Answer
The solution to the linear system is $\left( \left\{ 15,-12,-4 \right\} \right)$.
Work Step by Step
The matrix corresponding to the system of equations is given as follows:
$\left[ \begin{matrix}
-1 & -1 & -1 & 1 \\
4 & 5 & 0 & 0 \\
0 & 1 & -3 & 0 \\
\end{matrix} \right]$
Using the elementary row transformation, we will find the echelon form of the matrix.
Apply the row operation ${{R}_{1}}\to \left( -1 \right){{R}_{1}}$:
$\left[ \begin{matrix}
1 & 1 & 1 & -1 \\
4 & 5 & 0 & 0 \\
0 & 1 & -3 & 0 \\
\end{matrix} \right]$
Apply the row operation ${{R}_{2}}\to {{R}_{2}}-4{{R}_{1}}$:
$\left[ \begin{matrix}
1 & 1 & 1 & -1 \\
0 & 1 & -4 & 4 \\
0 & 1 & -3 & 0 \\
\end{matrix} \right]$
$\begin{align}
& {{R}_{1}}\to {{R}_{1}}-{{R}_{2}} \\
& {{R}_{3}}\to {{R}_{3}}-{{R}_{2}} \\
\end{align}$
$\left[ \begin{matrix}
1 & 0 & 5 & -5 \\
0 & 1 & -4 & 4 \\
0 & 0 & 1 & -4 \\
\end{matrix} \right]$
$\begin{align}
& {{R}_{1}}\to {{R}_{1}}-5{{R}_{3}} \\
& {{R}_{2}}\to {{R}_{4}}+4{{R}_{3}} \\
\end{align}$
$\left[ \begin{matrix}
1 & 0 & 0 & 15 \\
0 & 1 & 0 & -12 \\
0 & 0 & 1 & -4 \\
\end{matrix} \right]$
Thus, the solution set for the system of linear equations is $\left( \left\{ 15,-12,-4 \right\} \right)$.
