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If `"A"=[[2,1],[1,1]] " then show taht " A^2-3A+I=0`
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Solution
Given
`A=[[2,1],[1,1]]`
`A^2=A A=[[2,1],[1,1]] [[2,1],[1,1]]`
`=[[4+1,2+1],[2+1,1+1]]`
`=[[5,3],[3,2]]`
`A^2-3A+I=[[5,3],[3,2]]-3[[2,1],[1,1]]+[[1,0],[0,1]]`
`=[[5-6+1,3-3+0],[3-3+0,2-3+1]]`
`=[[0,0],[0,0]]`
`=0`
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