Answer
The solution for the provided system of equations is $\left\{ \left( z-2,\ z-2,\ z \right) \right\}$.
Work Step by Step
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix}
5 & -11 & 6 \\
-1 & 3 & -2 \\
3 & -5 & 2 \\
\end{matrix} \right|\begin{matrix}
12 \\
-4 \\
4 \\
\end{matrix} \right]$
Solve this matrix as below:
$\left[ \left. \begin{matrix}
5 & -11 & 6 \\
-1 & 3 & -2 \\
0 & 4 & -4 \\
\end{matrix} \right|\begin{matrix}
12 \\
-4 \\
-8 \\
\end{matrix} \right]$ $ By,\ 3{{R}_{2}}+{{R}_{3}}\to {{R}_{3}}$
$\left[ \left. \begin{matrix}
5 & -11 & 6 \\
0 & 4 & -4 \\
0 & 4 & -4 \\
\end{matrix} \right|\begin{matrix}
12 \\
-8 \\
-8 \\
\end{matrix} \right]$ $ By,\ 5{{R}_{2}}+{{R}_{1}}\to {{R}_{2}}$
$\left[ \left. \begin{matrix}
5 & -11 & 6 \\
0 & 4 & -4 \\
0 & 0 & 0 \\
\end{matrix} \right|\begin{matrix}
12 \\
-8 \\
0 \\
\end{matrix} \right]$ $ By,\ {{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$
Convert the last row into equation form:
$\begin{align}
0x+0y+0z=0 & \\
0=0 & \\
\end{align}$
The last row does not add any information about the variables; therefore drop it.
$\left[ \left. \begin{align}
& \begin{matrix}
5 & -11 & 6 \\
\end{matrix} \\
& \begin{matrix}
0 & 4 & -4 \\
\end{matrix} \\
\end{align} \right|\begin{matrix}
14 \\
-8 \\
\end{matrix} \right]$
Convert this system into linear equations.
$\begin{align}
5x-11y+6z=12 & \\
4y-4z=-8 & \\
\end{align}$
Write the value of x and y in terms of z and solve the equation, $\begin{align}
& x=z-2 \\
& y=z-2
\end{align}$
Hence the numbers satisfying the set of solutions $\left\{ \left( z-2,\ z-2,\ z \right) \right\}$
