Let A=begin{bmatrix} frac{1}{sqrt{2}} & -2 0 | vedclass

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(A) Given $P = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$ is an orthogonal matrix,so $P^T P = I$ and $P^T

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$65$

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(A) Given $P = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$ is an orthogonal matrix,so $P^T P = I$ and $P^T = P^{-1}$.Given $B = P A P^T$.Then $C = P^T B^{10} P = P^T (P A P^T)^{10} P = P^T (P A^{10} P^T) P = (P^T P) A^{10} (P^T P) = I A^{10} I = A^{10}$.Thus,the sum of the diagonal elements of $C$ is the trace of $A^{10}$.$A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \ 0 & 1 \end{bmatrix}$.For a matrix $A = \begin{bmatrix} a & b \ 0 & d \end{bmatrix}$,$A^n = \begin{bmatrix} a^n & b(a^{n-1} + a^{n-2}d + \dots + d^{n-1}) \ 0 & d^n \end{bmatrix}$.The diagonal elements of $A^{10}$ are $a^{10}$ and $d^{10}$.Here $a = \frac{1}{\sqrt{2}}$ and $d = 1$.Trace$(A^{10}) = a^{10} + d^{10} = (\frac{1}{\sqrt{2}})^{10} + 1^{10} = \frac{1}{2^5} + 1 = \frac{1}{32} + 1 = \frac{33}{32}$.Given $\frac{m}{n} = \frac{33}{32}$ with $\operatorname{gcd}(33, 32) = 1$,so $m = 33$ and $n = 32$.Therefore,$m + n = 33 + 32 = 65$.

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