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If (i) A=[cosalphasinalpha-sinalphacosal...

If (i) `A=[cosalphasinalpha-sinalphacosalpha]` , then verify that `AprimeA" "=" "I` . (ii) `A=[sinalphacosalpha-cosalphasinalpha]` , then verify that `AprimeA" "=" "I` .

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The correct Answer is:
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` A=[{:(cosalpha ,sin alpha),(-sin alpha,cosalpha ):}],`
`implies A'[{:(cos alpha ,-sin alpha),(sinalpha ,cosalpha):}]`
`therefore A'A =[{:(cosalpha,-sinalpha),(sinalpha ,cos alpha ):}][{:(cos alpha, sinalpha),(-sinalpha , cos alpha ):}]`
`=[{:(cos ^(2)alpha +sin^(2)alpha ,sinalpha cos alpha -sin slpha cos alpha ),( sin alpha cos alpha -sinalpha cos alpha ,sin^(2) alpha +cos^(2)alpha):}]`
`=[{:(1,0),(0,1):}]=I` hence proved
`(ii) if A=[{:(sinalpha ,cosalpha),(-cos alpha,sinalpha ):}],`
`implies A'=[{:(sin alpha ,-cos alpha),(cos alpha , sinalpha ):}]`
`therefore A'A =[{:(sin alpha ,-cosalpha ),(cosalpha ,sin alpha ):}][{:(sin alpha ,cosalpha ),(-cos alpha , sin alpha ):}]`
`=[{:(sin^(2)alpha +cos^(2)alpha ,sinalpha cos alpha - sin alpha cos alpha ),( sin alpha cos alpha -sin alpha cos alpha ,cos^(2)alpha +sin^(2)alpha ):}]`
`=[{:(1,0),(0,1):}]=I` hence proved .
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