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 2023-24

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

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Exercise 7.1 Solutions

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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

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

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: