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If A A=[3-2 4-2]and I=[1 0 0 1], find k ...

If A `A=[3-2 4-2]`and `I=[1 0 0 1]`, find k so that `A^2=k A-2I`.

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The correct Answer is:
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` A=[{:(3,-2),(4,-2):}] `
`therefore A^(2)=A.A =[{:(3,-2),(4,-2):}][{:(3,-2),(4,-4):}]`
`=[{:(9-8,-6+4),(12-8,-8+4):}]=[{:(1,-2),(4,-4):}]`
Given that ,`A^(2)=Ka-2l`
`implies A^(2)+2l=KA`
`[{:(1,-2),(4,-4):}]+2[{:(1,0),(0,1):}]=k[{:(3,-2),(4,-2):}]`
`implies [{:(1,-2),(4,-4):}]+[{:(2,0),(0,2):}]=K[{:(3,-2),(4,-2):}]`
`implies[{:(3,-2),(4,-2):}]=K[{:(3,-2),(4,-2):}]`
`implies K=1`
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