Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.1 | Set 1

Question 1. Find the adjoint of the following matrices:

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

(i)  

Solution:

Here, A = 

Cofactors of A are:

C₁₁ = 4      C₁₂ = -2

C₂₁ = -5    C₂₂ = -3

adj A =

(adj A) = 

= 

To Prove, (adj A)A = |A|I = A(adj A)

(adj A)A =  

|A|I =  =  

A(adj A) = 

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

Hence Proved

(ii) 

Solution:

Here, A = 

Cofactors of A are:

C₁₁ = d      C₁₂ = -c

C₂₁ = -b    C₂₂ = a

(adj A) =

= 

To Prove, (adj A)A = |A|I = A(adj A)

(adj A)A = 

|A|I =

A(adj A) =

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

Hence Proved

(iii) 

Solution:

Here, A = 

Cofactors of A are:

C₁₁ = cos α     C₁₂ = -sin α

C₂₁ = -sin α    C₂₂ = cos α

(adj A) =

=

To Prove, (adj A)A = |A|I = A(adj A)

(adj A)A = 

=

=

|A|I = 

=

=

=

A(adj A) = 

=

=

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

Hence Proved

(iv) 

Solution:

Here, A = 

Cofactors of A are:

C₁₁ = 1    C₁₂ = -(-tan α/2) = tan α/2

C₂₁ = -tan α/2    C₂₂ = 1

adj A = 

=

To Prove, (adj A)A = |A|I = A(adj A)

|A| = 

= 1 + tan² α/2

= sec²α/2

(adj)A = 

=

=

|A|I = (sec²α/2)

=

A(adj A) = 

=

=

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

Hence Proved

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

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

(i) 

Solution:

Here, A = 

Cofactors of A are

C₁₁ =  = -3

C₂₁ =  = 2

C₃₁ =  = 2

C₁₂ =  = 2

C₂₂ = =-3

C₃₂ =  = 2

C₁₃ =  = 2

C₂₃ =  = 2

C₃₃ =  = -3

adj A = 

=

=

To Prove, (adj A)A = |A|I = A(adj A)

|A| = -3 + 4 + 4 = 5

(adj A)A = 

|A|I= (5) = 

A(adj A) = 

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

Hence Proved

(ii) 

Solution:

Here, A = 

Cofactors of A are

C₁₁ =  = 2

C₁₂ =  = -3

C₁₃ =  = 5

C₂₁ =  = 3

C₂₂ =  = 6

C₂₃ =  = -3

C₃₁ =  = -13

C₃₂ =  = 9

C₃₃ =  = -1

adj A = 

= 

=

To Prove, (adj A)A = |A|I = A(adj A)

|A| = 1(3 – 1) – 2(2 + 1) + 5(2 + 3)

= 2 – 6 + 25 = 21

(adj A)A = 

|A|I = (21)

A(adj A) = 

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

Hence Proved

(iii) 

Solution:

Here, A = 

Cofactors of A are

C₁₁ =  = -22

C₁₂ = – = 4

C₁₃ =  = 16

C₂₁ = – = 11

C₂₂ =  = -2

C₂₃ = – = -8

C₃₁ =  = -11

C₃₂ = – = 2

C₃₃ =  = 8

adj A = 

= 

To Prove, (adj A)A = |A|I = A(adj A)

|A| = 2(-2 – 20) + 1(-4 – 0) + 3(16 – 0)

= -44 – 4 + 48 = 0

(adj A)A = 

|A|I = 

A(adj A) = 

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

Hence Proved

(iv) 

Solution:

Here, A =  

Cofactors of the A are

C₁₁ =  = 3

C₁₂ = – = -15

C₁₃ =  = 4

C₂₁ =  = -1

C₂₂ =  = 7

C₂₃ =  = -2

C₃₁ =  = 1

C₃₂ =  = -5

C₃₃ = = 2

adj A = 

=

To Prove, (adj A)A = |A|I = A(adj A)

|A| = 2(3 – 0) – 0(15 – 0) – 1(5 – 1)

= 6 – 4 = 2

(adj A)A = 

|A|I = (2)

A (adj A) = 

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

Hence Proved

Question 3. For the matrix A =, show that A(adj A) = O.

