Inverse of a Matrix

Exercise 7.1 Solutions

Exercise 7.2 Solutions

### Exercise 7.1 Page No: 7.22

1. Find the adjoint of each of the following matrices:

Verify that (adj A) A = |A| I = A (adj A) for the above matrices.

Solution:

(i) Let

A =

Cofactors of A are

C11 = 4

C12 = – 2

C21 = – 5

C22 = – 3

(ii) Let

A =

Therefore cofactors of A are

C11 = d

C12 = – c

C21 = – b

C22 = a

(iii) Let

A =

Therefore cofactors of A are

C11 = cos α

C12 = – sin α

C21 = – sin α

C22 = cos α

(iv) Let

A =

Therefore cofactors of A are

C11 = 1

C12 = tan α/2

C21 = – tan α/2

C22 = 1

2. Compute the adjoint of each of the following matrices.

Solution:

(i) Let

A =

Therefore cofactors of A are

C11 = – 3

C21 = 2

C31 = 2

C12 = 2

C22 = – 3

C23 = 2

C13 = 2

C23 = 2

C33 = – 3

(ii) Let

A =

Cofactors of A

C11 = 2

C21 = 3

C31 = – 13

C12 = – 3

C22 = 6

C32 = 9

C13 = 5

C23 = – 3

C33 = – 1

(iii) Let

A =

Therefore cofactors of A

C11 = – 22

C21 = 11

C31 = – 11

C12 = 4

C22 = – 2

C32 = 2

C13 = 16

C23 = – 8

C33 = 8

(iv) Let

A =

Therefore cofactors of A

C11 = 3

C21 = – 1

C31 = 1

C12 = – 15

C22 = 7

C32 = – 5

C13 = 4

C23 = – 2

C33 = 2

Solution:

Given

A =

Therefore cofactors of A

C11 = 30

C21 = 12

C31 = – 3

C12 = – 20

C22 = – 8

C32 = 2

C13 = – 50

C23 = – 20

C33 = 5

Solution:

Given

A =

Cofactors of A

C11 = – 4

C21 = – 3

C31 = – 3

C12 = 1

C22 = 0

C32 = 1

C13 = 4

C23 = 4

C33 = 3

Solution:

Given

A =

Cofactors of A are

C11 = – 3

C21 = 6

C31 = 6

C12 = – 6

C22 = 3

C32 = – 6

C13 = – 6

C23 = – 6

C33 = 3

Solution:

Given

A =

Cofactors of A are

C11 = 9

C21 = 19

C31 = – 4

C12 = 4

C22 = 14

C32 = 1

C13 = 8

C23 = 3

C33 = 2

7. Find the inverse of each of the following matrices:

Solution:

(i) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

Now, |A| = cos θ (cos θ) + sin θ (sin θ)

= 1

Hence, A – 1 exists.

Cofactors of A are

C11 = cos θ

C12 = sin θ

C21 = – sin θ

C22 = cos θ

(ii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

Now, |A| = – 1 ≠ 0

Hence, A – 1 exists.

Cofactors of A are

C11 = 0

C12 = – 1

C21 = – 1

C22 = 0

(iii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

(iv) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

Now, |A| = 2 + 15 = 17 ≠ 0

Hence, A – 1 exists.

Cofactors of A are

C11 = 1

C12 = 3

C21 = – 5

C22 = 2

8. Find the inverse of each of the following matrices.

Solution:

(i) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| =

= 1(6 – 1) – 2(4 – 3) + 3(2 – 9)

= 5 – 2 – 21

= – 18≠ 0

Hence, A – 1 exists

Cofactors of A are

C11 = 5

C21 = – 1

C31 = – 7

C12 = – 1

C22 = – 7

C32 = 5

C13 = – 7

C23 = 5

C33 = – 1

(ii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| =

= 1 (1 + 3) – 2 (– 1 + 2) + 5 (3 + 2)

= 4 – 2 + 25

= 27≠ 0

Hence, A – 1 exists

Cofactors of A are

C11 = 4

C21 = 17

C31 = 3

C12 = – 1

C22 = – 11

C32 = 6

C13 = 5

C23 = 1

C33 = – 3

(iii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| =

= 2(4 – 1) + 1(– 2 + 1) + 1(1 – 2)

