Answer
The system of equations $\begin{align}
& 2x+4y+5z=2 \\
& x+y+2z=1 \\
& 3x+5y+7z=4
\end{align}$ has no solution.
Work Step by Step
Consider the following system of equations:
$\begin{align}
& 2x+4y+5z=2 \\
& x+y+2z=1 \\
& 3x+5y+7z=4
\end{align}$
Express the above system in the form of a matrix as follows:
$\left[ \begin{matrix}
2 & 4 & 5 & 2 \\
1 & 1 & 2 & 1 \\
3 & 5 & 7 & 4 \\
\end{matrix} \right]$
Using the elementary row operations we will obtain the echelon form of a matrix:
${{R}_{1}}\to \frac{1}{2}{{R}_{1}}$ gives,
$\left[ \begin{matrix}
1 & 2 & \frac{5}{2} & 1 \\
1 & 1 & 2 & 1 \\
3 & 5 & 7 & 4 \\
\end{matrix} \right]$
${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},\text{ }{{\text{R}}_{3}}\to {{R}_{3}}+\left( -3 \right){{R}_{1}}$ gives,
$\left[ \begin{matrix}
1 & 2 & \frac{5}{2} & 1 \\
0 & -1 & -\frac{1}{2} & 0 \\
0 & -1 & -\frac{1}{2} & 1 \\
\end{matrix} \right]$
${{R}_{2}}\to \frac{1}{-1}{{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & 2 & \frac{5}{2} & 1 \\
0 & 1 & \frac{1}{2} & 0 \\
0 & -1 & -\frac{1}{2} & 1 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+1{{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & 2 & \frac{5}{2} & 1 \\
0 & -1 & -\frac{1}{2} & 0 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]$
Since, $0{\lt}1$, this implies that the system is inconsistent, and has no solution.
