#### Answer

The solution set of an inconsistent system of equations is $\phi $.

#### Work Step by Step

A system in which there is no particular solution set for the variables is called an inconsistent system.
So, in the Gaussian elimination method, a system is known to be inconsistent if there are 0's left of the line and a nonzero value to the right.
Example:
$\begin{align}
& 3x+y+3z=14 \\
& 7x+5y+8z=32 \\
& x+3y+2z=9\,
\end{align}$
The matrix corresponding to the system of equations is as follows:
$\left[ \begin{matrix}
3 & 1 & 3 & 14 \\
7 & 5 & 3 & 32 \\
1 & 3 & 2 & 9 \\
\end{matrix} \right]$
To obtain the desired results, using elementary row transformation, we will find the echelon form of the matrix.
Apply ${{R}_{1}}\leftrightarrow {{R}_{3}}$:
$\left[ \begin{matrix}
1 & 3 & 2 & 9 \\
7 & 5 & 8 & 32 \\
3 & 1 & 3 & 9 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$ and ${{R}_{2}}\to {{R}_{2}}-7{{R}_{1}}$:
$\left[ \begin{matrix}
1 & 3 & 2 & 9 \\
0 & -16 & -6 & -31 \\
0 & -8 & -3 & -13 \\
\end{matrix} \right]$
Apply ${{R}_{3}}\to {{R}_{3}}-\frac{1}{2}{{R}_{2}}$:
$\left[ \begin{matrix}
1 & 3 & 2 & 9 \\
0 & -16 & -6 & -31 \\
0 & 0 & 0 & \frac{5}{2} \\
\end{matrix} \right]$
Apply ${{R}_{2}}\to -\frac{1}{16}{{R}_{2}}$:
$\left[ \begin{matrix}
1 & 3 & 2 & 9 \\
0 & 1 & \frac{3}{8} & \frac{31}{16} \\
0 & 0 & 0 & \frac{5}{2} \\
\end{matrix} \right]$
From the above matrix:
$0x+0y+0z=\frac{5}{2}$, which is not possible since for any values of $x,\ y$, and $z$, $0\ne \frac{5}{2}$.
Thus, there is no solution to the given problem. Therefore, the solution set of an inconsistent system of equations is $\phi $.