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

### Question 1.

Solution:

Here, A =

A = AI

Using elementary row operation

â‡’

R1 -> 1/7R1

â‡’

R2 -> R2 – 4R1

â‡’

R2 -> (-7/25)R2

â‡’

R1 -> R1 – 1/7R2

â‡’

Therefore, A-1 =

### Question 2.

Solution:

Here, A =

A = AI

Using elementary row operation

â‡’

R1 -> 1/5R1

â‡’

R1 -> R2 – 2R1

â‡’

R2 -> 5R2

â‡’

R1 -> R1 – 2/5R2

â‡’

Therefore, A-1

### Question 3.

Solution:

Here, A =

A = AI

Using elementary row operation

â‡’

R2 -> R2 + 3R1

â‡’

R2 -> 1/23R2

â‡’

R1 -> R1 – 6R1

â‡’

Therefore, A-1

### Question 4.

Solution:

Here,

A = AI

Using elementary row operation

â‡’

R1 -> 1/2R1

â‡’

R2 -> R2 – R1

â‡’

R2 -> 2R2

â‡’

R1 -> R1 – 5/2R2

â‡’

Therefore, A-1

### Question 5.

Solution:

Here, A =

A = AI

â‡’

R1 -> 1/3R1

R2 -> R2 – 2R1

R2 -> 3R2

R1 -> R1 – 10/3R2

â‡’

Therefore, A-1

### Question 6.

Solution:

Here, A =

A = IA

â‡’

R1 â†” R2

â‡’

R3 -> R3 – 3R1

â‡’

R1 -> R1 – 2R2, R3 -> R3 + 5R2

â‡’

R3 -> R3/2

â‡’

R1 -> R1 + R3, R2 -> R2 – 2R3

â‡’

Therefore, A-1 =

### Question 7.

Solution:

Here, A =

A = IA

â‡’

R1 -> R1/2

â‡’

R2 -> R2 – 5R1

â‡’

R3 -> R3 – R2

â‡’

R3 -> 2R3

â‡’

R1 -> R1 + 1/2R3, R2 -> R2 – 5/2R3

â‡’

Therefore, A-1

### Question 8.

Solution:

Here, A =

A = IA

â‡’

R1 -> 1/2R1

â‡’

R2 -> R2 – 2R1, R3 -> R3 – 3R1

â‡’

R1 -> R1 – 3/2R2, R3 -> R3 – 5/2R2

â‡’

R3 -> 2R3

â‡’

R1 -> R1 – 1/2R3

â‡’

Therefore, A-1

### Question 9.

Solution:

Here, A =

A = IA

â‡’

R1 -> 1/3R1

â‡’

R2 -> R2 – 2R1

â‡’

R2 -> (-1)R2

â‡’

R1 -> R1 + R2, R3 -> R3 + R2

â‡’

R3 -> (-3)R3

â‡’

R2 -> R2 + 4/3R3

â‡’

Therefore, A-1

### Question 10.

Solution:

Here, A =

â‡’

R2 -> R2 – 2R1, R3 -> R3 – R1

â‡’

R2 -> (-1)R2

â‡’

R1 -> R1 – 2R2, R3 -> R3 + 3R2

â‡’

R3 -> R3/6

â‡’

R1 -> R1 + 2R3, R2 -> R2 – R3

â‡’

Therefore, A-1

### Question 11.

Solution:

Here, A =

A = IA

â‡’

R1 -> R1/2

â‡’

R2 -> R2 – R1, R3 -> R3 – 3R1

â‡’

R2 -> (2/5)R2

â‡’

R1 -> R1 + 1/2 R2, R3 -> R3 – 5/2R2

â‡’

R3 -> R3/-6

â‡’

R2 -> R2 – R3, R1 -> R1 – 2R3

â‡’

Therefore, A-1

### Question 12.

Solution:

Here, A =

A = IA

â‡’

R2 -> R2 – 3R1, R3 -> R3 – 2R1

â‡’

R2 -> R2/(-2)

â‡’

R1 -> R1 – R2, R3 -> R3 – R2

â‡’

R3 -> (-2/11)R3

â‡’

R1 -> R1 + 1/2R3, R2 -> R2 – 5/2R3

â‡’

Therefore, A-1

### Question 13.

Solution:

Here, A =

A = IA

â‡’

R1 -> 1/2R1

â‡’

R2 -> R2 – 4R1, R3 -> R3 – 3R1

â‡’

R2 -> 1/2R2

â‡’

R1 -> R1 + 1/2R2, R3 -> R3 + 1/2R2

â‡’

R3 -> (-2)R3

â‡’

R1 -> R1 – 1/2R3, R2 -> R2 + 3R3

â‡’

Therefore, A-1

### Question 14.

Solution:

Here, A =

A = IA

â‡’

R1 -> (1/3)R1

â‡’

R2 -> R2 – 2R1

â‡’

R2 -> (1/3)R2

â‡’

R3 -> R3 – 4R2

â‡’

R3 -> 9R3

â‡’

R1 -> R1 + 1/3R3, R2 -> R2 – 2/9R3

â‡’

Therefore, A-1

### Question 15.

Solution:

Here, A =

A = IA

â‡’

R2 -> 3R1 + R2, R3 -> R3 – 2R1

â‡’

R1 -> R1 – 3R2, R3 -> R3 + 5R2

â‡’

R2 -> R2 + 5/9R3, R1 -> R1 + 1/3R3

â‡’

Therefore, A-1

### Question 16.

Solution:

Here, A=

A = IA

â‡’

R1 -> (-1)R1

â‡’

R2 -> R2 – R1, R3 -> R3 – 3R1

â‡’

R2 -> R2/3

â‡’

R1 -> R1 + R2, R3 -> R3 – 4R2

â‡’

R3 -> R3/3

â‡’

R1 -> R1 + 1/3R3, R2 -> R2 – 5/3R3

â‡’

Therefore, A-1

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