Solve the shown using matrices: begin{cases}2x-y-2z=-1 x-2y-z=1 x+y+z=1 end{cases}

Solve the shown using matrices:
$\left\{\begin{array}{l}2x-y-2z=-1\\ x-2y-z=1\\ x+y+z=1\end{array}$
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Step 1
given equations are
$2x-y-2z=-1$
$x-2y-z=1$
$x+y+z=1$
We have to solve it by matrices
In matrix form
AX=B
$\left[\begin{array}{ccc}2& -1& -2\\ 1& -2& -1\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]$
augmented matrix
$\left[\begin{array}{ccccc}2& -1& -2& |& -1\\ 1& -2& -1& |& 1\\ 1& 1& 1& |& 1\end{array}\right]$
${R}_{3}\to {R}_{2}-{R}_{3}$
$\left[\begin{array}{ccccc}2& -1& -2& |& -1\\ 1& -2& -1& |& 1\\ 0& -3& -2& |& 0\end{array}\right]$
${R}_{2}\to 2{R}_{2}-{R}_{1}$
$\left[\begin{array}{ccccc}2& -1& -2& |& -1\\ 0& -3& 0& |& 3\\ 0& -3& -2& |& 0\end{array}\right]$
${R}_{1}\to {R}_{1}-{R}_{3},{R}_{2}\to \frac{{R}_{2}}{-3}$
$\left[\begin{array}{ccccc}2& 2& 0& |& -1\\ 0& 1& 0& |& -1\\ 0& -3& -2& |& 0\end{array}\right]$
${R}_{1}\to \frac{{R}_{1}}{2}$
$\left[\begin{array}{ccccc}1& 1& 0& |& -\frac{1}{2}\\ 0& 1& 0& |& -1\\ 0& -3& -2& |& 0\end{array}\right]$
${R}_{1}\to {R}_{1}-{R}_{2},{R}_{3}\to {R}_{3}+3{R}_{2}$
$\left[\begin{array}{ccccc}1& 0& 0& |& \frac{1}{2}\\ 0& 1& 0& |& -1\\ 0& 0& -2& |& -3\end{array}\right]$
${R}_{3}\to \frac{{R}_{3}}{-2}$
$\left[\begin{array}{ccccc}1& 0& 0& |& \frac{1}{2}\\ 0& 1& 0& |& -1\\ 0& 0& 1& |& \frac{3}{2}\end{array}\right]$
$⇒x=\frac{1}{2},y=-1,z=\frac{3}{2}$
Jeffrey Jordon