Question
Find $x, y, z$ if $\left\{\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 1 & 1\end{array}\right]-3\left[\begin{array}{cc}2 & 1 \\ 3 & -2 \\ 1 & 3\end{array}\right]\right\}\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}x-1 \\ y+1 \\ 2 z\end{array}\right]$

Answer

$\left\{5\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 1 & 1\end{array}\right]-3\left[\begin{array}{cc}2 & 1 \\ 3 & -2 \\ 1 & 3\end{array}\right]\right\}\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}x-1 \\ y+1 \\ 2 z\end{array}\right]$
$\begin{array}{ll}\therefore {\left[\left[\begin{array}{cc}0 & 5 \\ 5 & 0 \\ 5 & 5\end{array}\right]-\left[\begin{array}{cc}6 & 3 \\ 9 & -6 \\ 3 & 9\end{array}\right]\right\}\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}x-1 \\ y+1 \\ 2 z\end{array}\right]} \end{array} $
$ \therefore {\left[\begin{array}{cc}-6 & 2 \\ -4 & 6 \\ 2 & -4\end{array}\right]\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}x-1 \\ y+1 \\ 2 z\end{array}\right]}$
$\begin{aligned} & \therefore\left[\begin{array}{c}-12+2 \\ -8+6 \\ 4-4\end{array}\right]=\left[\begin{array}{c}x-1 \\ y+1 \\ 2 z\end{array}\right]\end{aligned}  $
$ \therefore\left[\begin{array}{c}-10 \\ -2 \\ 0\end{array}\right]=\left[\begin{array}{c}x-1 \\ y+1 \\ 2 z\end{array}\right]$
$\therefore$ By equality of matrices, we get
$x-1=-10 \therefore x=-9$
$y+1=-2 \therefore y=-3$
$2 z=0 \therefore z=0$

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