<s>[INST] Find all complex values for $a$ where there's no solution for the non homogeneous system 
Find all complex values for $a$ where there's no solution for the non homogeneous system

$$\left\{ {\begin{array}{*{20}{c}}
   {x + (1 - a)y + z = 1}  \\
   {2x - y + z = 3}  \\
   {3x - ay + (a - 1)z = 4}  \\
\end{array}} \right.$$
After moving to a matrix representation and reduction, I have:
$$\left( {\begin{array}{*{20}{c}}
   1 & {1 - a} & 1 & | & 1  \\
   0 & {2a - 3} & { - 1} & | & 1  \\
   0 & 0 & {a - 1} & | & 0  \\
\end{array}} \right)$$
Now, I don't see an $a$ that answering this request, and I double-checked my reduction process. 
EDIT:
followed by @DonAntonio suggestion:
defining $a= x + yi$
Developing the determinant by first column:  
$$\begin{array}{l}
 \left| {\begin{array}{*{20}{c}}
   {2a - 3} & { - 1}  \\
   0 & {a - 1}  \\
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
   {2(x + yi) - 3} & { - 1}  \\
   0 & {x + yi - 1}  \\
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
   {2x - 3 + 2yi} & { - 1}  \\
   0 & {x - 1 + yi}  \\
\end{array}} \right| =  \\ 
 (2x - 3 + 2yi) \cdot (x - 1 + yi) = 2{x^2} - 2x + 2xyi - 3x + 3 - 3yi + 2xyi - 2yi + 2{y^2}{i^2} \\ 
  = 2{x^2} - 5x - 5yi + 4xyi - 2{y^2} \\ 
 \end{array}$$
Comparing the result to $0$ we get:  
$$\left\{ {\begin{array}{*{20}{c}}
   {2{x^2} - 5x - 2{y^2} = 0}  \\
   { - 5y + 4xy = 0}  \\
\end{array}} \right.$$
Am I doing it right?
 [/INST] If you rref you get this 
$ [1 \ 0 \ 0 \ \frac{3a - 4}{2a - 3} \\ \ 0 \ 1 \ 0 \ \frac{1}{2a - 3} \\ \ 0 \ 0 \ 1 \ \ \ \ \ \  0 \ \ ] $
So $$ 2a - 3 \ne 0 $$
$$ a \ne \frac{3}{2}$$
Now you still need to check a = 1 because if your calculations are correct , (I didn't check them) to get a 1 in the last row you have to divide by $a - 1$. If a = 1 you are dividing by 0
I checked a = 1 and it works. Putting a = 1 leads to the system
$[1 \ 0 \ 0 \ 1 \\ 0 \ 1 \ 0 -1 \\ 0 \ 0 \ 1 \ 0 \ ]$
And this works by substituting a = 1 , x = 1 , y = -1 z = 0 in the OP.
UPDATE: I got it to this
$[1 \ (1-a) \  \ 1 \ 1 \\ 0 \ (2a - 3) \ \ -1 \ \ 1 \\ 0 \ \ \ 0 \ (a-3) \ \ 0 \ \ ] $
So even if a = 3 it will not produce 'no solution' because of 0 in last entry of row 3 , we would then have 0/0 which leads to a system with a row of 0's
I invite you to check a = 3 in the OP.
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