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Ex 4.4, 16 If A = [■8(2&−1&1@−1&2&−1@1&−1&2)] verify that A3 − 6A2 + 9A − 4I = O and hence find A−1 Calculating A2 A2 = A.A = [■8(2&−1&1@−1&2&−1@1&−1&2)] [■8(2&−1&1@−1&2&−1@1&−1&2)] =[■8(2(2)+(−1)(−1)+1(1)&2(−1)+(−1)(2)+1(−1)&2(1)+(−1)(−1)+1(2)@−1(2)+2(−1)+(−1)(1)&−1(−1)+2(2)+(−1)(−1)&−1(1)+2(−1)+(−1)(2)@1(2)+(−1)(−1)+2(1)&1(−1)+(−1)(2)+2(−1)&1(1)+(−1)(−1)+2(2))] = [■8(4+1+1&−2−2−1&2+1+2@−2−2−1&1+4+1&−1−2−2@2+1+2&−1−2−2&1+1+4)] = [■8(6&−5&5@−5&6&−5@5&−5&6)] Calculating A3 A3 = A2 . A = [■8(6&−5&5@−5&6&−5@5&−5&6)] [■8(2&−1&1@−1&2&−1@1&−1&2)] = [■8(6(2)+(−5)(−1)+5(1)&6(−1)+(−5)(2)+5(−1)&6(1)+(−5)(−1)+5(2)@−5(2)+6(−1)+(−5)(1)&−5(−1)+6(2)+(−5)(−1)&−5(1)+6(−1)+(−5)(2)@5(2)+(−5)(−1)+6(1)&5(−1)+(−5)(2)+6(−1)&5(1)+(−5)(−1)+6(2))] =[■8(12+5+5&−6−10−5&6+5+10@−10−6−5&5+12+5&−5−6−10@10+5+6&−5−10−6&5+5+12)] = [■8(22&−21&21@−21&22&−21@21&−21&22)] Now, putting values in A3 – 6A2 +9A – 4I = [■8(22&−21&21@−21&22&−22@21&−21&22)] – 6[■8(6&−5&5@−5&6&−5@5&−5&6)] + 9 [■8(2&−1&1@−1&2&−1@1&−1&2)] – 4 [■8(1&0&0@0&1&0@0&0&1)] = [■8(22&−21&21@−21&22&−22@21&−21&22)] – [■8(6(6)&6(−5)&6(5)@6(−5)&6(6)&6(−5)@6(5)&6(−5)&6(6))] + [■8(9(2)&9(−1)&9(1)@9(−1)&9(2)&9(−1)@9(1)&9(−1)&9(2))] – [■8(4(1)&0&0@0&4(1)&0@0&0&4(1))] =[■8(22&−21&21@−21&22&−22@21&−21&22)] – [■8(36&−30&30@−30&36&−30@30&−30&30)] + [■8(18&−9&9@−9&18&−9@9&−9&18)] – [■8(4&0&0@0&4&0@0&0&4)] = [■8(22−36+18+4&−21−(−30)+(−9)+0&21−30+9+0@−21−(−30)+(−9)60&22−36+18+4&−22−(−30)+(−9)+0@21−30+9+0&−21−(−30)+(−9)+0&22−36+18+4)] = [■8(36−36&−21+30−9&30−30@−21+30−9&36−36&−21+30−9@30−30&−21+30−9&36−36)] = [■8(0&0&0@0&0&0@0&0&0)] = O Hence proved Now calculating A-1 using A3 – 6A2 + 9A – 4I = 0 Post multiplying by A-1 both side (A3 – 6A2 + 9A – 4I ) A-1 = 0.A-1 A3 . A-1 – 6A2 . A + 9AA-1 – 4IA-1 = 0 A2 . AA-1 – 6A. A-1 A + 9AA-1 – 4IA-1 = 0 A2 I – 6AI + 9I – 4IA-1 = 0 A2 – 6A + 9I – 4A-1 = 0 4A-1 = A2 – 6A + 9I A-1 = 1/4 (A2 – 6A + 9I) Putting value A-1 = 1/4 ([■8(6&−5&5@−5&6&−5@5&−5&6)]" − 6 " [■8(2&−1&1@−1&2&−1@1&−1&2)]" + 9 " [■8(1&0&0@0&1&0@0&0&1)]) = 1/4 ([■8(6&−5&5@−5&6&−5@5&−5&6)]" − " [■8((6)2&6(−1)&6(1)@(6)(−1)&6(2)&6(−1)@6(1)&6(−1)&6(2))]"+ " [■8(9(1)&0&0@0&9(1)&0@0&0&9(1))]) = 1/4 ([■8(6&−5&5@−5&6&−5@5&−5&6)]" − " [■8(12&−6&6@−6&12&−6@6&−6&12)]" + " [■8(9&0&0@0&9&0@0&0&9)]) = 1/4 ([■8(6−12+9&−5+6+0&5−6+0@−5+6+0&6−12+9&−5+6+0@5−6+0&−5+6+0&6−12+9)]" " ) = 1/4 [■8(3&1&−1@1&3&1@−1&1&3)] Hence, A – 1 = 𝟏/𝟒 [■8(𝟑&𝟏&−𝟏@𝟏&𝟑&𝟏@−𝟏&𝟏&𝟑)]

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.