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

Class 12: Maths Chapter 7 solutions. Complete Class 12 Maths Chapter 7 Notes. RD Sharma Solutions for Class 12 Maths Chapter 7–Adjoint and Inverse of a Matrix RD Sharma 12th Maths Chapter 7, Class 12...

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Class 12: Maths Chapter 7 solutions. Complete Class 12 Maths Chapter 7 Notes.

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

RD Sharma 12th Maths Chapter 7, Class 12 Maths Chapter 7 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)

We have already know that[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

RD Sharma Solutions for Class 12 Maths Chapter 7 Adjoint and Inverse of a Matrix Image 130

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:

RD Sharma Solutions for Class 12 Maths Chapter 7: Download PDF

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

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Chapterwise RD Sharma Solutions for Class 12 Maths :

About RD Sharma

RD Sharma isn't the kind of author you'd bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like 'series solution of linear differential equations'. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it's only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. "I like to spend all my time thinking and writing about maths problems. I find it relaxing," he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government's Guru Nanak Dev Institute of Technology.

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