# Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.2

### Find the inverse of each of the following matrices by using elementary row transformation(Questions 1- 16):

### Question 1.

**Solution:**

Here, A =

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Using elementary row operation

⇒

R

_{1 }-> 1/7R_{1}⇒

R

_{2 }-> R_{2 }– 4R_{1}⇒

R

_{2 }-> (-7/25)R_{2}⇒

R

_{1 }-> R_{1 }– 1/7R_{2}⇒

Therefore, A

^{-1 }=

### Question **2. **

**Solution:**

Here, A =

A = AI

Using elementary row operation

⇒

R

_{1 }-> 1/5R_{1}⇒

R

_{1 }-> R_{2 }– 2R_{1}⇒

R

_{2 }-> 5R_{2}⇒

R

_{1 }-> R_{1 }– 2/5R_{2}⇒

Therefore, A

^{-1}=

### Question 3.

**Solution:**

Here, A =

A = AI

Using elementary row operation

⇒

R

_{2 }-> R_{2 }+ 3R_{1}⇒

R

_{2 }-> 1/23R_{2}⇒

R

_{1 }-> R_{1 }– 6R_{1}⇒

Therefore, A

^{-1}=

### Question 4.

**Solution:**

Here,

A = AI

Using elementary row operation

⇒

R

_{1 }-> 1/2R_{1}⇒

R

_{2 }-> R_{2 }– R_{1}⇒

R

_{2 }-> 2R_{2}⇒

R

_{1 }-> R_{1 }– 5/2R_{2}⇒

Therefore, A

^{-1 }=

### Question 5.

**Solution:**

Here, A =

A = AI

⇒

R

_{1 }-> 1/3R_{1}R

_{2 }-> R_{2 }– 2R_{1}R

_{2 }-> 3R_{2}R

_{1 }-> R_{1 }– 10/3R_{2}⇒

Therefore, A

^{-1 }=

### Question 6.

**Solution:**

Here, A =

A = IA

⇒

R

_{1 }↔ R_{2}⇒

R

_{3 }-> R_{3 }– 3R_{1}⇒

R

_{1 }-> R_{1 }– 2R_{2}, R_{3 }-> R_{3 }+ 5R_{2}⇒

R

_{3}-> R_{3}/2⇒

R

_{1}-> R_{1}+ R_{3}, R_{2}-> R_{2}– 2R_{3}⇒

Therefore, A

^{-1}=

### Question 7.

**Solution:**

Here, A =

A = IA

⇒

R

_{1 }-> R_{1}/2⇒

R

_{2 }-> R_{2 }– 5R_{1}⇒

R

_{3 }-> R_{3 }– R_{2}⇒

R

_{3 }-> 2R_{3}⇒

R

_{1 }-> R_{1 }+ 1/2R_{3}, R_{2 }-> R_{2 }– 5/2R_{3}⇒

Therefore, A

^{-1 }=

### Question 8.

**Solution:**

Here, A =

A = IA

⇒

R

_{1 }-> 1/2R_{1}⇒

R

_{2 }-> R_{2 }– 2R_{1}, R_{3 }-> R_{3 }– 3R_{1}⇒

R

_{1 }-> R_{1 }– 3/2R_{2}, R_{3 }-> R_{3 }– 5/2R_{2}⇒

R

_{3 }-> 2R_{3}⇒

R

_{1 }-> R_{1 }– 1/2R_{3}⇒

Therefore, A

^{-1 }=

### Question 9.

**Solution:**

Here, A =

A = IA

⇒

R

_{1 }-> 1/3R_{1}⇒

R

_{2 }-> R_{2 }– 2R_{1}⇒

R

_{2 }-> (-1)R_{2}⇒

R

