# Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.1 | Set 2

• Last Updated : 28 Mar, 2021

### (i) A = and B =

Solution:

To prove (AB)-1= B-1A-1

We take LHS

AB =

|AB| = 18 × 52 – 22 × 43

= 936 – 946 = -10

adj(AB) =

AB-1= adj(AB)/|AB| =

Now,

A =

|A| = 15 – 14 = 1

adj A =

Therefore, A-1 = adj A/|A| =

B =

|B| = 8 – 18 = -10

adj B =

Therefore, B-1= adj B/|B| =

Now, we take RHS

B-1A-1

LHS = RHS

Hence, Proved

### (ii) A = and B =

Solution:

To prove (AB)-1 = B-1A-1

We take LHS

AB =

|AB| = 11 × 27 – 29 × 14

= 407 – 406 = 1

adj(AB) =

AB-1= adj(AB)/|AB| =

=

Now,

A =

|A| = 6 – 5 = 1

adj A=

Therefore, A-1 = adj A/|A| =

B =

|B| = 16 – 15 = 1

adj B =

Therefore, B-1= adj B/|B| =

Now, we take RHS

B-1A-1

LHS = RHS

Hence, Proved

### Question 11.  Let A = and B = . Find (AB)-1.

Solution:

AB =

=

|AB| = 34 × 94 – 39 × 82 = -2

adj(AB) =

AB-1 = adj(AB)/|AB| =

### Question 12. Given A = , Compute A-1 and show that 2A-1 = 9I – A.

Solution:

A =

|A| = 14 – 12 = 2

adj A =

Therefore, A-1 = adj A/|A| =

To show 2A-1 = 9I – A.

LHS = 2 × (1/2)

Now we take RHS

= 9I – A

–

=

LHS = RHS

Hence Proved

### Question 13. If A = , then show that A – 3I = 2(I + 3A-1).

Solution:

Here, A =

|A| = 4 – 10 = -6

adj A =

Therefore, A-1 = adj A/|A| =

To show, A – 3I = 2(I + 3A-1)

Now we take LHS

= A – 3I

= – 3

=

Now we take RHS

= 2I + 6A-1

= 2 + 6 × (1/6)

LHS = RHS

Hence Proved

### Question 14. Find the inverse of the matrix A = and show that aA-1 = (a2 + bc + 1)I – aA.

Solution:

Here, A =

|A| = (a + abc)/a – bc = 1

Therefore, 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

To show that

aA-1 = (a2 + bc + 1)I – aA.

LHS = aA-1

= a

RHS = (a2 + bc + 1)I – aA

– a

–

LHS = RHS

Hence Proved

### Question 15. Given A = , B-1 = , Compute (AB)-1.

Solution:

We know (AB)-1 = B-1A-1

Here, A =

|A| = 5(3 – 4) + 4(4 – 3) = -5 + 4 = -1

Co-factors of A are:

C11 = -1      C12 = 0      C13 = 1

C21 = 8       C22 = 1      C23 = -10

C31 = -12    C32 = -2    C33 = 15

adj A =

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

Hence, A-1

(AB)-1 = B-1A-1

### (i) [F(α)]-1 = F(-α)

Solution:

We have F(α) =

|F(α)| = cos2α + sin2α = 1

Therefore, inverse of F(α) exists

Cofactors of F(α) are:

C11 = cosα    C12 = -sinα     C13 = 0

C21 = sinα     C22 = cosα     C23 = 0

C31 = 0          C32 = 0           C33 = 1

Adj F(α) =

=

=

[F(α)]-1 = 1/|F(α)|. adj F(α)

Hence, [F(α)]-1 = 1/1

=

Now, F(-α) =

=

So, [F(α)]-1 = F(-α)

Hence, Proved

### (ii) [G(β)]-1 = G(-β)

Solution:

We have G(β) =

|G(β)| = cos2β + sin2β = 1

Therefore, inverse of G(β) exists

Cofactors of G(β) are:

C11 = cosβ   C12 = 0      C13 = sinβ

C21 = 0         C22 = 1      C23 = 0

C31 = -sinβ   C32 = 0     C33 = sinβ

Adj G(β) =

=

=

[G(β)]-1 = 1/|G(β)|. adj G(β)

