$begin{cases}x-ty = 1\(t-1)x -2y = 1end{cases} x,y,t in mathbb{R} $

So the corresponding matrix is

$begin{bmatrix}1 & -t & 1 \t-1 & -2 & 1 end{bmatrix}$

I know by equaling (I) and (II) and substituting (where one has to divide by $t-2$ and $-1-t$) that for $t=2$ and $t=-1$ the system has no (unique) solution.

Now I want to solve the system by gaussian elimination and get a general answer for x,y dependent on t.

Problem: If I want to eliminate an coefficient, I need to multiply (I) by $t-1$ where $tneq 1$ or (II) by $t$ where $tneq0$.

But this adds an “extra case” because multiplying by $0$ is not allowed and my general solutions will not work only for this specific choice of $t$.

Can somebody just tell me, what i need to do in order to achieve general solutions with no special case for t by gaussian elimination ? Or am I just getting something wrong ?

Thanks in advance !