Find the inverse of 7 1 4 - 3 using elementary row transformations.

Given that A = 7 1 4 - 3 We know A = I A ⇒ 7 1 4 - 3 = 1 0 0 1 A Apply R 1 → 1 7 R 1 1 1 7 4 - 3 = 1 7 0 0 1 = A Apply R 2 → R 2 - 4 R 1 1 1 7 0 - 25 7 = 1 7 0 - 4 7 1 A Apply R 2 → - 7 25 R 2 1 1 7 0 1 = 1 7 0 4 25 - 7 25 A Apply R 1 → R 1 - 1 7 R 2 1 0 0 1 = 21 175 1 25 4 25 - 7 25 A Therefore, A - 1 = 21 175 1 25 4 25 - 7 25

Find the inverse of 2 - 1 4 4 0 2 3 - 2 7 using elementary row transformations.

Given that A = 2 - 1 4 4 0 2 3 - 2 7 We Know A = I A ⇒ 2 - 1 4 4 0 2 3 - 2 7 = 1 0 0 0 1 0 0 0 1 A Apply R 1 → 1 2 R 1 1 - 1 2 2 4 0 2 3 - 2 7 = 1 2 0 0 0 1 0 0 0 1 A Apply R 2 → R 2 - 4 R 1 and R 3 → R 3 - 3 R 1 1 - 1 2 2 0 2 - 6 0 - 1 2 1 = 1 2 0 0 - 2 1 0 - 3 2 0 1 A Apply R 2 → 1 2 R 2 1 - 1 2 2 0 1 - 3 0 - 1 2 1 = 1 2 0 0 - 1 1 2 0 - 3 2 0 1 A Apply R 1 → R 1 + 1 2 R 2 and R 3 → R 3 + 1 2 R 2 1 0 1 2 0 1 - 3 0 0 - 1 2 = 0 1 4 0 - 1 1 2 0 - 2 1 4 1 A Apply R 3 → - 2 R 3 1 0 1 2 0 1 - 3 0 0 1 = 0 1 4 0 - 1 1 2 0 4 - 1 2 - 2 A Apply R 1 → R 1 - 1 2 R 3 and R 2 → R 2 + 3 R 3 1 0 0 0 1 0 0 0 1 = - 2 1 2 1 11 - 1 - 6 4 - 1 2 - 2 A Therefore, A - 1 = - 2 1 2 1 11 - 1 - 6 4 - 1 2 - 2

Find the inverse of 3 0 - 1 2 3 0 0 4 1 using elementary row transformations.

Given that A = 3 0 - 1 2 3 0 0 4 1 We know A = I A ⇒ 3 0 - 1 2 3 0 0 4 1 = 1 0 0 0 1 0 0 0 1 A Apply R 1 → 1 3 R 2 1 0 - 1 3 2 3 0 0 4 1 = 1 3 0 0 0 1 0 0 0 1 Apply R 2 → R 2 - 2 R 1 1 0 - 1 3 0 3 2 3 0 4 1 = 1 3 0 0 - 2 3 1 0 0 0 1 A Apply R 2 → 1 3 R 2 1 0 - 1 3 0 1 2 9 0 4 1 = 1 3 0 0 - 2 9 1 3 0 0 0 1 A Apply R 3 → R 3 - 4 R 2 1 0 - 1 3 0 1 2 9 0 0 1 9 = 1 3 0 0 - 2 9 1 3 0 8 9 - 4 3 1 A Apply R 2 → R 2 - 2 9 R 3 and R 1 → R 1 + 1 3 R 3 1 0 0 0 1 0 0 0 1 = 3 - 4 3 - 2 3 - 2 8 - 12 9 A Therefore, A - 1 = 3 - 4 3 - 2 3 - 2 8 - 12 9

Find the inverse of 1 3 - 2 - 3 0 1 2 1 0 using elementary row transformations.

Given that A = 1 3 - 2 - 3 0 - 1 2 1 0 We know A = I A ⇒ 1 3 - 2 - 3 0 - 1 2 1 0 = 1 0 0 0 1 0 0 0 1 A ⇒ 1 3 - 2 0 9 - 7 0 - 5 4 = 1 0 0 3 1 0 - 2 0 1 A Applying R 2 → R 2 + 3 R 1 and R 3 → R 3 - 2 R 1 ⇒ 1 3 - 2 0 1 - 7 9 0 - 5 4 = 1 0 0 1 3 1 9 0 - 2 0 1 A Applying R 2 → 1 9 R 2 ⇒ 1 0 1 3 0 1 - 7 9 0 0 1 9 = 0 - 1 3 0 1 3 1 9 0 - 1 3 5 9 1 A Applying R 1 → R 1 - 3 R 2 and R 3 → R 3 + 5 R 2 ⇒ 1 0 1 3 0 1 - 7 9 0 0 1 = 0 - 1 3 0 1 3 1 9 0 - 3 5 9 A Applying R 3 → 9 R 3 ⇒ 1 0 0 0 1 0 0 0 1 = 1 - 2 - 3 - 2 4 7 - 3 5 9 A Applying R 2 → R 2 + 7 9 R 3 and R 1 → R 1 - 1 3 R 3 ⇒ A - 1 = 1 - 2 - 3 - 2 4 7 - 3 5 9

Find the inverse of - 1 1 2 1 2 3 3 1 1 using elementary row transformations.

Given that A = - 1 1 2 1 2 3 3 1 1 We know that A = I A ⇒ - 1 1 2 1 2 3 3 1 1 = 1 0 0 0 1 0 0 0 1 A 1 - 1 - 2 1 2 3 3 1 1 = - 1 0 0 0 1 0 0 0 1 A Applying R 1 → ( - 1 ) R 1 1 - 1 - 2 0 3 5 0 4 7 = - 1 0 0 1 1 0 3 0 1 A Applying R 2 → R 2 - R 1 and R 3 → R 3 - 3 R 1 1 - 1 - 2 0 1 5 3 0 4 7 = - 1 0 0 1 3 1 3 0 3 0 1 A Applying R 2 → 1 3 R 2 1 0 - 1 3 0 1 5 3 0 0 1 3 = - 2 3 1 3 0 1 3 1 3 0 5 3 - 4 3 1 A Applying R 3 → R 3 - 4 R 2 and R 1 → R 1 + R 2 1 0 - 1 3 0 1 5 3 0 0 1 = - 2 3 1 3 0 1 3 1 3 0 5 - 4 3 A [ Applying R 3 → 3 R 3 ] 1 0 0 0 1 0 0 0 1 = 1 - 1 1 - 8 7 - 5 5 - 4 3 A Applying R 2 → R 2 - 5 3 R 3 and R 1 → R 1 + 1 3 R 3 Therefore, A - 1 = 1 - 1 1 - 8 7 - 5 5 - 4 3

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MATHEMATICS CLASS XII VOLUME-1>Chapter 7 - Adjoint and Inverse of a Matrix>EXERCISE

R D Sharma Solutions for Chapter: Adjoint and Inverse of a Matrix, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Adjoint and Inverse of a Matrix, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 7: Adjoint and Inverse of a Matrix, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.