Answer
The system solution set is $\phi $.
Work Step by Step
Consider the system of equations:
$\left\{ \begin{align}
& 2w+x-2y-z=3 \\
& w-2x+y+z=4 \\
& -w-8x+7y+5z=13 \\
& 3w+x-2y+2z=6
\end{align} \right.$
The given system of equations can be written in matrix form as below:
$\left[ \begin{matrix}
2 & 1 & -2 & -1 & 3 \\
1 & -2 & 1 & 1 & 4 \\
-1 & -8 & 7 & 5 & 13 \\
3 & 1 & -2 & 2 & 6 \\
\end{matrix} \right]$
Using the elementary row transformation we will find the echelon form of the matrix.
Interchange ${{R}_{1}}\leftrightarrow {{R}_{2}}$:
$\left[ \begin{matrix}
1 & -2 & 1 & 1 & 4 \\
2 & 1 & -2 & -1 & 3 \\
-1 & -8 & 7 & 5 & 13 \\
3 & 1 & -2 & 2 & 6 \\
\end{matrix} \right]$
Now, apply ${{R}_{3}}\to {{R}_{3}}+{{R}_{1}}$, ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$
$\left[ \begin{matrix}
1 & -2 & 1 & 1 & 4 \\
0 & 5 & -4 & -3 & -5 \\
0 & -10 & 8 & 6 & 17 \\
3 & 1 & -2 & 2 & 6 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to {{R}_{3}}+2{{R}_{2}}$
$\left[ \begin{matrix}
1 & -2 & 1 & 1 & 4 \\
0 & 5 & -4 & -3 & -5 \\
0 & 0 & 0 & 0 & 7 \\
3 & 1 & -2 & 2 & 6 \\
\end{matrix} \right]$
In third row, after back substitution,
$0w+0x+0y+0z=7$
Which is not possible; therefore the linear system has no solution.
Hence, the system solution set is $\phi $.
