Ex 4.4, 13 - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Finding inverse when Equation of matrice given
Finding inverse when Equation of matrice given
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)]