_{1 }-> R_{1 }+ R_{2}, R_{3 }-> R_{3 }+ R_{2}⇒

R

_{3 }-> (-3)R_{3}⇒

R

_{2 }-> R_{2 }+ 4/3R_{3}⇒

Therefore, A

^{-1 }=

### Question 10.

**Solution:**

Here, A =

⇒

R

_{2 }-> R_{2 }– 2R_{1}, R_{3 }-> R_{3 }– R_{1}⇒

R

_{2 }-> (-1)R_{2}⇒

R

_{1 }-> R_{1 }– 2R_{2}, R_{3 }-> R_{3 }+ 3R_{2}⇒

R

_{3 }-> R_{3}/6⇒

R

_{1 }-> R_{1 }+ 2R_{3}, R_{2 }-> R_{2 }– R_{3}⇒

Therefore, A

^{-1 }=

### Question 11.

**Solution:**

Here, A =

A = IA

⇒

R

_{1 }-> R_{1}/2⇒

R

_{2 }-> R_{2 }– R_{1}, R_{3 }-> R_{3 }– 3R_{1}⇒

R

_{2 }-> (2/5)R_{2}⇒

R

_{1 }-> R_{1 }+ 1/2 R_{2}, R_{3 }-> R_{3 }– 5/2R_{2}⇒

R

_{3 }-> R_{3}/-6⇒

R

_{2 }-> R_{2 }– R_{3}, R_{1 }-> R_{1 }– 2R_{3}⇒

Therefore, A

^{-1}=

### Question 12.

**Solution:**

Here, A =

A = IA

⇒

R

_{2 }-> R_{2 }– 3R_{1}, R_{3 }-> R_{3 }– 2R_{1}⇒

R

_{2 }-> R_{2}/(-2)⇒

R

_{1 }-> R_{1 }– R_{2}, R_{3 }-> R_{3 }– R_{2}⇒

R

_{3 }-> (-2/11)R_{3}⇒

R

_{1 }-> R_{1 }+ 1/2R_{3}, R_{2 }-> R_{2 }– 5/2R_{3}⇒

Therefore, A

^{-1 }=

### Question 13.

**Solution:**

Here, A =

A = IA

⇒

R

_{1 }-> 1/2R_{1}⇒

R

_{2 }-> R_{2 }– 4R_{1}, R_{3 }-> R_{3 }– 3R_{1}⇒

R

_{2 }-> 1/2R_{2}⇒

R

_{1 }-> R_{1 }+ 1/2R_{2}, R_{3 }-> R_{3 }+ 1/2R_{2}⇒

R

_{3 }-> (-2)R_{3}⇒

R

_{1 }-> R_{1 }– 1/2R_{3}, R_{2 }-> R_{2 }+ 3R_{3}⇒

Therefore, A

^{-1 }=

### Question 14.

**Solution:**

Here, A =

A = IA

⇒

R

_{1}-> (1/3)R_{1}⇒

R

_{2}-> R_{2 }– 2R_{1}⇒

R

_{2}-> (1/3)R_{2}⇒

R

_{3}-> R_{3 }– 4R_{2}⇒

R

_{3 }-> 9R_{3}⇒

R

_{1 }-> R_{1 }+ 1/3R_{3}, R_{2 }-> R_{2 }– 2/9R_{3}⇒

Therefore, A

^{-1 }=

### Question 15.

**Solution:**

Here, A =

A = IA

⇒

R

_{2 }-> 3R_{1 }+ R_{2}, R_{3 }-> R_{3 }– 2R_{1}⇒

R

_{1 }-> R_{1 }– 3R_{2}, R_{3 }-> R_{3 }+ 5R_{2}⇒

R

_{2 }-> R_{2 }+ 5/9R_{3}, R_{1 }-> R_{1 }+ 1/3R_{3}⇒

Therefore, A

^{-1 }=

### Question 16.

**Solution:**

Here, A=

A = IA

⇒

R

_{1 }-> (-1)R_{1}⇒

R

_{2 }-> R_{2 }– R_{1}, R_{3 }-> R_{3 }– 3R_{1}⇒

R

_{2 }-> R_{2}/3⇒

R

_{1 }-> R_{1 }+ R_{2}, R_{3 }-> R_{3 }– 4R_{2}⇒

R

_{3 }-> R_{3}/3⇒

R

_{1 }-> R_{1 }+ 1/3R_{3}, R_{2 }-> R_{2 }– 5/3R_{3}⇒

Therefore, A

^{-1}=