Why do we require $ps-qr \neq 0$ in order to be able to solve the simultaneous equations $c_1p + c_2q = \alpha, c_1r + c_2s = \beta$?

Question

Suppose we have the below system of (simultaneous) linear equations $$ c_1p + c_2q = \alpha \\ c_1r + c_2s = \beta $$ I know that, in order for this to be solvable, we must have $ps-qr \neq 0$. On an intuitive level, why is this? I am struggling to visulaize the practical reason.

I am aware that, to be solvable, the equations in a system such as this must be linearly dependent, and that to do this, to find the determinant of the matrix formed by the coefficients $p,q,r,s$. I am not interested in this kind of explaination. What I'm trying to do right now is visualise why the above is the case, without creating a rigorous proof.

Answer 1

This is because the condition $pq=rs$ means the l.h.s. of the equations are proportional, so that, if you want the system to be solvable, the r.h.s. also have to be proportional with the same coefficient of proportionality.

Answer 2

Taken as equations in $c_1$ and $c_2$ the two equations represent lines in the plane. Normally two lines will intersect in a point, but this is not true if they are parallel. In this case the two lines will either be the same line written differently or different parallel lines which never meet.

They will be parallel if their gradients are equal, which will happen if $\cfrac pq=\cfrac rs$. Clearing denominators and taking all the terms to the same side this becomes the condition $ps-qr=0$.

If $ps-qr\neq 0$ the lines are therefore not parallel and will intersect.