If P = begin{bmatrix} frac{sqrt{3}}{2} & frac | vedclass

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(A) Given $Q = PAP^T$. Note that $P$ is an orthogonal matrix,so $PP^T = I$ and $P^T = P^{-1}$.We want to compute $X = P^T Q^{2005} P$.Since $Q = PAP^T

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$\begin{bmatrix} 1 & 2005 \ 0 & 1 \end{bmatrix}$

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(A) Given $Q = PAP^T$. Note that $P$ is an orthogonal matrix,so $PP^T = I$ and $P^T = P^{-1}$.We want to compute $X = P^T Q^{2005} P$.Since $Q = PAP^T$,we have $Q^n = (PAP^T)(PAP^T)...(PAP^T) = PA^n P^T$.Substituting this into the expression:$X = P^T (PA^{2005}P^T) P$$X = (P^T P) A^{2005} (P^T P)$Since $P^T P = I$,we get $X = I A^{2005} I = A^{2005}$.Given $A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$,we use the property of this specific matrix: $A^n = \begin{bmatrix} 1 & n \ 0 & 1 \end{bmatrix}$.Thus,$A^{2005} = \begin{bmatrix} 1 & 2005 \ 0 & 1 \end{bmatrix}$.Therefore,the correct option is $A$.

If $P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$,$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $Q = PAP^T$,then $P^T(Q^{2005})P$ is equal to

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