Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 4.4
Ex 4.4, 2
Ex 4.4, 3 Important
Ex 4.4, 4 Important
Ex 4.4, 5
Ex 4.4, 6 Important
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Ex 4.4, 10 Important
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Ex 4.4, 14 Important
Ex 4.4, 15 Important You are here
Ex 4.4, 16
Ex 4.4, 17 (MCQ) Important
Ex 4.4, 18 (MCQ) Important
Last updated at June 11, 2023 by Teachoo
Ex 4.4, 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)]