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

The solutions for the exercise wise answers are prepared by experts at BYJU’S in the best possible way and 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 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 Chapter 7 Adjoint and Inverse of a Matrix Exercise 7.1:-Download PDF Here

Access another exercise of RD Sharma Solutions For Class 12 Chapter 7 – Adjoint and Inverse of a Matrix

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

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

Hence, (adj A) A = |A|I = A(adj A)

Solution:

Given

A = 

Therefore cofactors of A

C₁₁ = 30

C₂₁ = 12

C₃₁ = – 3

C₁₂ = – 20

C₂₂ = – 8

C₃₂ = 2

C₁₃ = – 50

C₂₃ = – 20

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

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

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

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

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

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

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

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

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

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

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

(ii) Given

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

Cofactors of A are

C₁₁ = cos α

C₂₁ = sin α

C₃₁ = 0

C₁₂ = – sin α

C₂₂ = cos α

C₃₂ = 0

C₁₃ = 0

C₂₃ = – 10

C₃₃ = 1

(ii) We have

|G (β)| = cos² β + sin² β = 1

Cofactors of A are

C₁₁ = cos β

C₂₁ = sin α

C₃₁ = sin β

C₁₂ = 0

C₂₂ = 1

C₃₂ = 0

C₁₃ = sin β

C₂₃ = 0

C₃₃ = cos β

(iii) Now we have to show that

We have already know that

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