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Question
If A = `[(2,-1,1),(-1,2,-1),(1,-1,2)]` verify that A3 − 6A2 + 9A − 4I = 0 and hence find A−1.
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Solution
We have A = `[(2,-1,1),(-1,2,-1),(1,-1,2)]`
A2 = AA
= `[(2,-1,1),(-1,2,-1),(1,-1,2)] [(2,-1,1),(-1,2,-1),(1,-1,2)]`
= `[(4+1+1,-2-2-1,2+1+2),(-2-2-1,1+4+1,-1-2-2),(2+1+2,-1-2-2,1+1+4)]`
= `[(6,-5,5),(-5,6,-5),(5,-5,6)]`
A3 = A2A
= `[(6,-5,5),(-5,6,-5),(5,-5,6)] [(2,-1,1),(-1,2,-1),(1,-1,2)]`
= `[(12+5+5,-6-10-5,6+5+10),(-10-6-5,5+12+5,-5-6-10),(10+5+6,-5-10-6,5+5+12)]`
= `[(22,-21,21),(-21,22,-21),(21,-21,22)]`
Now, 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
Hence A3 − 6A2 + 9A − 4I = 0
Now, A3 − 6A2 + 9A − 4I = 0
⇒ 4I = A3 − 6A2 + 9A
Multiplying both sides by A−1, we get
⇒ `A^(−1) = 1/4 A^2 - 6/4 A + 9/4I`
= `1/4[(6,-5,5),(-5,6,-5),(5,-5,6)] - 6/4 [(2,-1,1),(-1,2,-1),(1,-1,2)] + 9/4 [(1,0,0),(0,1,0),(0,0,1)]`
= `1/4 [(6-12+9,-5+6+0,5-6+0),(-5+6+0,6-12+9,-5+6+0),(5-6+0,-5+6+0,6-12+9)]`
= `1/4 [(3,1,-1),(1,3,1),(-1,1,3)]`
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