Check sibling questions

Ex 4.5, 16 - Verify A3 - 6A2 + 9A -4I = O and hence find A-1 - Finding inverse when Equation of matrice given

Ex 4.5, 16 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.5, 16 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.5, 16 - Chapter 4 Class 12 Determinants - Part 4
Ex 4.5, 16 - Chapter 4 Class 12 Determinants - Part 5
Ex 4.5, 16 - Chapter 4 Class 12 Determinants - Part 6

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Transcript

Ex4.5, 16 If A = 2﷮−1﷮1﷮−1﷮2﷮−1﷮1﷮−1﷮2﷯﷯ verify that A3 − 6A2 + 9A − 4I = O and hence find A−1 Calculating A2 A2 = A.A = 2﷮−1﷮1﷮−1﷮2﷮−1﷮1﷮−1﷮2﷯﷯ 2﷮−1﷮1﷮−1﷮2﷮−1﷮1﷮−1﷮2﷯﷯ = 2 2﷯+ −1﷯ −1﷯+1(1)﷮2 −1﷯+ −1﷯ 2﷯+1(−1)﷮2 1﷯+ −1﷯ −1﷯+1(2)﷮−1 2﷯+2 −1﷯+(−1)(1)﷮−1 −1﷯+2 2﷯+(−1)(−1)﷮−1 1﷯+2 −1﷯+(−1)(2)﷮1 2﷯+ −1﷯ −1﷯+2(1)﷮1 −1﷯+ −1﷯ 2﷯+2(−1)﷮1 1﷯+ −1﷯ −1﷯+2(2)﷯﷯ = 4+1+1﷮−2−2−1﷮2+1+2﷮−2−2−1﷮1+4+1﷮−1−2−2﷮2+1+2﷮−1−2−2﷮1+1+4﷯﷯ = 6﷮−5﷮5﷮−5﷮6﷮−5﷮5﷮−5﷮6﷯﷯ Hence A2 = 6﷮−5﷮5﷮−5﷮6﷮−5﷮5﷮−5﷮6﷯﷯ Calculating A3 A3 = A2 . A = 6﷮−5﷮5﷮−5﷮6﷮−5﷮5﷮−5﷮6﷯﷯ 2﷮−1﷮1﷮−1﷮2﷮−1﷮1﷮−1﷮2﷯﷯ = 6 2﷯+ −5﷯ −1﷯+5(1)﷮6 −1﷯+ −5﷯ 2﷯+5(−1)﷮6 1﷯+ −5﷯ −1﷯+5(2)﷮−5 2﷯+6 −1﷯+(−5)(1)﷮−5 −1﷯+6 2﷯+(−5)(−1)﷮−5 1﷯+6 −1﷯+(−5)(2)﷮5 2﷯+ −5﷯ −1﷯+6(1)﷮5 −1﷯+ −5﷯ 2﷯+6(−1)﷮5 1﷯+ −5﷯ −1﷯+6(2)﷯﷯ = 12+5+5﷮−6−10−5﷮6+5+10﷮−10−6−5﷮5+12+5﷮−5−6−10﷮10+5+6﷮−5−10−6﷮5+5+12﷯﷯= 22﷮−21﷮21﷮−21﷮22﷮−21﷮21﷮−21﷮22﷯﷯ Hence, A3 = 22﷮−21﷮21﷮−21﷮22﷮−21﷮21﷮−21﷮22﷯﷯ Now, putting values in A3 – 6A2 +9A – 4I = 22﷮−21﷮21﷮−21﷮22﷮−22﷮21﷮−21﷮22﷯﷯ – 6 