= 6 – 2

= – 4≠ 0

Hence, A – 1 exists

Cofactors of A are

C11 = 3

C21 = 1

C31 = – 1

C12 = + 1

C22 = 3

C32 = 1

C13 = – 1

C23 = 1

C33 = 3

(iv) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| =

= 2(3 – 0) – 0 – 1(5)

= 6 – 5

= 1≠ 0

Hence, A – 1 exists

Cofactors of A are

C11 = 3

C21 = – 1

C31 = 1

C12 = – 15

C22 = 6

C32 = – 5

C13 = 5

C23 = – 2

C33 = 2

(v) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| =

= 0 – 1 (16 – 12) – 1 (– 12 + 9)

= – 4 + 3

= – 1≠ 0

Hence, A – 1 exists

Cofactors of A are

C11 = 0

C21 = – 1

C31 = 1

C12 = – 4

C22 = 3

C32 = – 4

C13 = – 3

C23 = 3

C33 = – 4

(vi) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| =

= 0 – 0 – 1(– 12 + 8)

= 4≠ 0

Hence, A – 1 exists

Cofactors of A are

C11 = – 8

C21 = 4

C31 = 4

C12 = 11

C22 = – 2

C32 = – 3

C13 = – 4

C23 = 0

C33 = 0

(vii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.

|A| =
– 0 + 0

= – (cos2 α – sin2 α)

= – 1≠ 0

Hence, A – 1 exists

Cofactors of A are

C11 = – 1

C21 = 0

C31 = 0

C12 = 0

C22 = – cos α

C32 = – sin α

C13 = 0

C23 = – sin α

C33 = cos α

9. Find the inverse of each of the following matrices and verify that A-1A = I3.

Solution:

(i) We have

|A| =

= 1(16 – 9) – 3(4 – 3) + 3(3 – 4)

= 7 – 3 – 3

= 1≠ 0

Hence, A – 1 exists

Cofactors of A are

C11 = 7

C21 = – 3

C31 = – 3

C12 = – 1

C22 = 1

C32 = 0

C13 = – 1

C23 = 0

C33 = 1

(ii) We have

|A| =

= 2(8 – 7) – 3(6 – 3) + 1(21 – 12)

= 2 – 9 + 9

= 2≠ 0

Hence, A – 1 exists

Cofactors of A are

C11 = 1

C21 = 1

C31 = – 1

C12 = – 3

C22 = 1

C32 = 1

C13 = 9

C23 = – 5

C33 = – 1

10. For the following pair of matrices verify that (AB)-1 = B-1A-1.

Solution:

(i) Given

Hence, (AB)-1 = B-1A-1

(ii) Given

Hence, (AB)-1 = B-1A-1

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Solution:

Given

A =
and B – 1 =

Here, (AB) – 1 = B – 1 A – 1

|A| = – 5 + 4 = – 1

Cofactors of A are

C11 = – 1

C21 = 8

C31 = – 12

C12 = 0

C22 = 1

C32 = – 2

C13 = 1

C23 = – 10

C33 = 15

(i) [F (α)]-1 = F (-α)

(ii) [G (β)]-1 = G (-β)

(iii) [F (α) G (β)]-1 = G (-β) F (-α)

Solution:

(i) Given

F (α) =

|F (α)| = cos2 α + sin2 α = 1≠ 0

Cofactors of A are

C11 = cos α

C21 = sin α

C31 = 0

C12 = – sin α

C22 = cos α

C32 = 0

C13 = 0

C23 = 0

C33 = 1

(ii) We have

|G (β)| = cos2 β + sin2 β = 1

Cofactors of A are

C11 = cos β

C21 = 0

C31 = -sin β

C12 = 0

C22 = 1

C32 = 0

C13 = sin β

C23 = 0

C33 = cos β

(iii) Now we have to show that

[F (α) G (β)] – 1 = G (– β) F (– α)

[G (β)] – 1 = G (– β)

[F (α)] – 1 = F (– α)

And LHS = [F (α) G (β)] – 1

= [G (β)] – 1 [F (α)] – 1

= G (– β) F (– α)

Hence = RHS

Solution:

Consider,

Solution:

Given

Solution:

Given

### Exercise 7.2 Page No: 7.34

Find the inverse of the following matrices by using elementary row transformations:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

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