I am working on some problems for practice and I have come across a type of problem I hadn't seen before. The question asks to find the values of $t$ and $s$, such that the system has (unique, infinite, no..) solutions. I was used to these questions when it was only giving one value i.e. just $t$ or just $s$, but i'll post below my work and the question.
The system is as follows,
$$tx+6y=2$$
$$x+(t+1)y=2s$$
I put that into the following matrix,
$$ \begin{pmatrix} t & 6 & 2\\ 1 &t+1 & 2s \end{pmatrix} $$
Then I did some row operations to obtain the following matrix,
$$ \begin{pmatrix} 1 & t+1& 2s \\ 0 & 6-t^2-t & 2-2st \end{pmatrix}$$
So I can see that if $6-t^2-t=0$ (i.e., $t=-3$ or $t=2$) and $2-2st=0$ then the system will have infinite solutions and if 2-2st $\neq$ 0, then there are no solutions. I.e. if $t=-3$ and $s=\frac{-1}3$, infinite solutions and similar if $t= 2$ and $s= \frac 12$ no solution.
First question is, is this a correct approach to this problem? Am i missing anything important and is there a more structured or organized way to do this?
Thanks all