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Question
If `A=[[1,2,2],[2,1,2],[2,2,1]]` ,then show that `A^2-4A-5I=0` and hence find A-1.
Solution
`A=[[1,2,2],[2,1,2],[2,2,1]]`
`A^2=[[1,2,2],[2,1,2],[2,2,1]][[1,2,2],[2,1,2],[2,2,1]]`
`=[[1xx1+2xx2+2xx2,1xx2+2xx2+2xx2,1xx2+2xx2+2xx1],[2xx1+1xx2+2xx2,2xx2+1xx1+2xx2,2xx2+1xx2+2xx1],[2xx1+2xx2+1xx2,2xx2+2xx1+1xx2,2xx2+2xx2+1xx1]]`
`=[[1+4+4,2+2+4,2+4+2],[2+2+4,4+1+4,4+2+2],[2+4+2,4+2+2,4+4+1]]`
`=[[9,8,8],[8,9,8],[8,8,9]]`
consider A2-4A-5I
`=[[9,8,8],[8,9,8],[8,8,9]]-4[[1,2,2],[2,1,2],[2,2,1]]-5[[1,0,0],[0,1,0],[0,0,1]]`
`=[[9,8,8],[8,9,8],[8,8,9]]-[[4,8,8],[8,4,8],[8,8,4]]-[[5,0,0],[0,5,0],[0,0,5]]`
`=[[9-9,8-8,8-8],[8-8,9-9,8-8],[8-8,8-8,9-9]]`
`=[[0,0,0],[0,0,0],[0,0,0]]`
Now
A2-4A-5I=0
A2-4A=5I
`A^2A^(-1)-4A.A^(-1)=5IA^(-1)` (Postmultiply by A-1)
A-4I=5A-1
`[[1,2,2],[2,1,2],[2,2,1]]-[[4,0,0],[0,4,0],[0,0,4]]=5A^-1`
`[[-3,2,2],[2,-3,2],[2,2,-3]]=5A^(-1)`
`A^-1 =[[-3/5,2/5,2/5],[2/5,-3/5,2/5],[2/5,2/5,-3/5]]`
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