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Question
If A = `[(2,-3,5),(3,2,-4),(1,1,-2)]` find A−1. Using A−1 solve the system of equations
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
Solution
`abs A = abs ((2,-3,5),(3,2,-4),(1,1,-2))`
`= 2 [2 xx (-2) - 1 xx (-4)] - (-3) [3 xx (-2) - (1) xx (- 4)] + 5 [ 3 xx 1 - 1 xx 2]`
`= 2 [-4 + 4] + 3 [-6 + 4] + 5 [3 - 2]`
`= 0 + 3 xx (-2) + 5 xx 1`
`= -6 + 5 = -1 ne 0`
`therefore A^-1` can be known,
Cofactors of the elements of `abs A`
`A_11 = abs ((2,-4),(1,-2)) = -4 + 4 = 0`
`A_12 = - abs ((3,-4),(1,-2)) = - (-6 + 4) = 2`
`A_13 = abs ((3,2),(1,1)) = 3 - 2 = 1`
`A_21 = - abs ((-3,5),(1,-2)) = - (6 - 5) = -1`
`A_22 = abs ((2,5),(1,-2)) = -4 - 5 = -9`
`A_23 = - abs ((2,-3),(1,1)) = - (2 + 3) = - 5`
`A_31 = abs ((-3,5),(2,-4)) = 12 - 10 = 2`
`A_32 = - abs ((2,5),(3,-4)) = - (-8 - 15) = 23`
`A_33 = abs ((2,-3),(3,2)) = 4 + 9 = 13`
The cofactor matrix of the elements of `therefore abs A` is C = `[(0,2,1),(-1,-9,-5),(2,23,13)]`
`therefore adj A = [(0,2,1),(-1,-9,-5),(2,23,13)] [(0,-1,2),(2,-9,23),(1,-5,13)]`
`therefore A^-1 = adj A/abs A`
`= - [(0,-1,2),(2,-9,23),(1,-5,13)] = [(0,1,-2),(-2,9,-23),(-1,5,-13)]`
Writing the given equation in the form AX = B,
Or `A = [(2,-3,5),(3,2,-4),(1,1,-2)], X = [(x),(y),(z)], B = [(11),(-5),(-3)]`
`therefore X = A^-1 B`
`[(x),(y),(z)] = [(0,1,-2),(-2,9,-23),(-1,5,-13)] [(11),(-5),(-3)] = [(0 - 5 + 6),(-22 - 45 + 69),(-11 - 25 + 39)] = [(1),(2),(3)]`
`=> x = 1, y = 2, z = 3`
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