Answer
The system solution set is $\left( 1,3,2,1 \right)$.
Work Step by Step
Consider the given system of the equation:
$\left\{ \begin{align}
& 2w+x-y=3 \\
& w-3x+2y=-4 \\
& 3w+x-3y+z=1 \\
& w+2x-4y-z=-2
\end{align} \right.$
The matrix corresponding to the system of equations is as follows:
$\left[ \begin{matrix}
2 & 1 & -1 & 0 & 3 \\
1 & -3 & 2 & 0 & -4 \\
3 & 1 & -3 & 1 & 1 \\
1 & 2 & -4 & -1 & -2 \\
\end{matrix} \right]$
Using the elementary row transformation we will find the echelon form of the matrix.
Interchange ${{R}_{1}}\leftrightarrow {{R}_{2}}$
$\left[ \begin{matrix}
1 & -3 & 2 & 0 & -4 \\
2 & 1 & -1 & 0 & 3 \\
3 & 1 & -3 & 1 & 1 \\
1 & 2 & -4 & -1 & -2 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$, ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$
$\left[ \begin{matrix}
1 & -3 & 2 & 0 & -4 \\
0 & 7 & -5 & 0 & 11 \\
0 & 10 & -9 & 1 & 13 \\
1 & 2 & -4 & -1 & -2 \\
\end{matrix} \right]$
Apply ${{R}_{4}}\to {{R}_{4}}-{{R}_{1}}$
$\left[ \begin{matrix}
1 & -3 & 2 & 0 & -4 \\
0 & 7 & -5 & 0 & 11 \\
0 & 10 & -9 & 1 & 13 \\
0 & 5 & -6 & -1 & 2 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to {{R}_{3}}-2{{R}_{4}}$
$\left[ \begin{matrix}
1 & -3 & 2 & 0 & -4 \\
0 & 7 & -5 & 0 & 11 \\
0 & 0 & 3 & 3 & 9 \\
0 & 5 & -6 & -1 & 2 \\
\end{matrix} \right]$
Interchange ${{R}_{3}}\leftrightarrow {{R}_{4}}$
$\left[ \begin{matrix}
1 & -3 & 2 & 0 & -4 \\
0 & 7 & -5 & 0 & 11 \\
0 & 5 & -6 & -1 & 2 \\
0 & 0 & 3 & 3 & 9 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to {{R}_{3}}-\frac{5}{7}{{R}_{2}}$
$\left[ \begin{matrix}
1 & -3 & 2 & 0 & -4 \\
0 & 7 & -5 & 0 & 11 \\
0 & 0 & -\frac{17}{7} & -1 & -\frac{41}{7} \\
0 & 0 & 3 & 3 & 9 \\
\end{matrix} \right]$
Apply ${{R}_{4}}\to {{R}_{4}}+\frac{21}{17}{{R}_{3}}$
$\left[ \begin{matrix}
1 & -3 & 2 & 0 & -4 \\
0 & 7 & -5 & 0 & 11 \\
0 & 0 & -\frac{17}{7} & -1 & -\frac{41}{7} \\
0 & 0 & 0 & \frac{30}{17} & \frac{30}{17} \\
\end{matrix} \right]$
Now use back-substitution to express $ w,x,y,z $.
$\begin{align}
& w-3x+2y=-4 \\
& 7x-5y=11 \\
& -\frac{17}{7}y-z=-\frac{41}{7} \\
& z=1
\end{align}$
Substitute $ z=1$ in the third equation to get:
$\begin{align}
& -\frac{17}{7}y-1=-\frac{41}{7} \\
& -\frac{17}{7}y=-\frac{41}{7}+1 \\
& -\frac{17}{7}y=-\frac{34}{7} \\
& y=2
\end{align}$
Substitute $ y=2$ in the second equation to get:
$\begin{align}
& 7x-5\left( 2 \right)=11 \\
& 7x=11+10 \\
& x=3
\end{align}$
Substitute $ x=3$ and $ y=2$ in the first equation to get:
$\begin{align}
& w-3\left( 3 \right)+2\left( 2 \right)=-4 \\
& w-9+4=-4 \\
& w=-4+5 \\
& w=1
\end{align}$
Hence, the system’s solution set is $\left( 1,3,2,1 \right)$.
