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 = 

A = AI 

Using elementary row operation

⇒

R₁ -> 1/7R₁

⇒

R₂ -> R₂ - 4R₁

⇒

R₂ -> (-7/25)R₂

⇒

R₁ -> R₁ - 1/7R₂

⇒

Therefore, A^-1 =

Question 2. 

Solution:

Here, A = 

A = AI

Using elementary row operation

⇒

R₁ -> 1/5R₁

⇒

R₁ -> R₂ - 2R₁

⇒

R₂ -> 5R₂

⇒

R₁ -> R₁ - 2/5R₂

⇒

Therefore, A^-1= 

Question 3. 

Solution:

Here, A = 

A = AI

Using elementary row operation

 ⇒

R₂ -> R₂ + 3R₁

⇒

R₂ -> 1/23R₂

⇒

R₁ -> R₁ - 6R₁

⇒

Therefore, A^-1 = 

Question 4. 

Solution:

Here, 

A = AI

Using elementary row operation

⇒

R₁ -> 1/2R₁

⇒

R₂ -> R₂ - R₁

⇒

R₂ -> 2R₂

⇒

R₁ -> R₁ - 5/2R₂

⇒

Therefore, A^-1 = 

Question 5. 

Solution:

Here, A = 

A = AI

⇒

R₁ -> 1/3R₁

R₂ -> R₂ - 2R₁

R₂ -> 3R₂

R₁ -> R₁ - 10/3R₂

⇒

Therefore, A^-1 = 

Question 6. 

Solution:

Here, A = 

A = IA

⇒

R₁ ↔ R₂

⇒

R₃ -> R₃ - 3R₁

⇒

R₁ -> R₁ - 2R₂, R₃ -> R₃ + 5R₂

⇒

R₃ -> R₃/2

⇒

R₁ -> R₁ + R₃, R₂ -> R₂ - 2R₃

⇒

Therefore, A^-1 =

Question 7. 

Solution:

Here, A = 

A = IA

⇒

R₁ -> R₁/2

⇒

R₂ -> R₂ - 5R₁

⇒

R₃ -> R₃ - R₂

⇒

R₃ -> 2R₃

⇒

R₁ -> R₁ + 1/2R₃, R₂ -> R₂ - 5/2R₃

⇒

Therefore, A^-1 = 

Question 8. 

Solution:

Here, A = 

A = IA

⇒

R₁ -> 1/2R₁

⇒

R₂ -> R₂ - 2R₁, R₃ -> R₃ - 3R₁

 ⇒

R₁ -> R₁ - 3/2R₂, R₃ -> R₃ - 5/2R₂

⇒

R₃ -> 2R₃

⇒

R₁ -> R₁ - 1/2R₃

⇒

Therefore, A^-1 = 

Question 9. 

Solution:

Here, A = 

A = IA

⇒

R₁ -> 1/3R₁

⇒

R₂ -> R₂ - 2R₁

⇒

R₂ -> (-1)R₂

⇒

R₁ -> R₁ + R₂, R₃ -> R₃ + R₂

⇒

R₃ -> (-3)R₃

⇒

R₂ -> R₂ + 4/3R₃

⇒

Therefore, A^-1 = 

Question 10. 

Solution:

Here, A = 

⇒

R₂ -> R₂ - 2R₁, R₃ -> R₃ - R₁

⇒

R₂ -> (-1)R₂

⇒

R₁ -> R₁ - 2R₂, R₃ -> R₃ + 3R₂

⇒

R₃ -> R₃/6

⇒

R₁ -> R₁ + 2R₃, R₂ -> R₂ - R₃

⇒

Therefore, A^-1 = 

Question 11. 

Solution:

Here, A = 

A = IA

⇒

R₁ -> R₁/2

⇒

R₂ -> R₂ - R₁, R₃ -> R₃ - 3R₁

⇒

R₂ -> (2/5)R₂

⇒

R₁ -> R₁ + 1/2 R₂, R₃ -> R₃ - 5/2R₂

⇒

R₃ -> R₃/-6

⇒

R₂ -> R₂ - R₃, R₁ -> R₁ - 2R₃

⇒

Therefore, A^-1 = 

Question 12. 

Solution:

Here, A = 

A = IA

⇒

R₂ -> R₂ - 3R₁, R₃ -> R₃ - 2R₁

⇒

R₂ -> R₂/(-2)

⇒

R₁ -> R₁ - R₂, R₃ -> R₃ - R₂

⇒

R₃ -> (-2/11)R₃

⇒

R₁ -> R₁ + 1/2R₃, R₂ -> R₂ - 5/2R₃

⇒

Therefore, A^-1 = 

Question 13. 

Solution:

Here, A = 

A = IA

⇒

R₁ -> 1/2R₁

⇒

R₂ -> R₂ - 4R₁, R₃ -> R₃ - 3R₁

⇒

R₂ -> 1/2R₂

⇒

R₁ -> R₁ + 1/2R₂, R₃ -> R₃ + 1/2R₂

⇒

R₃ -> (-2)R₃

⇒

R₁ -> R₁ - 1/2R₃, R₂ -> R₂ + 3R₃

⇒

Therefore, A^-1 = 

Question 14. 

Solution:

Here, A = 

A = IA

⇒

R₁ -> (1/3)R₁

⇒

R₂ -> R₂ - 2R₁

⇒

R₂ -> (1/3)R₂

⇒

R₃ -> R₃ - 4R₂

⇒

R₃ -> 9R₃

⇒

R₁ -> R₁ + 1/3R₃, R₂ -> R₂ - 2/9R₃

⇒

Therefore, A^-1 = 

Question 15. 

Solution:

Here, A = 

A = IA

⇒

R₂ -> 3R₁ + R₂, R₃ -> R₃ - 2R₁

⇒

R₁ -> R₁ - 3R₂, R₃ -> R₃ + 5R₂

⇒

R₂ -> R₂ + 5/9R₃, R₁ -> R₁ + 1/3R₃

⇒

Therefore, A^-1 = 

Question 16. 

Solution:

Here, A= 

A = IA

⇒

R₁ -> (-1)R₁

⇒

R₂ -> R₂ - R₁, R₃ -> R₃ - 3R₁

⇒

R₂ -> R₂/3

⇒

R₁ -> R₁ + R₂, R₃ -> R₃ - 4R₂

⇒

R₃ -> R₃/3

⇒

R₁ -> R₁ + 1/3R₃, R₂ -> R₂ - 5/3R₃

⇒

Therefore, A^-1 = 

Summary

Exercise 7.2 typically focuses on the properties and applications of inverse matrices. Key topics usually include:

• Properties of inverse matrices
• Solving matrix equations using inverse matrices
• Finding unknown elements in a matrix given its inverse
• Verifying inverse matrix properties
• Applications of inverse matrices in solving systems of linear equations