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If P = [[cos frac(pi)(6), sin frac(pi)(6...

If `P = [[cos frac(pi)(6), sin frac(pi)(6) ],[-sinfrac(pi)(6),cosfrac(pi)(6)]], A = [[1,1],[0,1]] then P+ A`

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Verified by Experts

The correct Answer is:
B
We have, `P^(T) = P^(-1) " " [ because PP^(T) =I]`
Now, ,,`Q= PAP^(T) = PAP^(-1)`
`therefore Q^(2007) = PA^(2007) P^(-1) `
`therefore P^(T) Q^(2007 ) P= P^(-1) (PA^(2007)P^(-1)) P`
`= A^(2007 = [[1,2007],[0,1]][becauseA^(2)=[[1,2],[0,1]],A^(3) = [[1,3],[0,1]],...]`
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