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Ex 4.5, 16 - Chapter 4 Class 12 Determinants (Term 1)
Last updated at May 29, 2018 by Teachoo
Finding inverse when Equation of matrice given
Chapter 4 Class 12 Determinants
Concept wise
- Finding determinant of a 2x2 matrix
- Evalute determinant of a 3x3 matrix
- Area of triangle
- Equation of line using determinant
- Finding Minors and cofactors
- Evaluating determinant using minor and co-factor
- Find adjoint of a matrix
- Finding Inverse of a matrix
- Inverse of two matrices and verifying properties
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Finding inverse when Equation of matrice given
- Checking consistency of equations
- Find solution of equations- Equations given
- Find solution of equations- Statement given
- Verifying properties of a determinant
- Two rows or columns same
- Whole row/column zero
- Whole row/column one
- Making whole row/column one and simplifying
- Proving Determinant 1 = Determinant 2
- Solving by simplifying det.
- Using Property 5 (Determinant as sum of two or more determinants)
Transcript
Ex4.5, 16 If A = 2−11−12−11−12 verify that A3 − 6A2 + 9A − 4I = O and hence find A−1 Calculating A2 A2 = A.A = 2−11−12−11−12 2−11−12−11−12 = 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+1−2−2−12+1+2−2−2−11+4+1−1−2−22+1+2−1−2−21+1+4 = 6−55−56−55−56 Hence A2 = 6−55−56−55−56 Calculating A3 A3 = A2 . A = 6−55−56−55−56 2−11−12−11−12 = 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+5−6−10−56+5+10−10−6−55+12+5−5−6−1010+5+6−5−10−65+5+12= 22−2121−2122−2121−2122 Hence, A3 = 22−2121−2122−2121−2122 Now, putting values in A3 – 6A2 +9A – 4I = 22−2121−2122−2221−2122 – 6 6−55−56−55−56 + 9 2−11−12−11−12 – 4 100010001 = 22−2121−2122−2221−2122 – 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)0004(1)0004(1) = 22−2121−2122−2221−2122 – 36−3030−3036−3030−3030 + 18−99−918−99−918 – 400040004 = 22−36+18+4−21− −30+ −9+021−30+9+0−21− −30+ −96022−36+18+4−22− −30+ −9+021−30+9+0−21− −30+ −9+022−36+18+4 = 36−36−21+30−930−30−21+30−936−36−21+30−930−30−21+30−936−36 = 000000000 = 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 = 14 (A2 – 6A + 9I) Putting value A-1 = 14 6−55−56−55−56 − 6 2−11−12−11−12 + 9 100010001 = 14 6−55−56−55−56 − (6)26(−1)6(1)(6)(−1)6(2)6(−1)6(1)6(−1)6(2)+ 9(1)0009(1)0009(1) = 14 6−55−56−55−56 − 12−66−612−66−612 + 900090009 = 14 6−12+9−5+6+05−6+0−5+6+06−12+9−5+6+05−6+0−5+6+06−12+9 = 14 31−1131−113 Hence, A – 1 = 𝟏𝟒 𝟑𝟏−𝟏𝟏𝟑𝟏−𝟏𝟏𝟑
