Answer
The system solution set is $\left( 1,-2,1,1 \right)$.
Work Step by Step
Consider the given system of equations:
$\left\{ \begin{align}
& w-3x+y-4z=4 \\
& -2w+x+2y=-2 \\
& 3w-2x+y-6z=2 \\
& -w+3x+2y-z=-6
\end{align} \right.$
The matrix corresponding to the system of equations is as follows:
$\left[ \begin{matrix}
1 & -3 & 1 & -4 & 4 \\
-2 & 1 & 2 & 0 & -2 \\
3 & -2 & 1 & -6 & 2 \\
-1 & 3 & 2 & -1 & -6 \\
\end{matrix} \right]$
Use the elementary row transformation to find the echelon form of the matrix.
Apply ${{R}_{2}}\to {{R}_{2}}+2{{R}_{1}}$, ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$
$\left[ \begin{matrix}
1 & -3 & 1 & -4 & 4 \\
0 & -5 & 4 & -8 & 6 \\
0 & 7 & -2 & 6 & -10 \\
-1 & 3 & 2 & -1 & -6 \\
\end{matrix} \right]$
Now, apply ${{R}_{3}}\to {{R}_{3}}+{{R}_{1}}$
$\left[ \begin{matrix}
1 & -3 & 1 & -4 & 4 \\
0 & -5 & 4 & -8 & 6 \\
0 & 7 & -2 & 6 & -10 \\
0 & 0 & 3 & -5 & -2 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to {{R}_{3}}+\frac{7}{5}{{R}_{2}}$
$\left[ \begin{matrix}
1 & -3 & 1 & -4 & 4 \\
0 & -5 & 4 & -8 & 6 \\
0 & 0 & \frac{18}{5} & -\frac{26}{5} & -\frac{8}{5} \\
0 & 0 & 3 & -5 & -2 \\
\end{matrix} \right]$
Apply ${{R}_{4}}\to {{R}_{4}}-\frac{15}{18}{{R}_{3}}$
$\left[ \begin{matrix}
1 & -3 & 1 & -4 & 4 \\
0 & -5 & 4 & -8 & 6 \\
0 & 0 & \frac{18}{5} & -\frac{26}{5} & -\frac{8}{5} \\
0 & 0 & 0 & -\frac{2}{3} & -\frac{2}{3} \\
\end{matrix} \right]$
Now, use back-substitution to express $ w,x,y,z $.
$\begin{align}
& w-3x+1y-4z=4 \\
& -5x+4y-8z=6 \\
& \frac{18}{5}y-\frac{26}{5}z=-\frac{8}{5} \\
& z=1
\end{align}$
Substitute $ z=1$ in the third equation as follows;
$\begin{align}
& \frac{18}{5}y-\frac{26}{5}\left( 1 \right)=-\frac{8}{5} \\
& \frac{18}{5}y=-\frac{8}{5}+\frac{26}{5} \\
& \frac{18}{5}y=\frac{18}{5} \\
& y=1
\end{align}$
Substitute $ y=1$, $ z=1$ in the second equation as follows:
$\begin{align}
& -5x+4\left( 1 \right)-8\left( 1 \right)=6 \\
& -5x=6+4 \\
& x=-2
\end{align}$
Substitute $ x=-2$, $ y=1$, $ z=1$ in the first equation as follows:
$\begin{align}
& w-3\left( -2 \right)+1\left( 1 \right)-4\left( 1 \right)=4 \\
& w+6-3=4 \\
& w=4-3 \\
& w=1
\end{align}$
Hence, the system’s solution set is $\left( 1,-2,1,1 \right)$.
