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Solution
Here,
\[\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}2 \\ - 1 \\ 3\end{bmatrix} \]
\[ \Rightarrow \begin{bmatrix}x \\ z \\ y\end{bmatrix} = \begin{bmatrix}2 \\ - 1 \\ 3\end{bmatrix}\]
\[ \therefore x = 2, y = 3\text{ and }z = - 1\]
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