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If $\mathrm{AX}=\mathrm{B}$, where $\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 2 & 1\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right], \mathrm{B}=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$, then $2 x+y-z=0$
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$\begin{aligned} & {\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 2 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]} \\ & \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \text { and } \mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_1 \\ & {\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 3 & -5 \\ 0 & 2 & 0\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}4 \\ -8 \\ -2\end{array}\right]}\end{aligned}$
$\begin{array}{lll}\therefore & y=-1 & \ldots[\operatorname{from}(3)] \\ \therefore & -3-5 z=-8 \quad \Rightarrow z=1 & \ldots[\operatorname{from}(2)] \\ \therefore & x-(-1)+1=4 \Rightarrow x=2 & \ldots[\operatorname{from}(1)] \\ \therefore & 2 x+y-z=2(2)-1-1=2 & \end{array}$
$\begin{array}{lll}\therefore & y=-1 & \ldots[\operatorname{from}(3)] \\ \therefore & -3-5 z=-8 \quad \Rightarrow z=1 & \ldots[\operatorname{from}(2)] \\ \therefore & x-(-1)+1=4 \Rightarrow x=2 & \ldots[\operatorname{from}(1)] \\ \therefore & 2 x+y-z=2(2)-1-1=2 & \end{array}$
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