Answer
The solution for the equations is $\left\{ \left( 2,-1,3 \right) \right\}$.
Work Step by Step
Consider the following equations,
$\begin{align}
& 2x-y-2z=-1 \\
& x-2y-z=1 \\
& x+y+z=1
\end{align}$
To solve the provided equations, we use the elementary row transformation until the augmented matrix will be converted into the identity matrix.
The augmented matrix of the provided equations will be,
$\left[ \begin{matrix}
2 & -1 & -2 & -1 \\
2 & -2 & -1 & 1 \\
1 & 1 & 1 & 4 \\
\end{matrix} \right]$
Use the row transformation $\frac{1}{2}{{R}_{1}}$ ,
$\left[ \begin{matrix}
1 & -\frac{1}{2} & -1 & -\frac{1}{2} \\
1 & -2 & -1 & 1 \\
1 & 1 & 1 & 4 \\
\end{matrix} \right]$
Use the row transformation $-{{R}_{1}}+{{R}_{2}}$ ,
$\left[ \begin{matrix}
1 & -\frac{1}{2} & -1 & -\frac{1}{2} \\
1 & -\frac{3}{2} & 0 & \frac{3}{2} \\
1 & 1 & 1 & 4 \\
\end{matrix} \right]$
Use the row transformation $-{{R}_{1}}+{{R}_{3}}$ ,
$\left[ \begin{matrix}
1 & -\frac{1}{2} & -1 & -\frac{1}{2} \\
0 & -\frac{3}{2} & 0 & \frac{3}{2} \\
0 & \frac{3}{2} & 2 & \frac{9}{2} \\
\end{matrix} \right]$
Use the row transformation ${{R}_{2}}+{{R}_{3}}$ ,
$\left[ \begin{matrix}
1 & -\frac{1}{2} & -1 & -\frac{1}{2} \\
0 & -\frac{3}{2} & 0 & \frac{3}{2} \\
0 & 0 & 2 & 6 \\
\end{matrix} \right]$
Use the row transformation $-\frac{2}{3}{{R}_{2}}$ ,
$\left[ \begin{matrix}
1 & -\frac{1}{2} & -1 & -\frac{1}{2} \\
0 & -\frac{3}{2} & 0 & \frac{3}{2} \\
0 & 0 & 2 & 6 \\
\end{matrix} \right]$
Use the row transformation $-\frac{2}{3}{{R}_{2}}$ ,
Use the row transformation $\frac{1}{2}{{R}_{3}}$ ,
$\left[ \begin{matrix}
1 & -\frac{1}{2} & -1 & -\frac{1}{2} \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 3 \\
\end{matrix} \right]$
Use the row transformation $\frac{1}{2}{{R}_{2}}+{{R}_{1}}$ ,
$\left[ \begin{matrix}
1 & 0 & -1 & -1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 3 \\
\end{matrix} \right]$
Use the row transformation ${{R}_{3}}+{{R}_{1}}$ ,
$\left[ \begin{matrix}
1 & 0 & 0 & 2 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 3 \\
\end{matrix} \right]$
Thus, it has been converted into the identity matrix.
Therefore, the solution is $\left\{ \left( 2,-1,3 \right) \right\}$.
