Answer
$\Rightarrow \left\{\begin{matrix}
\;w +2x -y =\;\;2 \\
\; \;\;x + y -2 z=-3 \\
\;\; \; \;\;\;\;\;\;\; y -z =-2\\
\; \;\;\;\;\;\;\;\;\;\;\;\;\; z=\;\; 3
\end{matrix}\right.$
$\Rightarrow \{(-1,2,1,3)\}$.
Work Step by Step
The given augmented matrix is
$\Rightarrow \left[\begin{array}{cccc|c}
1 & 2 & -1&0& 2\\
0&1&1&-2& -3 \\
0 & 0& 1& -1&-2\\
0&0&0&1&3
\end{array}\right]$
The above matrix is the reduced row-echelon form.
Convert the matrix into system of equations as shown below.
$\Rightarrow \left\{\begin{matrix}
1w +2x -1y +0z =\;2 \\
0w +1x +1 y -2 z=\;-3 \\
0w +0 x +1 y -1z =-2\\
0 w +0 x +0 y +1 z=\; 3
\end{matrix}\right.$
or we can write.
$\Rightarrow \left\{\begin{matrix}
\;w +2x -y =\;\;2 \\
\; \;\;x + y -2 z=-3 \\
\;\; \; \;\;\;\;\;\;\; y -z =-2\\
\; \;\;\;\;\;\;\;\;\;\;\;\;\; z=\;\; 3
\end{matrix}\right.$
Substitute back the value of $z$ into third equation.
$\Rightarrow y-(3)=-2$
Simplify.
$\Rightarrow y-3=-2$
Solve for $y$.
$\Rightarrow y=1$.
Substitute back the value of $z$ and $y$ into the second equation.
$\Rightarrow x + (1) -2 (3)=-3$
Simplify.
$\Rightarrow x +1 - 6=-3$
Solve for $x$.
$\Rightarrow x=2$.
Substitute back the value of $z,y$ and $x$ into the first equation.
$\Rightarrow w +2(2) -(1) =2$
Simplify.
$\Rightarrow w +4 -1 =2$
Solve for $w$.
$\Rightarrow w=-1$.
The solution set is $\{(w,x,y,z)\}=\{(-1,2,1,3)\}$.
