# RD Sharma Class 12 Ex 7.1 Solutions Chapter 7 Adjoint and Inverse of a Matrix

Here we provide RD Sharma Class 12 Ex 7.1 Solutions Chapter 7 Adjoint and Inverse of a Matrix for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 7.1 Solutions Chapter 7 Adjoint and Inverse of a Matrix book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 7.1 Solutions Chapter 7 Adjoint and Inverse of a Matrix

### (i)

Solution:

Here, A =

Cofactors of A are:

C11 = 4      C12 = -2

C21 = -5    C22 = -3

|A|I =  =

Hence Proved

### (ii)

Solution:

Here, A =

Cofactors of A are:

C11 = d      C12 = -c

C21 = -b    C22 = a

|A|I =

Hence Proved

### (iii)

Solution:

Here, A =

Cofactors of A are:

C11 = cos α     C12 = -sin α

C21 = -sin α    C22 = cos α

=

=

=

|A|I =

=

=

=

=

=

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

=

|A| =

= 1 + tan2 α/2

= sec2α/2

=

=

|A|I = (sec2α/2)

=

=

=

Hence Proved

### (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

=

=

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

|A|I= (5) =

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

=

|A| = 1(3 – 1) – 2(2 + 1) + 5(2 + 3)

= 2 – 6 + 25 = 21

|A|I = (21)

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

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

= -44 – 4 + 48 = 0

|A|I =

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

=

|A| = 2(3 – 0) – 0(15 – 0) – 1(5 – 1)

= 6 – 4 = 2

|A|I = (2)

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

=

=

= 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

=

### Question 5. If A = , show that adj A = 3AT.

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

=

AT=

Now, 3AT = 3 =

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

=

=

= 25

= 25I3

### (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

=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

### (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

= 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

### (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

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

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

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

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

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

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

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

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

Hence, A-1 =

To verify A-1A =

=

### (i) A = and B =

Solution:

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

We take LHS

AB =

|AB| = 18 × 52 – 22 × 43

= 936 – 946 = -10

Now,

A =

|A| = 15 – 14 = 1

Therefore, A-1 = adj A/|A| =

B =

|B| = 8 – 18 = -10

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

=

Now,

A =

|A| = 6 – 5 = 1

Therefore, A-1 = adj A/|A| =

B =

|B| = 16 – 15 = 1

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

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

Solution:

A =

|A| = 14 – 12 = 2

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

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

= 1/1

To show that

aA-1 = (a+ bc + 1)I – aA.

LHS = aA-1

= a

RHS = (a+ 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

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

=

=

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

=

=

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

A– 4A + I = O

– +

Hence, =

Now, A– 4A + I = O

A– 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

A+ 4A – 42I =  +  –

=

Hence,

Now, A+ 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 =

A

=

Now, A– 5A + 7I =  + 5 + 7

=

=

Now, A– 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, A– 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

A– 9A + 14I = 0

⇒ 9A = A+ 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 A= λA – 2I

λA = A+ 2I

⇒ λ  =

⇒ λ  =

⇒ λ = 1

Now, A= 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, A– 3A – 7=

=

We have, A– 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, A– 12A + I = –

=

We have, A– 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 =

A

A3 =

A– 6A2 + 5A + 11I

– 6

=

We have, A– 6A2 + 5A + 11I = O.

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

⇒ A– 6A + 5I = -11A-1

⇒ -11A-1 = (A– 6A + 5I)

=

=

=

=

Therefore, A-1 =

### Question 25. Show that the matrix A =  satisfies the equation A3 – A2 – 3A – I3 = 0. Hence, find A-1.

Solution:

Here, A =

A2 =

A3 =

Now A– A– 3A – I

=

So, A– A– 3A – I= 0

⇒ A-1(AAA) – A-1(AA) – 3A-1A – A-1I = 0

⇒ A– A – 3I – A-1 = 0

⇒ A-1 = A– A – 3I

Therefore, A-1 =

### Question 26. If A = . Verify that A3 – 6A2 + 9A – 4I = O and hence find A-1.

Solution:

Here, A =

A2 =

=

A3 = A2A =

Now, A– 6A+ 9A – 4I

So, A– 6A+ 9A – 4I = O

Multiplying both side by A-1

⇒ A-1(AAA) – 6A-1(AA) 9A-1A – 4A-1I = O

⇒ AAI – 6AI + 9I = 4A-1

⇒ 4A-1 = A2I – 6AI + 9I

=

=

=

⇒ A-1

### Question 27. If A =, prove that A-1 = AT.

Solution:

Here, A =

AT =

Now, Finding A-1

|A| = 1/9[-8(16 + 56) – 1(16 – 7) + 4(-32 – 4)]

= -81

Therefore, inverse of A exists

Cofactors of A are:

