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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

Serial order wise

Last updated at March 16, 2023 by Teachoo

Ex 4.5, 13 If A = [■8(3&[email protected]−1&2)] show that A2 – 5A + 7I = O. Hence find A–1. Calculating A2 A2 = A.A = [■8(3&[email protected]−1&2)] [■8(3&[email protected]−1&2)] =[■8(3(3)+1(−1)&3(1)+1(2)@−1(3)+2(−1)&−1(1)+2(2))] = [■8(9−1&[email protected]−3−2&−1+4)] = [■8(8&[email protected]−5&3)] Solving L.H.S A2 – 5A + 7I = [■8(8&[email protected]−5&3)] – 5 [■8(3&[email protected]−1&2)] + 7 [■8(1&[email protected]&1)] = [■8(8&[email protected]−5&3)] – [■8(5(3)&5(1)@5(−1)&5(2))] + [■8(7(1)&7(0)@7(0)&7(1))] = [■8(8&[email protected]−5&3)] – [■8(15&[email protected]−5&10)] + [■8(7&[email protected]&7)] = [■8(8−15+7&5−[email protected]−5−(−5)+0&3−10+7)] = [■8(8−15+7&5−[email protected]−5+5+0&3−10+7)] = [■8(0&[email protected]&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&[email protected]&1)]−[■8(3&[email protected]−1&2)]) = 1/7 ([■8(5&[email protected]&5)]−[■8(3&[email protected]−1&2)]) = 1/7 ([■8(5−3&0−[email protected]−(−1)&5−2)]) = 1/7 [■8(2&−[email protected]&3)] Thus, A-1 = 𝟏/𝟕 [■8(𝟐&−𝟏@𝟏&𝟑)]