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Question
if `A = [(1,0,2),(0,2,1),(2,0,3)]` , prove that `A^2 - 6A^2 + 7A + 2I = 0`
Solution
Given: A = `[(1,0,2),(0,2,1),(2,0,3)]`
`therefore "A"^2 = "A" xx "A" = [(1,0,2),(0,2,1),(2,0,3)] [(1,0,2),(0,2,1),(2,0,3)]`
`= [(1 + 0 + 4, 0 + 0 + 0, 2 + 0 + 6),(0 + 0 + 2, 0 + 4 + 0, 0 + 2 + 3),(2 + 0 + 6, 0 + 0 + 0, 4 + 0 + 9)]`
`= [(5,0,8),(2,4,5),(8,0,13)]`
`"A"^3 = "A"^2. "A" = [(5,0,8),(2,4,5),(8,0,13)] [(1,0,2),(0,2,1),(2,0,3)]`
`= [(5 + 0 + 16, 0 + 0 + 0, 10 + 0 + 24), (2 + 0 + 10, 0 + 8 + 0, 4 + 4 + 15), (8 + 0 + 26, 0 + 0 + 0, 16 + 0 + 39)]`
`= [(21,0,34),(12,8,23),(34,0,55)]`
`"A"^3 - 6"A"^2 + 7 "A" + 2 "I" = [(21,0,34),(12,8,23),(34,0,55)] - 6 [(5,0,8),(2,4,5),(8,0,13)] + 7 [(1,0,2),(0,2,1),(2,0,3)] + 2 [(1,0,0),(0,1,0),(0,0,1)]`
`= [(21,0,34),(12,8,23),(34,0,55)] - [(30,0,48),(12,24,30),(48,0,73)] + [(7,0,14),(0,14,7),(14,0,21)] + [(2,0,0),(0,2,0),(0,0,2)]`
`= [(0,0,0),(0,0,0), (0,0,0)]` = 0
`"A"^3 - 6 "A"^2 + 7 "A" + 2 "I" = 0`
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