Solve the system of linear equations for all values of a When solving the following matrix for $a$:
$$ \left[ 
\begin{array}{ccc|c}
  x&2y&az&3\\
  (2a-4)x&(2-a)y&(-2a+4)z&0\\
  4x&4y&2az&8
\end{array}
\right]
$$
When do I put restrictions on which values $a$ can take? Say I divide the second row by $(2-a)$, does this mean that $a$ cannot be 2, or can I only say something like that when the entire matrix is in row echelon form, and I would be left with something like:
$$ \left[ 
\begin{array}{ccc|c}
  1&0&0&\frac{2a}{a-2}\\
  0&1&0&\frac{4-5a}{a-2}\\
  0&0&1&\frac{a^2}{a-2}
\end{array}
\right]
$$

I didn't check the detail of calculation but, if it is correct, the solution you have found is valid for $a\neq 2$ (i.e. the system has exactly a solution $\forall a\neq2$). 
Now you need also to verify the solution/s for the case $a=2$ in the original system, that is
$$\left[ 
\begin{array}{ccc|c}
  1&2&2&3\\
  4&4&4&8
\end{array}
\right]\implies \left[ 
\begin{array}{ccc|c}
  1&2&2&3\\
  1&1&1&2
\end{array}
\right]\implies \left[ 
\begin{array}{ccc|c}
  1&2&2&3\\
  0&1&1&1
\end{array}
\right]\implies\left[ 
\begin{array}{ccc|c}
  1&0&0&1\\
  0&1&1&1
\end{array}
\right]$$
thus for a=2 the system has infinitely many solutions


*

*$x=1$

*$y+z=1$

