# Ex 4.5, 13 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 23, 2020 by Teachoo

Last updated at Jan. 23, 2020 by Teachoo

Transcript

Ex 4.5, 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) (AA-1 = I ) (IA = 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)]

Ex 4.5

Ex 4.5, 1

Ex 4.5, 2

Ex 4.5, 3 Important

Ex 4.5, 4 Important

Ex 4.5, 5

Ex 4.5, 6 Important

Ex 4.5, 7

Ex 4.5, 8

Ex 4.5, 9

Ex 4.5, 10 Important

Ex 4.5, 11 Important

Ex 4.5, 12

Ex 4.5, 13 You are here

Ex 4.5, 14 Important

Ex 4.5, 15 Important

Ex 4.5, 16

Ex 4.5, 17 (MCQ) Important

Ex 4.5, 18 (MCQ) Important

Chapter 4 Class 12 Determinants (Term 1)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.