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Question
If A = `[(3,1),(-1,2)]` show that A2 – 5A + 7I = O. Hence, find A–1.
Solution
LHS `= A^2 - 5A + 7I = A A - 5A + 7I`
`= [(3,1),(-1,2)] [(3,1),(-1,2)] - 5 [(3,1),(-1,2)] + 7 [(1,0),(0,1)]`
`= [(9 - 1,3 + 2),(-3 -2,-1 + 4)] - [(15,5),(-5,10)] + [(7,0),(0,7)]`
`= [(8 - 15 + 7,5 -5+0),(-5 +5+0,3 -10+7)] - [(0,0),(0,0)] =0`
`A^2 - 5A + 7I = 0`
`= A^2 - 5A = -7I`
= (A-1A)A - 5AA-1 + 7IA-1 = 0
`= - 7"IA"^-1`
`= 7A^-1 = 5I - AI`
`= 5 [(1,0),(0,1)] - [(3,1),(-1,2)]`
`= [(5,0),(0,5)] - [(3,1),(-1,2)] = [(2,-1),(1,3)]`
`A^-1 = 1/7 [(2,-1),(1,3)]`
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