Check sibling questions

Ex 4.5, 15 - Show that A3 - 6A2 + 5A + 11I = O, hence find A-1

Ex 4.5, 15 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.5, 15 - Chapter 4 Class 12 Determinants - Part 3 Ex 4.5, 15 - Chapter 4 Class 12 Determinants - Part 4 Ex 4.5, 15 - Chapter 4 Class 12 Determinants - Part 5


Transcript

Ex 4.5, 15 For the matrix A = [■8(1&1&[email protected]&2&−[email protected]&−1&3)] show that A3 − 6A2 + 5A + 11I = O. Hence, find A−1. Calculating A2 A2 = A.A = [■8(1&1&[email protected]&2&−[email protected]&−1&3)] [■8(1&1&[email protected]&2&−[email protected]&−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−[email protected]+2−6&1+4+3&1−6−[email protected]−1+6&2−2−3&2+3+9)] = [■8(4&2&[email protected]−3&8&−[email protected]&−3&14)] Now finding A3 A3 = A2 A A3 = [■8(4&2&[email protected]−3&8&−[email protected]&−3&14)] [■8(1&1&[email protected]&2&−[email protected]&−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−[email protected]−3+8−8&−3+16 −4&−3−24−[email protected]−3+28&7−6−14&7+9+42)] = [■8(8&7&[email protected]−23&27&−[email protected]&−13&58)] Now Putting value of A3 , A2 in A3 – 6A2 + 5A + 11I = [■8(8&7&[email protected]−23&27&−[email protected]&−13&58)] – 6 [■8(4&2&[email protected]−3&8&−[email protected]&−3&14)] + 5 [■8(1&1&[email protected]&2&−[email protected]&−1&3)] + 11 [■8(1&0&[email protected]&1&[email protected]&0&1)] = [■8(8&7&[email protected]−23&27&−[email protected]&−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&[email protected]&11(1)&[email protected]&0&11(1))] = [■8(8&7&[email protected]−23&27&−[email protected]&−13&58)] – [■8(24&12&[email protected]−18&48&−[email protected]&−18&84)] + [■8(5&5&[email protected]&10&−[email protected]&−5&15)] + [■8(11&0&[email protected]&11&[email protected]&0&11)] = [■8(8−24+5+11&7−12+5+0&1−[email protected]−23+18+5+0&27−48+10+11&−69+84−[email protected]−42+10+0&−13+18−5+0&58+84+15+11)] = [■8(24−24&12−12&6−[email protected]−23+23&−48+48&84−[email protected]−42+42&18−18&84−84)] = [■8(0&0&[email protected]&0&[email protected]&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&[email protected]−3&8&−[email protected]&−3&14)]" + 6 " [■8(1&1&[email protected]&2&−[email protected]&−1&3)]−"5 " [■8(1&0&[email protected]&1&[email protected]&0&1)]) = 1/11 ([■8(−4&−2&−[email protected]&−8&[email protected]−7&3&−14)]" + " [■8(6&6&[email protected]&12&−[email protected]&−6&18)]−[■8(5&0&[email protected]&5&[email protected]&0&5)]) = 1/11 ([■8(−4+6−5&−2+6+0&−[email protected]+6+0&−8+12−5&14−[email protected]−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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.