Ex 4.4

Chapter 4 Class 12 Determinants
Serial order wise

### Transcript

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)]

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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.