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Question
For the matrix A = `[(1,1,1),(1,2,-3),(2,-1,3)]` show that A3 − 6A2 + 5A + 11 I = O. Hence, find A−1.
Solution
A `= [(1,1,1),(1,2,-3),(2,-1,3)]`
`"A"^2 = [(1,1,1),(1,2,-3),(2,-1,3)] [(1,1,1),(1,2,-3),(2,-1,3)] = [(4,2,1),(-3,8,-14),(7,-3,14)]`
`"A"^3 = "A"^2 "A" = [(4,2,1),(-3,8,-14),(7,-3,14)] [(1,1,1),(1,2,-3),(2,-1,3)]`
`= [(8,7,1),(-23,27,-69),(32,-13,58)]`
`"LHS" = "A"^3 - 6"A"^2 + 5 "A" + 11 "I"`
`= [(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)] - [(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)]`
`= [(0,0,0),(0,0,0),(0,0,0)] = 0 ="RHS"`
`"A"^3 - 6"A"^2 + 5"A" + 11 "I" = 0`
`"A"^3 - 6"A"^2 + 5"A" = -11 "I"`
`"A"^2 "AA"^-1 = 6 "AAA"^-1 + 5 "AA"^-1 = 11"IA"^-1`
`11"A"^-1 = - "A"^2 + 6"A" - 5"I" = [(-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)]`
`= [(-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)]`
`= [(-3,4,5),(9,-1,-4),(5,-3,-1)]`
`"A"^-1 = 1/11 [(-3,4,5),(9,-1,-4),(5,-3,-1)]`
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