If matrix A = a i j 3 × 2 and a i j = 3 i - 2 j 2 of matrix A , then find A .

Given that A = a i j 3 × 2 ⇒ A = a 11 a 12 a 21 a 22 a 31 a 32 3 × 2 And also given that a i j = 3 i - 2 j 2 Now, a 11 = 3 1 - 2 1 2 = 3 - 2 2 = 1 a 12 = 3 1 - 2 2 2 = 3 - 4 2 = 1 a 21 = 3 2 - 2 1 2 = 6 - 2 2 = 16 a 22 = 3 2 - 2 2 2 = 6 - 4 2 = 4 a 31 = 3 3 - 2 1 2 = 9 - 2 2 = 49 a 32 = 3 3 - 2 2 2 = 9 - 4 2 = 25 Therefore, A = 1 1 16 4 49 25 3 × 2 .

Express 2 5 − 1 3 1 5 7 6 9 as sum of symmetric and skew-symmetric matrices.

Let A = 2 5 - 1 3 1 5 7 6 9 A T = 2 3 7 5 1 6 - 1 5 9 We know that A = 1 2 A + A T + 1 2 A - A T Now, 1 2 A + A T = 1 2 2 5 - 1 3 1 5 7 6 9 + 2 3 7 5 1 6 - 1 5 9 = 1 2 4 8 6 8 2 11 6 11 18 = 2 4 3 4 1 11 2 3 11 2 9 1 2 A - A T = 1 2 2 5 - 1 3 1 5 7 6 9 - 2 3 7 5 1 6 - 1 5 9 = 1 2 0 2 - 8 - 2 0 - 1 8 1 0 = 0 1 - 4 - 1 0 - 1 2 4 1 2 0 ⇒ 2 5 - 1 3 1 5 7 6 9 = 2 4 3 4 1 11 2 3 11 2 9 + 0 1 - 4 - 1 0 - 1 2 4 1 2 0 .

Using elementary transformations find inverse of 2 4 1 1 2 3 1 - 3 0

Let A = 2 4 1 1 2 3 1 - 3 0 Applying column operation to find A - 1 ∴ A = A I ⇒ 2 4 1 1 2 3 1 - 3 0 = A 1 0 0 0 1 0 0 0 1 ⇒ 1 2 1 2 1 2 3 1 - 3 0 = A 1 2 0 0 0 1 0 0 0 1 (Apply R 1 → R 1 2 ) ⇒ 1 2 1 2 0 0 5 2 1 - 3 0 = A 1 2 0 0 - 1 2 1 0 0 0 1 (Apply R 2 → R 2 − R 1 ) ⇒ 1 2 1 2 0 0 5 2 0 - 5 - 1 2 = A 1 2 0 0 - 1 2 1 0 - 1 2 0 1 (Apply R 3 → R 3 - R 1 ) ⇒ 1 2 1 2 0 - 5 - 1 2 0 0 5 2 = A 1 2 0 0 - 1 2 0 1 - 1 2 1 0 (Apply R 2 ↔ R 3 ) ⇒ 1 2 1 2 0 1 1 10 0 0 5 2 = A 1 2 0 0 1 10 0 - 1 5 - 1 2 1 0 (Apply R 2 → - R 2 5 ) ⇒ 1 2 1 2 0 1 1 10 0 0 1 = A 1 2 0 0 1 10 0 - 1 5 - 1 5 2 5 0 (Apply R 3 → 2 5 R 3 ) ⇒ 1 2 1 2 0 1 0 0 0 1 = A 1 2 0 0 3 25 - 1 25 - 1 5 - 1 5 2 5 0 (Apply R 2 → R 2 - R 3 10 ) ⇒ 1 2 0 0 1 0 0 0 1 = A 3 5 - 1 5 0 3 25 - 1 25 - 1 5 - 1 5 2 5 0 (Apply R 1 → R 1 - R 3 2 ) ⇒ 1 0 0 0 1 0 0 0 1 = A 9 5 - 3 5 2 5 3 25 - 1 25 - 1 5 - 1 5 2 5 0 (Apply R 1 → R 1 - 2 R 2 ) ∴ A - 1 = 9 5 - 3 5 2 5 3 25 - 1 25 - 1 5 - 1 5 2 5 0 .

If A = 2 1 3 - 5 and f x = x 2 - 2 x + 3 then find f A .

Given that: A = 2 1 3 - 5 and f x = x 2 - 2 x + 3 We have to find f A . f A = 2 1 3 - 5 2 - 2 2 1 3 - 5 + 3 1 0 0 1 f A = 2 1 3 - 5 × 2 1 3 - 5 - 4 2 6 - 10 + 3 0 0 3 f A = 7 - 3 - 9 28 - 4 2 6 - 10 + 3 0 0 3 ⇒ f A = 6 - 1 - 3 21

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 3 - Matrices>Punjab Board-2018

Embibe Experts Solutions for Chapter: Matrices, Exercise 1: Punjab Board-2019

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Matrices, Exercise 1: Punjab Board-2019

Attempt the free practice questions on Chapter 3: Matrices, Exercise 1: Punjab Board-2019 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.