Answer
$\Rightarrow \left\{\begin{matrix}
w -x +y +z =\;3 \\
\; \; \;\;\;x -2 y - z=\;0 \\
\;\; \;\;\; \;\;\;\;\;\; y + 6z =17\\
\;\; \;\;\; \;\;\;\;\;\;\;\;\;\;\;\; z=\; 3
\end{matrix}\right.$
$\Rightarrow \{(2,1,-1,3)\}$.
Work Step by Step
The given augmented matrix is
$\Rightarrow \left[\begin{array}{cccc|c}
1 & -1 & 1&1& 3\\
0&1&-2&-1& 0 \\
0 & 0& 1& 6&17\\
0&0&0&1&3
\end{array}\right]$
The above matrix is the reduced row-echelon form.
Convert the matrix into a system of equations as shown below.
$\Rightarrow \left\{\begin{matrix}
1w -1x +1y +1z =\;3 \\
0w +1x -2 y -1 z=\;0 \\
0w +0 x +1 y + 6z =17\\
0 w +0 x +0 y +1 z=\; 3
\end{matrix}\right.$
or we can write.
$\Rightarrow \left\{\begin{matrix}
w -x +y +z =\;3 \\
\; \; \;\;\;x -2 y - z=\;0 \\
\;\; \;\;\; \;\;\;\;\;\; y + 6z =17\\
\;\; \;\;\; \;\;\;\;\;\;\;\;\;\;\;\; z=\; 3
\end{matrix}\right.$
Substitute back the value of $z$ into the third equation.
$\Rightarrow y+6(3)=17$
Simplify.
$\Rightarrow y+18=17$
Solve for $y$.
$\Rightarrow y=-1$.
Substitute back the value of $z$ and $y$ into the second equation.
$\Rightarrow x -2 (-1) - (3)=0$
Simplify.
$\Rightarrow x +2 - 3=0$
Solve for $x$.
$\Rightarrow x=1$.
Substitute back the value of $z,y$ and $x$ into the first equation.
$\Rightarrow w -(1) +(-1) +(3) =3$
Simplify.
$\Rightarrow w -1 -1 +3 =3$
Solve for $x$.
$\Rightarrow w=2$.
The solution set is $\{(w,x,y,z)\}=\{(2,1,-1,3)\}$.