C11 = 72      C12 = -9       C13 = -36

C21 = -36    C22 = -36    C23 = -63

C31 = -9      C32 = 72      C33 = -36

=

=

Hence, A-1 =

= AT

Hence Proved

### Question 28. If A = , show that A-1 = A3.

Solution:

Here, A =

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

= 3 + 6 – 8

= 1

Therefore, inverse of A exists

Cofactors of A are:

C11 = 1      C12 = -2     C13 = -2

C21 = -1    C22 = 3      C23 = 3

C31 = 0     C32 = -4     C33 = -3

Now, A2 =

=

A3 =

=

= A-1

Hence Proved

### Question 29. If A =, show that A2 = A-1.

Solution:

Here, A =

LHS = A2

|A| = -1(1 – 0) – 2(-1 – 0) + 0

= 1

Therefore, inverse of A exists

Cofactors of A are:

C11 = -1     C12 = 0     C13 = -1

C21 = 0      C22 = 0     C23 = 1

C31 = 21    C32 = 1      C33 = 1

=

Hence, A-1 =

=

= A

Hence Proved

### Question 30. Solve the matrix equation, where X is a 2×2 matrix.

Solution:

We have,

Let A =  and B =

So. AX = B

⇒ X = A-1B

Now, |A| = 5 – 4 = 1

Co factors of A are:

C11 = 1         C12 = -1

C21 = -4      C22 = 5

=

A-1 =

Therefore, X =

X =

### Question 31. Find the matrix X satisfying the matrix equation: X.

Solution:

We have,

Let A =and B =

So, XA = B

XAA-1 = BA-1

XI = BA-1 ………..(i)

Now, |A| = -7

Co factors of A are:

C11 = -2       C12 = 1

C21 = -3      C22 = 5

=

=

Therefore, X =

=

X =

### Question 32. Find the matrix X for which: X

Solution:

Let, A =

B =

C =

Then the given equation becomes

A × B = C

⇒ X = A-1CB-1

Now |A| = 35 -14 = 21

|B| = -1 + 2 = 1

X = A-1 CB-1 =

### Question 33. Find the matrix X satisfying the equation:

Solution:

Let A =  B =

AXB = I

X = A-1B-1

|A| = 6 – 5 = 1

|B| = 10 – 9 = 1

A-1 = adj A /|A| =

X =

### Question 34. If A = , Find A-1 and prove that A2 – 4A – 5I = O.

Solution:

Here, A =

A2 =

Now, A+ 4A – 5I

Now, A– 4A – 5I = O

⇒ A-1AA – 4A-1A – 5A-1I = O

⇒ 5A-1 = [A – 4I]

A-1

### Question 35. If A is a square matrix of order n, prove that |A adj A| = |A|n.

Solution:

Given, |A adj A| = |A|n

Taking LHS = |A Adj A|

= |A| |A|n-1

= |A|n-1+1

= |A|n = RHS

Hence Proved

### Question 36. If A-1 =  and B = , find (AB)-1.

Solution:

Here, B =

|B| = 1(3 – 0) – 2(-1 – 0) – 2(2 – 0)

= 3 + 2 – 4 = 1

Therefore, inverse of B exists

Cofactors of B are:

C11 = 3       C12 = 1      C13 = 2

C21 = 2      C22 = 1      C23 = 2

C31 = 6     C32 = 2      C33 = 5

Therefore,

B-1

Hence, (AB)-1 = B-1A-1

### Question 37. If A = , find (AT)-1.

Solution:

Assuming B = AT =

|B| = 1(-1 – 8) – 0 – 2(-8 + 3)

= -9 + 10 = 1

Therefore, inverse of B exists

Cofactors of B are:

C11 = -9      C12 = 8      C13 = -5

C21 = -8     C22 = 7      C23 = -4

C31 = -2     C32 = 2      C33 = -1

=

B-1 =

or (AT)-1 =

### Question 38. Find the adjoint of the matrix A = and hence show that A (adj A) = |A|I3.

Solution:

Here, A =

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

= 3 + 12 + 12 = 27

Therefore, inverse of A exists

Cofactors of A are:

C11 = -3     C12 = -6      C13 = -6

C21 = 6      C22 = 3       C23 = -6

C31 = 6      C32 = -6      C33 = 3

or A (adj A) = 27

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

Hence Proved

### Question 39. If A = , A-1 and show that A-1 = 1/2(A2 – 3I).

Solution:

Here, A =

|A| = 0 – 1(0 – 1) + 1(1 – 0)

= 1 + 1 = 2

Therefore, inverse of A exists

Cofactors of A are:

C11 = -1     C12 = 1      C13 = 1

C21 = 1      C22 = -1    C23 = 1

C31 = 1      C32 = 1      C33 = -1

Hence, A-1 =

Now, A2 – 3I =

=

=

Hence, A-1 = 1/2(A2 – 3I)

Hence Proved

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