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If A=[3 1 1 2], show that A^2-5A+71=0. H...

If `A=[3 1 1 2]`, show that `A^2-5A+71=0`. Hence find `A^(-1)`.

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The correct Answer is:
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` A^(2)=A.A=[{:(3,1),(-1,2):}][{:(3,1),(-1,2):}] `
`=[{:(9-1,3+2),(-3-2,-1+4):}]=[{:(8,5),(-5,3):}]`
` Now, L.H.S =A^(2)-5A +7I_(2)`
`=[{:(8,5),(-5,3):}]-5[{:(3,1)(-1,2):}]+7[{:(1,0),(0,1):}]`
`=[{:(8,5),(-5,3):}]-[{:(15,5),(-5,10):}]+[{:(7,0),(0,7):}]`
`=[{:(8-15+7,5-5+0),(-5+50,0-3-10+7):}]`
`=[{:(0,0),(0,0):}]`
`=O=R.H.S.` hence proved .
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