If A=2,1,1,1 Then Show Taht A^2-3a+I=0 - Mathematics and Statistics

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Sum

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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2014-2015 (March)

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