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(A) Given $B = PAP^{-1}$, we can multiply by $P$ on the right to get $BP = PA$. Substituting the matrices: $\begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \

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$x = 1, y = 1$

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(A) Given $B = PAP^{-1}$, we can multiply by $P$ on the right to get $BP = PA$. Substituting the matrices: $\begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \ x & 0 & 0 \ 0 & 0 & y \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \ x & 0 & 0 \ 0 & 0 & y \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$ Calculating the products: Left side: $\begin{bmatrix} x & 0 & 0 \ 0 & 0 & y \ 0 & 0 & 0 \end{bmatrix}$ Right side: $\begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & x \ 0 & 0 & 0 \end{bmatrix}$ Equating the corresponding elements, we get $x = 1$ and $y = x$. Therefore, $x = 1$ and $y = 1$.

Let $A = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$, and $P = \begin{bmatrix} 0 & 1 & 0 \\ x & 0 & 0 \\ 0 & 0 & y \end{bmatrix}$ be an orthogonal matrix such that $B = PAP^{-1}$ holds. Then:

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