Answer
The system of equations has no solution.
Work Step by Step
Consider the following system of equations:
$\begin{align}
& 2x-2y+2z=5 \\
& x-y+z=2 \\
& 2x+y-z=1
\end{align}$
The matrix corresponding to the system of equations is as follows:
$\left[ \begin{matrix}
2 & -2 & 2 & 5 \\
1 & -1 & 1 & 2 \\
2 & 1 & -1 & 1 \\
\end{matrix} \right]$
Using the elementary row transformation we will obtain the echelon form of the matrix.
${{R}_{1}}\to \frac{1}{2}{{R}_{1}}$ gives,
$\left[ \begin{matrix}
1 & -1 & 1 & \frac{5}{2} \\
1 & -1 & 1 & 2 \\
2 & 1 & -1 & 1 \\
\end{matrix} \right]$
${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},\text{ }{{\text{R}}_{3}}\to {{R}_{3}}+\left( -2 \right){{R}_{1}}$ gives,
$\left[ \begin{matrix}
1 & -1 & 1 & \frac{5}{2} \\
0 & 0 & 0 & -\frac{1}{2} \\
0 & 3 & -3 & -4 \\
\end{matrix} \right]$
${{R}_{2}}\leftrightarrow {{R}_{3}}$ gives,
$\left[ \begin{matrix}
1 & -1 & 1 & \frac{5}{2} \\
0 & 3 & -3 & -4 \\
0 & 0 & 0 & -\frac{1}{2} \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+1{{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & 2 & \frac{5}{2} & 1 \\
0 & -1 & -\frac{1}{2} & 0 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]$
Since, in the last row $0\ne 1$, this implies that the system is inconsistent, and has no solution.
