# Ex 4.4, 13 - Chapter 4 Class 12 Determinants

Last updated at April 16, 2024 by Teachoo

Ex 4.4

Ex 4.4, 1

Ex 4.4, 2

Ex 4.4, 3 Important

Ex 4.4, 4 Important

Ex 4.4, 5

Ex 4.4, 6 Important

Ex 4.4, 7

Ex 4.4, 8

Ex 4.4, 9

Ex 4.4, 10 Important

Ex 4.4, 11 Important

Ex 4.4, 12

Ex 4.4, 13 You are here

Ex 4.4, 14 Important

Ex 4.4, 15 Important

Ex 4.4, 16

Ex 4.4, 17 (MCQ) Important

Ex 4.4, 18 (MCQ) Important

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at April 16, 2024 by Teachoo

Ex 4.4, 13 If A = [■8(3&1@−1&2)] show that A2 – 5A + 7I = O. Hence find A–1. Calculating A2 A2 = A.A = [■8(3&1@−1&2)] [■8(3&1@−1&2)] =[■8(3(3)+1(−1)&3(1)+1(2)@−1(3)+2(−1)&−1(1)+2(2))] = [■8(9−1&3+2@−3−2&−1+4)] = [■8(8&5@−5&3)] Solving L.H.S A2 – 5A + 7I = [■8(8&5@−5&3)] – 5 [■8(3&1@−1&2)] + 7 [■8(1&0@0&1)] = [■8(8&5@−5&3)] – [■8(5(3)&5(1)@5(−1)&5(2))] + [■8(7(1)&7(0)@7(0)&7(1))] = [■8(8&5@−5&3)] – [■8(15&5@−5&10)] + [■8(7&0@0&7)] = [■8(8−15+7&5−5+0@−5−(−5)+0&3−10+7)] = [■8(8−15+7&5−5+0@−5+5+0&3−10+7)] = [■8(0&0@0&0)] = O Thus, A2 – 5A + 7I = O Hence proved Finding A–1 A2 – 5A + 7I = O Pre multiplying A-1 both sides A-1 (A2 – 5A + 7I ) = A-1 O A-1 . A2 – 5A-1A + 7A-1 = O A-1 AA – 5(A-1 A) + 7A-1 = O (A-1A)A – 5 (A-1 A) + 7 (A-1 I) = O IA – 5I + 7A-1 = O A – 5I + 7 A-1 = 0 7A-1 = 5I – A A-1 = 𝟏/𝟕 (5I – A) Putting values A-1 = 1/7 (5[■8(1&0@0&1)]−[■8(3&1@−1&2)]) = 1/7 ([■8(5&0@0&5)]−[■8(3&1@−1&2)]) = 1/7 ([■8(5−3&0−1@0−(−1)&5−2)]) = 1/7 [■8(2&−1@1&3)] Thus, A-1 = 𝟏/𝟕 [■8(𝟐&−𝟏@𝟏&𝟑)]