# RD Sharma Solutions for Class 12 Maths Exercise 7.1 Chapter 7 Adjoint and Inverse of a Matrix

RD Sharma Solutions for Class 12 Maths Exercise 7.1 Chapter 7 Adjoint and Inverse of a Matrix is provided here. The solutions for the exercise-wise answers are prepared by the experts at BYJU’S in the best possible way so that they are easily understandable by students.

The PDF of RD Sharma Solutions for Class 12, Exercise 7.1 of Chapter 7 Adjoint and Inverse of a Matrix can be downloaded from the given links. This exercise consists of two levels according to the increasing order of difficulties. Let us have a look at the important topics covered in this exercise.

• 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

## RD Sharma Solutions for Class 12 Maths Chapter 7 Adjoint and Inverse of a Matrix Exercise 7.1

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Exercise 7.2 Solutions

### Access answers to Maths RD Sharma Solutions for Class 12 Chapter 7 – Adjoint and Inverse of a Matrix Exercise 7.1

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 criterion for the existence of an inverse matrix is that the determinant of a given matrix should not be 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 criterion for the existence of an inverse matrix is that the determinant of a given matrix should not be 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 criterion for the existence of an inverse matrix is that the determinant of a given matrix should not be equal to zero.

(iv) The criterion for the existence of an inverse matrix is that the determinant of a given matrix should not be 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 criterion for the existence of an inverse matrix is that the determinant of a given matrix should not  be 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 criterion for the existence of an inverse matrix is that the determinant of a given matrix should not be 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 criterion for the existence of an inverse matrix is that the determinant of a given matrix should not be 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 criterion for the existence of an inverse matrix is that the determinant of a given matrix should not be 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 criterion for the existence of an inverse matrix is that the determinant of a given matrix should not be 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 criterion for the existence of an inverse matrix is that the determinant of a given matrix should not be 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 criterion for the existence of an inverse matrix is that the determinant of a given matrix should not be 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