Answer
Dependent;
$x=2-3t$
$y=3-5t$
$z=t$
Work Step by Step
$ \begin{bmatrix}
2 & -3 & -9 & -5\\
1 & 0 & 3 & 2\\
-3 & 1 & -4 & -3
\end{bmatrix} $
Interchange the 1st and the 2nd row.
$ \begin{bmatrix}
1 & 0 & 3 & 2\\
2 & -3 & -9 & -5\\
-3 & 1 & -4 & -3
\end{bmatrix} $
Add -2 times the 1st row to the 2nd row to produce a new 2nd row.
Add 3 times the 1st row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1 & 0 & 3 & 2\\
0 & -3 & -15 & -9\\
0 & 1 & 5 & 3
\end{bmatrix} $
Interchange the 2nd row and the 3rd row.
$ \begin{bmatrix}
1 & 0 & 3 & 2\\
0 & 1 & 5 & 3\\
0 & -3 & -15 & -9
\end{bmatrix} $
Add 3 times the 2nd row to the 3rd row.
$ \begin{bmatrix}
1 & 0 & 3 & 2\\
0 & 1 & 5 & 3\\
0 & 0 & 0 & 0
\end{bmatrix} $ $\rightarrow$ Dependent system
From the last row we got the equation $0=0$ which is always true and therefore the system has infinitely many solutions but x,y and z cannot be any random numbers. To describe the complete solution of the system of equations, let $z=t$ where t is any real number.
Use back-substitution to find the solution.
$z=t$
$y+5z=3\rightarrow y=3-5t$
$x+3y=2 \rightarrow x=2-3t$
