Answer
The system solution set is $\left( -1,2,1,1 \right)$.
Work Step by Step
Consider the given system of equations, $\left\{ \begin{align}
& w-2x-y-3z=-9 \\
& w+x-y=0 \\
& 3w+4x+z=6 \\
& 2x-2y+z=3
\end{align} \right.$
The given system of equations can be written in matrix form as below:
$\left[ \begin{matrix}
1 & -2 & -1 & -3 & -9 \\
1 & 1 & -1 & 0 & 0 \\
3 & 4 & 0 & 1 & 6 \\
0 & 2 & -2 & 1 & 3 \\
\end{matrix} \right]$
Using the elementary row transformation we will find the echelon form of the matrix.
Apply ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$, ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$:
$\left[ \begin{matrix}
1 & -2 & -1 & -3 & -9 \\
0 & 3 & 0 & 3 & 9 \\
0 & 10 & 3 & 10 & 33 \\
0 & 2 & -2 & 1 & 3 \\
\end{matrix} \right]$
Now, apply ${{R}_{3}}\to {{R}_{3}}-5{{R}_{4}}$:
$\left[ \begin{matrix}
1 & -2 & -1 & -3 & -9 \\
0 & 3 & 0 & 3 & 9 \\
0 & 0 & 13 & 5 & 18 \\
0 & 2 & -2 & 1 & 3 \\
\end{matrix} \right]$
Apply ${{R}_{2}}\to \frac{1}{3}{{R}_{2}}:$
$\left[ \begin{matrix}
1 & -2 & -1 & -3 & -9 \\
0 & 1 & 0 & 1 & 3 \\
0 & 0 & 13 & 5 & 18 \\
0 & 2 & -2 & 1 & 3 \\
\end{matrix} \right]$
Apply ${{R}_{4}}\to {{R}_{4}}-2{{R}_{2}}:$
$\left[ \begin{matrix}
1 & -2 & -1 & -3 & -9 \\
0 & 1 & 0 & 1 & 3 \\
0 & 0 & 13 & 5 & 18 \\
0 & 0 & -2 & -1 & -3 \\
\end{matrix} \right]$
Apply ${{R}_{4}}\to {{R}_{4}}+\frac{2}{13}{{R}_{3}}:$
$\left[ \begin{matrix}
1 & -2 & -1 & -3 & -9 \\
0 & 1 & 0 & 1 & 3 \\
0 & 0 & 13 & 5 & 18 \\
0 & 0 & 0 & -\frac{3}{13} & -\frac{3}{13} \\
\end{matrix} \right]$
This gives the system as:
$\begin{align}
& w-2x-y-3z=-9 \\
& x+z=3 \\
& 13y+5z=18 \\
& z=1
\end{align}$
Substitute $ z=1$ in the third equation to get:
$\begin{align}
& 13y+5\left( 1 \right)=18 \\
& 13y=18-5 \\
& 13y=13 \\
& y=1
\end{align}$
Substitute $ z=1$ in the second equation to get:
$\begin{align}
& x+1=3 \\
& x=2
\end{align}$
Substitute $ z=1$, $ x=2$, and $ y=1$ in the first equation to get:
$\begin{align}
& w-2\left( 2 \right)-1-3\left( 1 \right)=-9 \\
& w-4-4=-9 \\
& w=-1
\end{align}$
The system’s solution set is: $\left( -1,2,1,1 \right)$.
