Question:
If $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}2 \\ -1 \\ 3\end{array}\right]$, find $x, y, z$
Solution:
Here,
$\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}2 \\ -1 \\ 3\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ z \\ y\end{array}\right]=\left[\begin{array}{c}2 \\ -1 \\ 3\end{array}\right]$
$\therefore x=2, y=3$ and $z=-1$
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