For What $a$ The Linear Equations Have Sloutions $2x+ay-z=-2$

$x-3z=-3$

$x+2y+az=-1$
I have thought about reducing a matrix so in the end I will have an equation with $a$ then I can determine for which $a$ the are one solution/infinite solutions/no solution, but it does not seems to work
I have row reduced the matrix and got
$$\left(\begin{array}{ccc|c} 
1 & 2 & a & -1 \\
0 & -2 & 3-a & 4 \\
0 & a-4 & -a-1& -1 
\end{array}\right)$$  
the solutions are for one solution: $a=5$,$a\neq 2$  no solution for: $a=-5,a=2$

Substitute the 2nd equation in the other two to get ay+5z=4 and 2y+(a+3)z=2. Take 2 x first - a x second to get $z=\frac{2(a-4)}{(a+5)(a-2)}$. If $a\ne 2\text{ or} -5$ then this gives a unique solution for x,y,z. If a=2, then the two equations above become 2y+5z=4 and 2y+5z=2, which is impossible. Similarly, if a=-5 they become y-z=-4/5 and 1, which is impossible. 
