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Question
If A = `[(2,-1,1),(-1,2,-1),(1,-1,2)]` verify that A3 − 6A2 + 9A − 4I = O and hence find A−1
Solution
`A = [(2,-1,1),(-1,2,-1),(1,-1,2)]`
`A^2 = [(2,-1,1),(-1,2,-1),(1,-1,2)] [(2,-1,1),(-1,2,-1),(1,-1,2)] = [(6,-5,5),(-5,6,-5),(5,-5,6)]`
`A^3 = A^2A = [(6,-5,5),(-5,6,-5),(5,-5,6)] [(2,-1,1),(-1,2,-1),(1,-1,2)] = [(22,-21,21),(-21,22,-21),(21,-21,22)]`
LHS = A3 - 6A2 + 9A - 4I
`= [(22,-21,21),(-21,22,-21),(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,-21),(21,-21,22)] - [(36,-30,30),(-30,36,-30),(30,-30,36)] + [(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 - 0, 22 - 36 + 18 - 4, -21 - 30 + 9 - 0),(21 - 30 + 9 - 0, -21 + 30 - 9 - 0,22 - 36 + 18 - 4)]`
`= [(0,0,0),(0,0,0),(0,0,0)] = 0 = "RHS"`
A3 - 6A2 + 9A - 4I = 0
A3 - 6A2 + 9A = 4I
A2 AA-1 - 6 AA-1 + 9 AA-1 = 4IA-1
4A-1 = A2 - 6A + 9I `= [(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)]`
`= 4 [(3,1,-1),(1,3,1),(-1,1,3)]`
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