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 =