Answer
The solution for the given system of equations is $\left\{ \left( -2z+5,\ -2+z,\ z \right) \right\}$.
Work Step by Step
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix}
8 & 5 & 11 \\
-1 & 4 & 2 \\
2 & -1 & 5 \\
\end{matrix} \right|\begin{matrix}
30 \\
3 \\
12 \\
\end{matrix} \right]$
Solve this matrix as below:.
$\left[ \left. \begin{matrix}
8 & 5 & 11 \\
-1 & 4 & 2 \\
0 & -9 & 9 \\
\end{matrix} \right|\begin{matrix}
30 \\
3 \\
18 \\
\end{matrix} \right]$ $ By,\ 4{{R}_{3}}-{{R}_{1}}\to {{R}_{3}}$
$\left[ \left. \begin{matrix}
8 & 5 & 11 \\
0 & -27 & 27 \\
0 & -9 & 9 \\
\end{matrix} \right|\begin{matrix}
30 \\
54 \\
18 \\
\end{matrix} \right]$ $ By,\ 8{{R}_{2}}+{{R}_{1}}\to {{R}_{2}}$
$\left[ \left. \begin{matrix}
8 & 5 & 11 \\
0 & -27 & 27 \\
0 & 0 & 0 \\
\end{matrix} \right|\begin{matrix}
30 \\
54 \\
0 \\
\end{matrix} \right]$ $ By,\ {{R}_{2}}-3{{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 we will drop it.
$\left[ \left. \begin{align}
& \begin{matrix}
8 & 5 & 11 \\
\end{matrix} \\
& \begin{matrix}
0 & -27 & 27 \\
\end{matrix} \\
\end{align} \right|\begin{matrix}
30 \\
54 \\
\end{matrix} \right]$
Convert this system into linear equations.
$\begin{align}
& 8x+5y+11z=30 \\
& -27y+27z=54 \\
\end{align}$
Now write the value of x and y in terms of z and solving the equation we get, $\begin{align}
& x=5-2z \\
& y=z-2 \\
\end{align}$
Hence, the solution for the given system of equations is $\left\{ \left( -2z+5,\ -2+z,\ z \right) \right\}$.
