system of equation with 3 unknown Solve $$\begin{matrix}i \\ ii \\ iii\end{matrix}\left\{\begin{matrix}x-y-az=1\\ -2x+2y-z=2\\ 2x+2y+bz=-2\end{matrix}\right.$$
For which $a$ does the equation have


*

*no solution

*one solution

*$\infty$ solutions


I did one problem like this and got a fantastic solution from @amzoti. Now, I think that if I see another example, I will really get it. 

EDIT 
Here is my attempt with rref and here with equations
Problems


*

*I don't know how to handle the $b$ in the end. 

*Does it ever lead to, speaking in "matrix terms", the case 0 0 0 | 0 so that I'll have $\infty$ number of solutions?


use Cramer's rule 
$$x=\frac{(2b+2)-8(a)}{4(-2+2a)}$$
so 
simplify $x=\frac{b+1-8a}{-4+4a}$
if $-4+4a=0 \Rightarrow a=1$  there is no answer for the system of equation
