# 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:

C

_{11 }= 4 C_{12 }= -2C

_{21 }= -5 C_{22 }= -3adj 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:

C

_{11 }= d C_{12 }= -cC

_{21 }= -b C_{22 }= 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:

C

_{11 }= cos α C_{12 }= -sin αC

_{21 }= -sin α C_{22 }= 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:

C

_{11 }= 1 C_{12 }= -(-tan α/2) = tan α/2C

_{21 }= -tan α/2 C_{22 }= 1adj A =

=

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

|A| =

= 1 + tan

^{2}α/2= sec

^{2}α/2(adj)A =

=

=

|A|I = (sec

^{2}α/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

C

_{11 }= = -3C

_{21 }= = 2C

_{31 }= = 2C

_{12 }= = 2C

_{22 }= =-3C

_{32 }= = 2C

_{13 }= = 2C

_{23 }= = 2C

_{33 }= = -3adj 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

C

_{11 }= = 2C

_{12 }= = -3C

_{13 }= = 5C

_{21 }= = 3C

_{22 }= = 6C

_{23 }= = -3C

_{31 }= = -13C

_{32 }= = 9C

_{33 }= = -1adj 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

C

_{11 }= = -22C

_{12 }= – = 4C

_{13 }= = 16C

_{21 }= – = 11C

_{22 }= = -2C

_{23 }= – = -8C

_{31 }= = -11C

_{32 }= – = 2C

_{33 }= = 8adj 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

C

_{11 }= = 3C

_{12 }= – = -15C

_{13 }= = 4C

_{21 }= = -1C

_{22 }= = 7C

_{23 }= = -2C

_{31 }= = 1C

_{32 }= = -5C

_{33 }= = 2adj 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,

C

_{11 }= 30 C_{12}= -20 C_{13 }= -50C

_{21 }= 12 C_{22}= -8 C_{23 }= -20C

_{31 }= -3 C_{32 }= 2 C_{33 }= 5adj A =

=

A(adj A) =

=

= 0

Hence Proved

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

**Solution:**

Here, A =

Cofactor of A are,

C

_{11 }= -4 C_{12}= 1 C_{13 }= 4C

_{21 }= -3 C_{22}= 0 C_{23 }= 4C

_{31 }= 4 C_{32 }= 4 C_{33 }= 3adj A =

=

Therefore, adj A = A

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

**Solution:**

Here, A =

Cofactor of A are,

C

_{11 }= -3 C_{12}= -6 C_{13 }= -6C

_{21 }= 6 C_{22}= 3 C_{23 }= -6C

_{31 }= 6 C_{32 }= -6 C_{33 }= 3adj 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,

C

_{11 }= 9 C_{12}= 4 C_{13 }= 8C

_{21 }= 19 C_{22}= 14 C_{23 }= 3C

_{31 }= -4 C_{32 }= 1 C_{33 }= 2adj A =

=

=

A(adj A) =

=

= 25

= 25I

_{3}

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

### (i)

**Solution:**

Here, A =

|A| = cos

^{2}θ + sin^{2}θ = 1Hence, inverse of A exist

Cofactors of A are,

Cofactor of A are,

C

_{11}= cos θ C_{12}= sin θC

_{21 }= -sin θ C_{22}= cos θadj A =

=

=

A

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

### (ii)

**Solution:**

Here, A =

|A| = -1

Hence, inverse of A exist

Cofactor of A are,

C

_{11 }= 0 C_{12}= -1C

_{21 }= -1 C_{22}= 0adj 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,

C

_{11 }= (1 + bc)/a C_{12}= -cC

_{21 }= -b C_{22}= aadj 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,

C

_{11 }= 1 C_{12}= 3C

_{21 }= -5 C_{22}= 2adj 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:

C

_{11 }= 5 C_{12}= -1 C_{13 }= -7C

_{21 }= -1 C_{22}= -7 C_{23 }= 5C

_{31 }= -7 C_{32 }= 5 C_{33 }= -1adj A =

=

=

A

^{-1 }= 1/|A|. adj AHence, 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:

C

_{11 }= 4 C_{12}= -1 C_{13 }= 5C

_{21 }= -17 C_{22}= -11 C_{23 }= 1C

_{31 }= 3 C_{32 }= 6 C_{33 }= -3adj A =

=

=

A

^{-1 }= 1/|A|. adj AHence, 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:

C

_{11 }= 3 C_{12}= 1 C_{13 }= -1C

_{21 }= 1 C_{22}= 3 C_{23 }= 1C

_{31 }= -1 C_{32 }= 1 C_{33 }= 3adj A =

=

=

A

^{-1 }= 1/|A|. adj AHence, 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:

C

_{11 }= 3 C_{12}= -15 C_{13 }= 5C

_{21 }= -1 C_{22}= 6 C_{23 }= -2C

_{31 }= 1 C_{32 }= -5 C_{33 }= 2adj A =

=

=

A

^{-1 }= 1/|A|. adj AHence, 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:

C

_{11 }= 0 C_{12 }= -4 C_{13 }= -3C

_{21 }= -1 C_{22}= 3 C_{23 }= 3C

_{31 }= 1 C_{32 }= -4 C_{33 }= -4adj A =

=

=

A

^{-1 }= 1/|A|. adj AHence, A

^{-1}==

### (vi)

**Solution:**

Here, A =

|A| = 0 – 0 – 1(-12 + 8)

= -1(-4) = 4

Therefore, inverse of A exists

Cofactors of A are:

C

_{11 }= -8 C_{12}= 11 C_{13 }= -4C

_{21 }= 4 C_{22}= -2 C_{23 }= 0C

_{31 }= 4 C_{32 }= -3 C_{33 }= 0adj A =

=

=

A

^{-1 }= 1/|A|. adj AHence, A

^{-1}==

### (vii)

**Solution:**

Here, A =

|A| = -cos

^{2}α – sin^{2}α= -(cos

^{2}α + sin^{2}α) = -1Therefore, inverse of A exists

Cofactors of A are:

C

_{11 }= -1 C_{12}= 0 C_{13 }= -0C

_{21 }= 0 C_{22}= -cosα C_{23 }= -sinαC

_{31 }= 0 C_{32 }= -sinα C_{33 }= cosαadj A =

=

=

A

^{-1 }= 1/|A|. adj AHence, 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:

C

_{11 }= 7 C_{12}= -1 C_{13 }= -1C

_{21 }= -3 C_{22}= 1 C_{23 }= 0C

_{31 }= -3 C_{32 }= 0 C_{33 }= 1adj A =

=

=

A

^{-1 }= 1/|A|. adj AHence, A

^{-1}= 1/1=

To verify A

^{-1}A ==

=

### (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:

C

_{11 }= 1 C_{12}= -3 C_{13 }= 9C

_{21 }= 1 C_{22}= 1 C_{23 }= -5C

_{31 }= -1 C_{32 }= 1 C_{33 }= -1adj A =

=

=

A

^{-1 }= 1/|A|. adj AHence, A

^{-1}=To verify A

^{-1}A ==

=