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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 =   


Article Tags :
Class 12
Mathematics
RD Sharma Solutions
School Learning
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