#### Answer

There is only one solution.

#### Work Step by Step

In this matrix, we will convert all the rows into a linear equation and solve them.
Translating every row of the matrix into equation form we get, $\begin{align}
x+y+2z=19 & \\
y+2z=13 & \\
z=5 & \\
\end{align}$
Substitute, $ z=5$ in $ y+2z=13$ to get, $\begin{align}
& y+2\times 5=13 \\
& y+10=13 \\
& y=13-10 \\
& y=3
\end{align}$
Substitute, $ z=5$, $ y=3$ in $ x+y+2z=19$ to get, $\begin{align}
& x+3+2\times 5=19 \\
& x+3+10=19 \\
& x=19-13 \\
& x=6
\end{align}$
So, the values are:
$\begin{align}
& x=6 \\
& y=3 \\
& z=5
\end{align}$
Hence, the system has one solution.