6﷮−5﷮5﷮−5﷮6﷮−5﷮5﷮−5﷮6﷯﷯ + 9 2﷮−1﷮1﷮−1﷮2﷮−1﷮1﷮−1﷮2﷯﷯ – 4 1﷮0﷮0﷮0﷮1﷮0﷮0﷮0﷮1﷯﷯ = 22﷮−21﷮21﷮−21﷮22﷮−22﷮21﷮−21﷮22﷯﷯ – 6(6)﷮6(−5)﷮6(5)﷮6(−5)﷮6(6)﷮6(−5)﷮6(5)﷮6(−5)﷮6(6)﷯﷯ + 9(2)﷮9(−1)﷮9(1)﷮9(−1)﷮9(2)﷮9(−1)﷮9(1)﷮9(−1)﷮9(2)﷯﷯ – 4(1)﷮0﷮0﷮0﷮4(1)﷮0﷮0﷮0﷮4(1)﷯﷯ = 22﷮−21﷮21﷮−21﷮22﷮−22﷮21﷮−21﷮22﷯﷯ – 36﷮−30﷮30﷮−30﷮36﷮−30﷮30﷮−30﷮30﷯﷯ + 18﷮−9﷮9﷮−9﷮18﷮−9﷮9﷮−9﷮18﷯﷯ – 4﷮0﷮0﷮0﷮4﷮0﷮0﷮0﷮4﷯﷯ = 22−36+18+4﷮−21− −30﷯+ −9﷯+0﷮21−30+9+0﷮−21− −30﷯+ −9﷯60﷮22−36+18+4﷮−22− −30﷯+ −9﷯+0﷮21−30+9+0﷮−21− −30﷯+ −9﷯+0﷮22−36+18+4﷯﷯ = 36−36﷮−21+30−9﷮30−30﷮−21+30−9﷮36−36﷮−21+30−9﷮30−30﷮−21+30−9﷮36−36﷯﷯ = 0﷮0﷮0﷮0﷮0﷮0﷮0﷮0﷮0﷯﷯ = O Hence proved Now calculating A-1 using A3 – 6A2 + 9A – 4I = 0 Post multiplying by A-1 both side (A3 – 6A2 + 9A – 4I ) A-1 = 0.A-1 A3 . A-1 – 6A2 . A + 9AA-1 – 4IA-1 = 0 A2 . AA-1 – 6A. A-1 A + 9AA-1 – 4IA-1 = 0 A2 I – 6AI + 9I – 4IA-1 = 0 A2 – 6A + 9I – 4A-1 = 0 4A-1 = A2 – 6A + 9I A-1 = 1﷮4﷯ (A2 – 6A + 9I) Putting value A-1 = 1﷮4﷯ 6﷮−5﷮5﷮−5﷮6﷮−5﷮5﷮−5﷮6﷯﷯ − 6 2﷮−1﷮1﷮−1﷮2﷮−1﷮1﷮−1﷮2﷯﷯ + 9 1﷮0﷮0﷮0﷮1﷮0﷮0﷮0﷮1﷯﷯﷯ = 1﷮4﷯ 6﷮−5﷮5﷮−5﷮6﷮−5﷮5﷮−5﷮6﷯﷯ − (6)2﷮6(−1)﷮6(1)﷮(6)(−1)﷮6(2)﷮6(−1)﷮6(1)﷮6(−1)﷮6(2)﷯﷯+ 9(1)﷮0﷮0﷮0﷮9(1)﷮0﷮0﷮0﷮9(1)﷯﷯﷯ = 1﷮4﷯ 6﷮−5﷮5﷮−5﷮6﷮−5﷮5﷮−5﷮6﷯﷯ − 12﷮−6﷮6﷮−6﷮12﷮−6﷮6﷮−6﷮12﷯﷯ + 9﷮0﷮0﷮0﷮9﷮0﷮0﷮0﷮9﷯﷯﷯ = 1﷮4﷯ 6−12+9﷮−5+6+0﷮5−6+0﷮−5+6+0﷮6−12+9﷮−5+6+0﷮5−6+0﷮−5+6+0﷮6−12+9﷯﷯ ﷯ = 1﷮4﷯ 3﷮1﷮−1﷮1﷮3﷮1﷮−1﷮1﷮3﷯﷯ Hence, A – 1 = 𝟏﷮𝟒﷯ 𝟑﷮𝟏﷮−𝟏﷮𝟏﷮𝟑﷮𝟏﷮−𝟏﷮𝟏﷮𝟑﷯﷯

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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.