Hence, [G(β)]-1 = 1/1

Now, G(-β) =

So, [G(β)]-1 = G(-β)

Hence, Proved

### (iii) [F(α)G(β)]-1 = F(-α)G(-β)

Solution:

We already know that S[G(β)]-1 = G(-β)

[F(α)]-1 = F(-α)

Taking LHS = [F(α)G(β)]-1

= [F(α)]-1[G(β)]-1

= F(-α)G(-β) = RHS

Hence, Proved

### Question 17. If A = , Verify that A2 – 4A + I = O, where I =  and O = , Hence, find A-1.

Solution:

Here, A =

A2

4A = 4

A2 – 4A + I = O

– +

Hence, =

Now, A2 – 4A + I = O

A2 – 4A = -I

Multiplying both side by A-1 both sides we get

A.A(A-1) – 4AA-1 = -IA-1

AI – 4I = -A-1

A-1 = 4I – AI

–

### Question 18.  Show that A =  satisfies the equation A2 + 4A – 42I = O. Hence, Find A-1.

Solution:

Here, A =

A2

=

=

4A = 4

A2 + 4A – 42I =  +  –

=

Hence,

Now, A2 + 4A – 42I = 0

⇒ A-1A.A + 4A-1.A – 42A-1I = 0

⇒ IA + 4I – 42A-1 = 0

⇒ A-1 = 1/42 [A + 4I]

⇒ A-1

### Question 19. If A = , show that A2 – 5A + 7I = O. Hence find A-1.

Solution:

Here, A =

A2

=

Now, A2 – 5A + 7I =  + 5 + 7

=

=

Now, A2 – 5A + 7I = O

Multiplying by A-1 both sides

⇒ A-1AA + 5AA – 1 + 7IA-1 = 0

⇒ A-1 = 1/7[5I – A]

⇒ A-1

⇒ A-1

### Question 20. If A = , find x and y such that A2 – xA + yI = O. Hence, evaluate A-1.

Solution:

Here, A =

A2

Now, A2 – xA + yI = O

⇒  –   +

⇒ 22 – 4x + y = 0

⇒ 4x – y = 22  ………(i)

or

18 – 2x = 0

⇒ x = 9

Putting x = 9 in eq (i)

⇒ y = 14

A2 – 9A + 14I = 0

⇒ 9A = A2 + 14I

⇒ 9A-1A = A-1AA + 14A-1

⇒ 9I = IA + 14A-1

⇒ A-1 = 1/14[9I – A] = 1/14()

⇒ A-1

### Question 21. If A = , find the value of λ so that A2 = λA – 2I. Hence, find A-1.

Solution:

Here, A =

A2

If A2 = λA – 2I

λA = A2 + 2I

⇒ λ  =

⇒ λ  =

⇒ λ = 1

Now, A2 = λA – 2I

Multiplying both side A-1

⇒ A-1AA = A-1A – 2A-1I

⇒ A = I – 2A-1

⇒ 2A-1 = I – A =

A-1

### Question 22. Show that A =  satisfies the equation x2 – 3x – 7 = 0. Thus, find A-1.

Solution:

Here, A =

A2

Now, A2 – 3A – 7=

=

We have, A2 – 3A – 7 = 0

⇒ A-1AA – 3A-1A – 7A-1 = 0

⇒ A-3I – 7A-1 = 0

⇒ 7A-1 = A – 3I

⇒ 7A-1 –

A-1

### Question 23. Show that A =  satisfies the equation x2 – 12x + 1 = 0. Thus, find A-1.

Solution:

Here, A =

A2

Now, A2 – 12A + I = –

=

We have, A2 – 12A + I = 0

⇒ A – 12I + A-1 = 0

⇒ A-1 = 12I – A

⇒ A-1

⇒ A-1

### Question 24. For the matrix A = show that A3 – 6A2 + 5A + 11I3 = O.  Hence, find A-1.

Solution:

Here, A =

A2

A3

A3 – 6A2 + 5A + 11I

– 6

=

We have, A3 – 6A2 + 5A + 11I = O.

⇒ A-1(AAA) – 6A-1(AA) + 5A-1A + 11IA-1 = 0

⇒ A2 – 6A + 5I = -11A-1

⇒ -11A-1 = (A2 – 6A + 5I)

=

=

=

=

Therefore, A-1

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