Answer
$(1,-2,3)$
Work Step by Step
To solve the system $\begin{cases} \hspace{5mm} 2y-z=-7 \\ x+4y+z=-4 \\ 5x-y+2z=13 \\ \end{cases},$ we perform elementary row operations on the corresponding augmented matrix to obtain an equivalent matrix with $1s$ along the main diagonal (if possible).
The corresponding augmented matrix is
$$\left[
\begin{array}{ccc|c}
0 & 2 & -1 & -7 \\
1 & 4 & 1 & -4\\
5 & -1 & 2 & 13\\
\end{array}
\right].$$
First, we switch Row_1 and Row_2 to obtain the equivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 4 & 1 & -4 \\
0 & 2 & -1 & -7\\
5 & -1 & 2 & 13\\
\end{array}
\right].$$
Next, we replace Row_3 with Row_3-5Row_1 to obtain the equaivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 4 & 1 & -4 \\
0 & 2 & -1 & -7\\
0 & -21 & -3 & 33\\
\end{array}
\right].$$
Thirdly, we multiply Row_2 by $\dfrac{1}{2}$ and multiply Row_3 by $-\dfrac{1}{3}$ to obtain the equaivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 4 & 1 & -4 \\
0 & 1 & -\frac{1}{2} & -\frac{7}{2}\\
0 & 7 & 1 & 11\\
\end{array}
\right].$$
Fourthly, we replace Row_3 with Row_3-7Row_2 to obtain the equaivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 4 & 1 & -4 \\
0 & 1 & -\frac{1}{2} & -\frac{7}{2}\\
0 & 0 & \frac{9}{2} & \frac{27}{2}\\
\end{array}
\right].$$
Fifthly, we multiply Row_3 by $\dfrac{2}{9}$ to obtain the equivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 4 & 1 & -4 \\
0 & 1 & -\frac{1}{2} & -\frac{7}{2}\\
0 & 0 & 1 & 3\\
\end{array}
\right].$$
Now, we form the system of equations corresponding to this matrix:
$$\begin{cases} x+4y+z=-4 \\ \hspace{5mm} y-\dfrac{1}{2}z=-\dfrac{7}{2} \\ \hspace{15mm} z=3 \\ \end{cases}.$$
We see $z=3$. To find $y$, we plug $z=3$ into our second equation and solve for $y$:
$$y-\dfrac{1}{2}(3)=-\dfrac{7}{2} \\ y-\dfrac{3}{2}=-\dfrac{7}{2} \\ y=-\dfrac{4}{2} \\ y=-2.$$
To find $x$, we plug $y=-2$ and $z=3$ into the first equation and solve for $x$:
$$x+4(-2)+3=-4 \\ x-8+3=-4 \\ x-5=-4 \\ x=1.$$
Thus we have exactly one solution, which is $(1,-2,3).$