# Ex 4.5, 13

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 4.5, 13 If A = 31−12 show that A2 – 5A + 7I = O. Hence find A–1. Calculating A2 A2 = A.A = 31−12 31−12 = 3 3+1(−1)3 1+1(2)−1 3+2(−1)−1 1+2(2) = 9−13+2−3−2−1+4 = 85−53 Taking L.H.S A2 – 5A + 7I = 85−53 – 5 31−12 + 7 1001 = 85−53 – 5(3)5(1)5(−1)5(2) + 7(1)7(0)7(0)7(1) = 85−53 – 155−510 + 7007 = 8−15+75−5+0−5−(−5)+0−5−10+7 = 8−15+75−5+0−5+5+0−5−10+7 = 0000 = 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 = 𝟏𝟕 (5I – A) Putting values A-1 = 17 5 1001− 31−12 = 17 5005− 31−12 = 17 5−30−10−(−1)5−2 = 17 2−113 Thus, A-1 = 17 2−113

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Davneet Singh

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