Answer
The solution of the given linear system of equations is $\left\{ \left( 0,2,2 \right) \right\}$.
Work Step by Step
A system of linear equations can be solved with the help of matrices. One of the methods (Gaussian elimination) involves the augmented matrix.
The augment matrix is,
$\left[ \begin{matrix}
3 & 1 & -1 & 0 \\
1 & 1 & 2 & 6 \\
2 & 2 & 3 & 10 \\
\end{matrix} \right]$
The Gaussian elimination method is used to solve the system of linear equations by turning the augmented matrix into row echelon form. Perform the operations as follows:
Replace ${{R}_{1}}\,\text{by}\,{{R}_{1}}-{{R}_{3}}$ to get,
$\left[ \begin{matrix}
1 & -1 & -4 & -10 \\
1 & 1 & 2 & 6 \\
2 & 2 & 3 & 10 \\
\end{matrix} \right]$
Replace ${{R}_{2}}\,\text{by}\,{{R}_{2}}-{{R}_{1}}$ to get,
$\left[ \begin{matrix}
1 & -1 & -4 & -10 \\
0 & 2 & 6 & 16 \\
2 & 2 & 3 & 10 \\
\end{matrix} \right]$
Replace ${{R}_{2}}\,\text{by}\,\frac{1}{2}{{R}_{2}}$ to get,
$\left[ \begin{matrix}
1 & -1 & -4 & -10 \\
0 & 1 & 3 & 8 \\
2 & 2 & 3 & 10 \\
\end{matrix} \right]$
Replace ${{R}_{3}}\,\text{by}\,{{R}_{3}}-2{{R}_{1}}$ to get,
$\left[ \begin{matrix}
1 & -1 & -4 & -10 \\
0 & 1 & 3 & 8 \\
0 & 4 & 11 & 30 \\
\end{matrix} \right]$
Replace ${{R}_{3}}\,\text{by}\,{{R}_{3}}-4{{R}_{2}}$ to get,
$\left[ \begin{matrix}
1 & -1 & -4 & -10 \\
0 & 1 & 3 & 8 \\
0 & 0 & -1 & -2 \\
\end{matrix} \right]$
Replace ${{R}_{3}}\,\text{by}\,-{{R}_{3}}$ to get,
$\left[ \begin{matrix}
1 & -1 & -4 & -10 \\
0 & 1 & 3 & 8 \\
0 & 0 & 1 & 2 \\
\end{matrix} \right]$
The system of equations for the above augmented matrix is given by:
$x-y-4z=-10$ ……. (I)
$y+3z=8$ …… (II)
$z=2$ …… (III)
Substitute the value of $z$ in equation(II) to get,
$\begin{align}
& y+6=8 \\
& y=2
\end{align}$
Substitute value of $y,z$ in (I) to get,
$\begin{align}
& x-2-8=-10 \\
& x=0
\end{align}$
Hence, the solution of the given linear system of equations is $\left\{ \left( 0,2,2 \right) \right\}$.