 Solution:

Cofactor of A are, 

C₁₁ = 30    C₁₂ = -20    C₁₃ = -50

C₂₁ = 12    C₂₂ = -8     C₂₃ = -20

C₃₁ = -3    C₃₂ = 2       C₃₃ = 5  

adj A = 

=

A(adj A) = 

=

= 0

Hence Proved

Question 4. If A =, show that adj A = A. 

Solution:

Here, A =

Cofactor of A are,

C₁₁ = -4    C₁₂ = 1     C₁₃ = 4

C₂₁ = -3    C₂₂ = 0    C₂₃ = 4

C₃₁ = 4    C₃₂ = 4     C₃₃ = 3  

adj A = 

=

Therefore, adj A = A

Question 5. If A = , show that adj A = 3A^T.

Solution:

Here, A = 

Cofactor of A are,

C₁₁ = -3    C₁₂ = -6    C₁₃ = -6

C₂₁ = 6    C₂₂ = 3      C₂₃ = -6

C₃₁ = 6    C₃₂ = -6    C₃₃ = 3  

adj A = 

=

A^T=

Now, 3A^T = 3 =

adj A = 3.A^T 

Hence Proved

Question 6. Find A(adj A) for the matrix A =.

Solution:

Here, A =

Cofactor of A are,

C₁₁ = 9    C₁₂ = 4    C₁₃ = 8

C₂₁ = 19    C₂₂ = 14    C₂₃ = 3

C₃₁ = -4    C₃₂ = 1    C₃₃ = 2

adj A = 

=

=

A(adj A) = 

= 

= 25

= 25I₃

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

(i) 

Solution:

Here, A = 

|A| = cos²θ + sin²θ = 1

Hence, inverse of A exist 

Cofactors of A are,

Cofactor of A are,

C₁₁ = cos θ     C₁₂ = sin θ

C₂₁ = -sin θ    C₂₂ = cos θ

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

=1/1.   

(ii)     

Solution:

Here, A =   

|A| = -1

Hence, inverse of A exist  

Cofactor of A are,

C₁₁ = 0      C₁₂ = -1

C₂₁ = -1    C₂₂ = 0

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

= 

= 

(iii) 

Solution:

Here, A =       

|A| = a(1 + bc)/a – bc = 1 + bc – bc = 1

Hence, inverse of A exists.  

Cofactor of A are,

C₁₁ = (1 + bc)/a     C₁₂ = -c

C₂₁ = -b                 C₂₂ = a

adj A =

= 

= 

A^-1 = 1/|A|. adj A

= 1/1 

= 

(iv) 

Solution:

Here, A = 

|A| = 2 + 15 = 17

Hence, inverse of A exists.  

Cofactor of A are,

C₁₁ = 1      C₁₂ = 3

C₂₁ = -5   C₂₂ = 2

adj A = 

=       

= 

A^-1 = 1/|A|. adj A

= 

= 

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

(i) 

Solution:

Here, A =

|A| = 1(6 – 1) – 2(4 – 3) + 3(2 – 9)

= 5 – 2 – 21 = -18

Therefore, inverse of A exists

Cofactors of A are:

C₁₁ = 5    C₁₂ = -1      C₁₃ = -7

C₂₁ = -1    C₂₂ = -7    C₂₃ = 5

C₃₁ = -7    C₃₂ = 5     C₃₃ = -1  

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

Hence, A^-1 = 

=  

(ii) 

Solution:

Here, A = 

|A| = 1(1 + 3) – 2(-1 + 2) + 5(3 + 2)

