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

Here we provide RD Sharma Class 12 Ex 7.2 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 7.2 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 7.2 Solutions Chapter 7 Adjoint and

### Question 1.

Solution:

Here, A =

A = AI

Using elementary row operation

R-> 1/7R1

R-> R– 4R1

R-> (-7/25)R2

R-> R– 1/7R2

Therefore, A-1 =

### Question 2.

Solution:

Here, A =

A = AI

Using elementary row operation

R-> 1/5R1

R-> R– 2R1

R-> 5R2

R-> R– 2/5R2

Therefore, A-1

### Question 3.

Solution:

Here, A =

A = AI

Using elementary row operation

⇒

R-> R+ 3R1

R-> 1/23R2

R-> R– 6R1

Therefore, A-1 =

### Question 4.

Solution:

Here,

A = AI

Using elementary row operation

R-> 1/2R1

R-> R– R1

R-> 2R2

R-> R– 5/2R2

Therefore, A-1

### Question 5.

Solution:

Here, A =

A = AI

R-> 1/3R1

R-> R– 2R1

R-> 3R2

R-> R– 10/3R2

Therefore, A-1

### Question 6.

Solution:

Here, A =

A = IA

R↔ R2

R-> R– 3R1

R-> R– 2R2, R-> R+ 5R2

R3 -> R3/2

R1 -> R1 + R3, R2 -> R2 – 2R3

Therefore, A-1 =

### Question 7.

Solution:

Here, A =

A = IA

R-> R1/2

R-> R– 5R1

R-> R– R2

R-> 2R3

R-> R+ 1/2R3, R-> R– 5/2R3

Therefore, A-1

### Question 8.

Solution:

Here, A =

A = IA

R-> 1/2R1

R-> R– 2R1, R-> R– 3R1

⇒

R-> R– 3/2R2, R-> R– 5/2R2

R-> 2R3

R-> R– 1/2R3

Therefore, A-1

### Question 9.

Solution:

Here, A =

A = IA

R-> 1/3R1

R-> R– 2R1

R-> (-1)R2

R-> R+ R2, R-> R+ R2

R-> (-3)R3

R-> R+ 4/3R3

Therefore, A-1

### Question 10.

Solution:

Here, A =

R-> R– 2R1, R-> R– R1

R-> (-1)R2

R-> R– 2R2, R-> R+ 3R2

R-> R3/6

R-> R+ 2R3, R-> R– R3

Therefore, A-1

### Question 11.

Solution:

Here, A =

A = IA

R-> R1/2

R-> R– R1, R-> R– 3R1

R-> (2/5)R2

R-> R+ 1/2 R2, R-> R– 5/2R2

R-> R3/-6

R-> R– R3, R-> R– 2R3

Therefore, A-1 =

### Question 12.

Solution:

Here, A =

A = IA

R-> R– 3R1, R-> R– 2R1

R-> R2/(-2)

R-> R– R2, R-> R– R2

R-> (-2/11)R3

R-> R+ 1/2R3, R-> R– 5/2R3

Therefore, A-1

### Question 13.

Solution:

Here, A =

A = IA

R-> 1/2R1

R-> R– 4R1, R-> R– 3R1

R-> 1/2R2

R-> R+ 1/2R2, R-> R+ 1/2R2

R-> (-2)R3

R-> R– 1/2R3, R-> R+ 3R3

Therefore, A-1

### Question 14.

Solution:

Here, A =

A = IA

R1 -> (1/3)R1

R2 -> R– 2R1

R2 -> (1/3)R2

R3 -> R– 4R2

R-> 9R3

R-> R+ 1/3R3, R-> R– 2/9R3

Therefore, A-1

### Question 15.

Solution:

Here, A =

A = IA

R-> 3R+ R2, R-> R– 2R1

R-> R– 3R2, R-> R+ 5R2

R-> R+ 5/9R3, R-> R+ 1/3R3

Therefore, A-1

### Question 16.

Solution:

Here, A=

A = IA

R-> (-1)R1

R-> R– R1, R-> R– 3R1

R-> R2/3

R-> R+ R2, R-> R– 4R2

R-> R3/3

R-> R+ 1/3R3, R-> R– 5/3R3

Therefore, A-1 =

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