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?

 A: The complete matrix of your system is
$$
\begin{bmatrix}
1 & -1 & -a & 1 \\
-2 & 2 & -1 & 2\\
2 & 2 & b & -2
\end{bmatrix}
$$
and with Gaussian elimination you get
$$
\begin{bmatrix}
1 & -1 & -a & 1 \\
0 & 0 & -1-2a & 4\\
0 & 4 & b+2a & -4
\end{bmatrix}
$$
(sum to the second row the first multiplied by $2$ and to the third row the first multiplied by $-2$). Now swap the second and third rows:
$$
\begin{bmatrix}
1 & -1 & -a & 1 \\
0 & 4 & b+2a & -4 \\
0 & 0 & -1-2a & 4
\end{bmatrix}
$$
You see that you have solutions if and only if $-1-2a\ne0$. Otherwise the last equation would become $0=4$ that obviously has no solution.
The solution is unique for $a\ne-1/2$ and does not exist for $a=-1/2$.
A: 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
A: Recalling Cramer's rule, we have 

$$ \Delta = 6a - b, \quad  \Delta_x = 5b+12, \quad \Delta_y=\dots,\quad \Delta_z=\dots\,.$$

where $\Delta$ is the determinant of the coefficient matrix. Now,
i) if $ \Delta \neq 0 $, then the system has a unique solution.
ii) if $ \Delta = 0 $ and $ \Delta_x,\Delta_y,\Delta_z \neq 0 $, then the system has infinite number of soltions.
iii) if $ \Delta = 0 $ and $ \Delta_x=0$  or $\Delta_y=0$ or $\Delta_z = 0 $, then the system has no solution.
Note: Check the augumented matrix.
