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Ex 4.5, 2
Ex 4.5, 3 Important
Ex 4.5, 4 Important
Ex 4.5, 5
Ex 4.5, 6 Important
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Ex 4.5, 9
Ex 4.5, 10 Important
Ex 4.5, 11 Important
Ex 4.5, 12
Ex 4.5, 13
Ex 4.5, 14 Important
Ex 4.5, 15 Important You are here
Ex 4.5, 16
Ex 4.5, 17 (MCQ) Important
Ex 4.5, 18 (MCQ) Important
Ex 4.5, 15 For the matrix A = [■8(1&1&1@1&2&−3@2&−1&3)] show that A3 − 6A2 + 5A + 11I = O. Hence, find A−1. Calculating A2 A2 = A.A = [■8(1&1&1@1&2&−3@2&−1&3)] [■8(1&1&1@1&2&−3@2&−1&3)] = [■8(1(1)+1(1)+1(2)&1(1)+1(2)+1(−1)&1(1)+1(−3)+1(3)@1(1)+2(1)+(−3)(2)&1(1)+2(2)+(−3)(−1)&1(1)+2(−3)+(−3)(3)@2(1)+(−1)(1)+3(2)&2(1)+(−1)(2)+3(−1)&2(1)+(−1)(−3)+3(3))] = [■8(1+1+2&1+2−1&1−3+3@1+2−6&1+4+3&1−6−9@2−1+6&2−2−3&2+3+9)] = [■8(4&2&1@−3&8&−14@7&−3&14)] Now finding A3 A3 = A2 A A3 = [■8(4&2&1@−3&8&−14@7&−3&14)] [■8(1&1&1@1&2&−3@2&−1&3)] =[■8(4(1)+2(1)+1(2)&4(1)+2(2)+1(−1)&4(1)+2(−3)+1(3)@−3(1)+8(1)+(−4)(2)&−3(1)+8(2)+(−4)(−1)&3(1)+8(−3)+(−4)(3)@7(1)+(−3)(1)+14(2)&7(1)+(−3)(2)+14(−1)&7(1)+(−3)(−3)+14(3))] = [■8(4+2+2&4+4−1&4−6+3@−3+8−8&−3+16 −4&−3−24−42@7−3+28&7−6−14&7+9+42)] = [■8(8&7&1@−23&27&−69@32&−13&58)] Now Putting value of A3 , A2 in A3 – 6A2 + 5A + 11I = [■8(8&7&1@−23&27&−69@32&−13&58)] – 6 [■8(4&2&1@−3&8&−14@7&−3&14)] + 5 [■8(1&1&1@1&2&−3@2&−1&3)] + 11 [■8(1&0&0@0&1&0@0&0&1)] = [■8(8&7&1@−23&27&−69@32&−13&58)] – [■8(6(4)&6(2)&6(1)@6(−3)&6(8)&6(−14)@6(7)&6(−3)&6(14))] + [■8(5(1)&5(1)&5(1)@5(1)&5(2)&5(−3)@5(2)&5(−1)&5(3))] + [■8(11(1)&0&0@0&11(1)&0@0&0&11(1))] = [■8(8&7&1@−23&27&−69@32&−13&58)] – [■8(24&12&6@−18&48&−84@42&−18&84)] + [■8(5&5&5@5&10&−15@10&−5&15)] + [■8(11&0&0@0&11&0@0&0&11)] = [■8(8−24+5+11&7−12+5+0&1−6+5+0@−23+18+5+0&27−48+10+11&−69+84−15+0@32−42+10+0&−13+18−5+0&58+84+15+11)] = [■8(24−24&12−12&6−6@−23+23&−48+48&84−84@−42+42&18−18&84−84)] = [■8(0&0&0@0&0&0@0&0&0)] = O Hence proved Finding A-1 A3 – 6A2 + 5A + 11I = O Post multiplying A-1 both sides (A3 – 6A2 + 5A +11I)A-1 = O A-1 A3 .A-1 – 6 A2. A-1 + 5AA-1 + 11IA-1 = O A2. A.A-1 – 6A.AA-1 + 5AA-1 + 11A-1 = O A2 (AA-1) – 6A(AA-1) + 5(AA-1)11A-1 = O A2 I – 6AI + 5I + 11A-1 = 0 A2 – 6A + 5I + 11A-1 = 0 11A-1 = A2 + 6A – 5I A-1 = 1/11 (– A2 + 6A – 5I) Putting values A-1 = 1/11 ([■8(4&2&1@−3&8&−14@7&−3&14)]" + 6 " [■8(1&1&1@1&2&−3@2&−1&3)]−"5 " [■8(1&0&0@0&1&0@0&0&1)]) = 1/11 ([■8(−4&−2&−1@3&−8&14@−7&3&−14)]" + " [■8(6&6&6@6&12&−18@12&−6&18)]−[■8(5&0&0@0&5&0@0&0&5)]) = 1/11 ([■8(−4+6−5&−2+6+0&−1+6+0@3+6+0&−8+12−5&14−18+0@−7+12+0&3−6+0&−14+18−5)]" " ) = 𝟏/𝟏𝟏 [■8(−𝟑&𝟒&𝟓@𝟗&−𝟏&−𝟒@𝟓&−𝟑&−𝟏)]