= 4 – 2 – 25 = 27

Therefore, inverse of A exists

Cofactors of A are:

C₁₁ = 4        C₁₂ = -1     C₁₃ = 5

C₂₁ = -17    C₂₂ = -11   C₂₃ = 1

C₃₁ = 3       C₃₂ = 6      C₃₃ = -3

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

Hence, A^-1 = 

= 

=   

(iii) 

Solution:

Here, A =

|A| = 2(4 – 1) – (-1)(-2 + 1) + 1(1 – 2)

= 6 – 1 – 1 = 4

Therefore, inverse of A exists

Cofactors of A are:

C₁₁ = 3    C₁₂ = 1      C₁₃ = -1

C₂₁ = 1    C₂₂ = 3     C₂₃ = 1

C₃₁ = -1    C₃₂ = 1    C₃₃ = 3

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

Hence, A^-1 = 

=    

(iv) 

Solution:

Here, A =

|A| = 2(3 – 0) – 0 + 1(5)

= 6 – 5 = 1

Therefore, inverse of A exists

Cofactors of A are:

C₁₁ = 3     C₁₂ = -15     C₁₃ = 5

C₂₁ = -1   C₂₂ = 6        C₂₃ = -2

C₃₁ = 1     C₃₂ = -5      C₃₃ = 2

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

Hence, A^-1 = 

=   

(v) 

Solution:

Here, A = 

|A| = 0 – 1(16 – 12) – 1(-12 + 9)

= -4 + 3 = -1

Therefore, inverse of A exists

Cofactors of A are:

C₁₁ = 0    C₁₂ = -4    C₁₃ = -3

C₂₁ = -1   C₂₂ = 3     C₂₃ = 3

C₃₁ = 1    C₃₂ = -4    C₃₃ = -4

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

Hence, A^-1 = 

=          

(vi) 

Solution:

Here, A = 

|A| = 0 – 0 – 1(-12 + 8)

= -1(-4) = 4

Therefore, inverse of A exists

Cofactors of A are:

C₁₁ = -8    C₁₂ = 11      C₁₃ = -4

C₂₁ = 4     C₂₂ = -2     C₂₃ = 0

C₃₁ = 4    C₃₂ = -3      C₃₃ = 0

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

Hence, A^-1 = 

=   

(vii) 

Solution:

Here, A = 

|A| = -cos²α – sin²α

= -(cos²α + sin²α) = -1

Therefore, inverse of A exists

Cofactors of A are:  

C₁₁ = -1     C₁₂ = 0           C₁₃ = -0

C₂₁ = 0      C₂₂ = -cosα   C₂₃ = -sinα

C₃₁ = 0      C₃₂ = -sinα     C₃₃ = cosα

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

Hence, A^-1 = 

=       

Question 9. (i) 

Solution:

Here, A = 

|A| = 1(16 – 9) – 3(4 – 3) + 3(3 – 4)

= 7 – 3 – 3 = 1

Therefore, inverse of A exists

Cofactors of A are:

C₁₁ = 7       C₁₂ = -1   C₁₃ = -1

C₂₁ = -3    C₂₂ = 1     C₂₃ = 0

C₃₁ = -3    C₃₂ = 0    C₃₃ = 1

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

Hence, A^-1 = 1/1

=

To verify A^-1A = 

= 

=         

(ii) 

Solution:

Here, A = 

|A| = 2(8 – 7) – 3(6 – 3) + 1(21 – 12)

= 2 – 3(3) + 1(9) = 2

Therefore, inverse of A exists

Cofactors of A are:

C₁₁ = 1      C₁₂ = -3    C₁₃ = 9

C₂₁ = 1     C₂₂ = 1      C₂₃ = -5

C₃₁ = -1   C₃₂ = 1      C₃₃ = -1

adj A = 

= 

= 

A^-1 = 1/|A|. adj A

Hence, A^-1 = 

To verify A^-1A = 

= 

=