Ex 4.5, 15 - Show that A3 - 6A2 + 5A + 11I = O, hence find A-1 - Finding inverse when Equation of matrice given

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.5, 15 For the matrix A = 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 = 1﷮1﷮1﷮1﷮2﷮−3﷮2﷮−1﷮3﷯﷯ 1﷮1﷮1﷮1﷮2﷮−3﷮2﷮−1﷮3﷯﷯ = 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)﷯﷯ = 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﷯﷯ = 4﷮2﷮1﷮−3﷮8﷮−14﷮7﷮−3﷮14﷯﷯ Now finding A3 A3 = A2 A A3 = 4﷮2﷮1﷮−3﷮8﷮−14﷮7﷮−3﷮14﷯﷯ 1﷮1﷮1﷮1﷮2﷮−3﷮2﷮−1﷮3﷯﷯ = 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)﷯﷯ = 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﷮7﷮1﷮−23﷮27﷮−69﷮32﷮−13﷮58﷯﷯ Now Putting value of A3 , A2 in A3 – 6A2 + 5A + 11I = 8﷮7﷮1﷮−23﷮27﷮−69﷮32﷮−13﷮58﷯﷯ – 6 4﷮2﷮1﷮−3﷮8﷮−14﷮7﷮−3﷮14﷯﷯ + 5 1﷮1﷮1﷮1﷮2﷮−3﷮2﷮−1﷮3﷯﷯ + 11 1﷮0﷮0﷮0﷮1﷮0﷮0﷮0﷮1﷯﷯ = 8﷮7﷮1﷮−23﷮27﷮−69﷮32﷮−13﷮58﷯﷯ – 6(4)﷮6(2)﷮6(1)﷮6(−3)﷮6(8)﷮6(−14)﷮6(7)﷮6(−3)﷮6(14)﷯﷯ + 5(1)﷮5(1)﷮5(1)﷮5 1﷯﷮5(2)﷮5(−3)﷮5(2)﷮5(−1)﷮5(3)﷯﷯ + 11(1)﷮0﷮0﷮0﷮11(1)﷮0﷮0﷮0﷮11(1)﷯﷯ = 8﷮7﷮1﷮−23﷮27﷮−69﷮32﷮−13﷮58﷯﷯ – 24﷮12﷮6﷮−18﷮48﷮−84﷮42﷮−18﷮84﷯﷯ + 5﷮5﷮5﷮5﷮10﷮−15﷮10﷮−5﷮15﷯﷯ + 11﷮0﷮0﷮0﷮11﷮0﷮0﷮0﷮11﷯﷯ = 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﷯﷯ = 24−24﷮12−12﷮6−6﷮−23+23﷮−48+48﷮84−84﷮−42+42﷮18−18﷮84−84﷯﷯ = 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﷯ 4﷮2﷮1﷮−3﷮8﷮−14﷮7﷮−3﷮14﷯﷯ + 6 1﷮1﷮1﷮1﷮2﷮−3﷮2﷮−1﷮3﷯﷯−5 1﷮0﷮0﷮0﷮1﷮0﷮0﷮0﷮1﷯﷯﷯ = 1﷮11﷯ −4﷮−2﷮−1﷮3﷮−8﷮14﷮−7﷮3﷮−14﷯﷯ + 6(1)﷮6(1)﷮6(1)﷮6(1)﷮6(2)﷮6(−3)﷮(2)﷮6(−1)﷮6(3)﷯﷯ + 5 5(1)﷮0﷮0﷮0﷮5(1)﷮0﷮0﷮0﷮5(1)﷯﷯﷯ = 1﷮11﷯ −4﷮−2﷮−1﷮3﷮−8﷮14﷮−7﷮3﷮−14﷯﷯ + 6﷮6﷮6﷮6﷮12﷮−18﷮12﷮−6﷮18﷯﷯− 5﷮0﷮0﷮0﷮5﷮0﷮0﷮0﷮5﷯﷯﷯ = 1﷮11﷯ −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﷯ −3﷮4﷮5﷮9﷮−1﷮−4﷮5﷮−3﷮−1﷯﷯ Thus, A-1 = 1﷮11﷯ −3﷮4﷮5﷮9﷮−1﷮−4﷮5﷮−3﷮−1﷯﷯

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