Answer
The solution to the system is $\left\{ \left( \frac{1}{3}t,\frac{2}{3}t,-\frac{1}{3}t,t \right) \right\}$.
Work Step by Step
Consider the given system of equations:
$\left\{ \begin{align}
& 2w-x+3y+z=0 \\
& 3w+2x+4y-z=0 \\
& 5w-2x-2y-z=0 \\
& 2w+3x-7y-5z=0
\end{align} \right.$
The matrix corresponding to the system of equations is as follows:
$\left[ \begin{matrix}
2 & -1 & 3 & 1 & 0 \\
3 & 2 & 4 & -1 & 0 \\
5 & -2 & -2 & -1 & 0 \\
2 & 3 & -7 & -5 & 0 \\
\end{matrix} \right]$
Use the elementary row transformation to find the echelon form of the matrix.
Apply ${{R}_{2}}\to {{R}_{2}}-{{R}_{4}}$
$\left[ \begin{matrix}
2 & -1 & 3 & 1 & 0 \\
1 & -1 & 11 & 4 & 0 \\
5 & -2 & -2 & -1 & 0 \\
2 & 3 & -7 & -5 & 0 \\
\end{matrix} \right]$
Interchange ${{R}_{1}}\leftrightarrow {{R}_{2}}$
$\left[ \begin{matrix}
1 & -1 & 11 & 4 & 0 \\
2 & -1 & 3 & 1 & 0 \\
5 & -2 & -2 & -1 & 0 \\
2 & 3 & -7 & -5 & 0 \\
\end{matrix} \right]$
Apply ${{R}_{4}}\to {{R}_{4}}-{{R}_{2}}$, ${{R}_{3}}\to {{R}_{3}}-5{{R}_{1}}$
$\left[ \begin{matrix}
1 & -1 & 11 & 4 & 0 \\
2 & -1 & 3 & 1 & 0 \\
0 & 3 & -57 & -21 & 0 \\
0 & 4 & -10 & -6 & 0 \\
\end{matrix} \right]$
Apply ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$
$\left[ \begin{matrix}
1 & -1 & 11 & 4 & 0 \\
0 & 1 & -19 & -7 & 0 \\
0 & 3 & -57 & -21 & 0 \\
0 & 4 & -10 & -6 & 0 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to {{R}_{3}}-3{{R}_{2}},{{R}_{4}}\to {{R}_{4}}-4{{R}_{2}}$
$\left[ \begin{matrix}
1 & -1 & 11 & 4 & 0 \\
0 & 1 & -19 & -7 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 66 & 22 & 0 \\
\end{matrix} \right]$
In third row, after back substitution,
$0w+0x+0y+0z=0$
This means the system has infinitely many solutions.
Express $ w,x,y\text{ and }z $ in terms of z:
From fourth row:
$\begin{align}
& 66y+22z=0 \\
& 3y+z=0 \\
& y=-\frac{1}{3}z
\end{align}$
From second row we get;
$\begin{align}
& x-19y-7z=0 \\
& x-19\left( -\frac{1}{3}z \right)-7z=0 \\
& x-\frac{2}{3}z=0 \\
& x=\frac{2}{3}z
\end{align}$
From first row we get:
$\begin{align}
& w-x+11y+4z=0 \\
& w-\frac{2}{3}z+11\left( -\frac{1}{3}z \right)+4z=0 \\
& w-\frac{1}{3}z=0 \\
& w=\frac{1}{3}z
\end{align}$
Let $ z=\text{ any variable say, }t $; the solution to the system is $\left\{ \left( \frac{1}{3}t,\frac{2}{3}t,-\frac{1}{3}t,t \right) \right\}$.
Thus, the solution to the system is $\left\{ \left( \frac{1}{3}t,\frac{2}{3}t,-\frac{1}{3}t,t \right) \right\}$.
