Show that if $ad-bc \ne 0$ then a system of equations has a unique solution. Know answer. Trouble understanding it. I am starting through the free Hefferon Linear Algebra book. I am stuck on the problem below. What I don't get is the "We take 3 cases" part right at the beginning. Why those 3 cases? What significance do those 3 cases have? What are they related to? Why are we just plugging in zero and non-zero, and not say every kind of number? This is my first real exposure to proofs and it shows. Any help with this is greatly appreciated!
Problem
Solution
 A: We can actually avoid all this splitting into cases. If we multiply the first equation by $d$ and the second by $b$, the equations become
$$\begin{cases}
  ad\,x+bd\,y=d\,j\\
  \kern3.8mu bc\,x+bd\,y=b\,k.
\end{cases}$$
Subtracting one from the other yields
$$(ad-bc)x = dj-bk\implies x=\frac{dj-bk}{ad-bc},$$
where you'll notice that I could divide by $ad-bc$ precisely since it wasn't zero.
By precisely similar reasoning (instead multiplying the first by $c$ and the second by $a$), we can subtract them and find the explicit unique solution $y=(cj-ak)/(ad-bc)$ for $y$.
On the other hand, if $ad-bc$ is zero, then the equation we got for $x$ above is just $0x=dj-bk$. This either has no solutions (if $dj-bk$ is non-zero, we want to find $x$ such that $0x=\text{a non-zero number}$, which is impossible), or infinitely many (if $dj-bk$ is zero, then the equation is $0x=0$, which is true for any $x$), and the exact same goes for the equation we would get for $y$.

With regards to all the casework, there is no reason why doing it that way is fundamentally important.  The simple fact is that if you want to divide by $a$, you need to assume that $a\neq 0$, and then think about the specific case when $a=0$ separately afterwards. Similarly for $c$. This is why the author decided to split into cases. But as I showed above, it isn't necessary to prove what you want. (Also as saulspatz pointed out, you usually realise you need to split in to cases half way through, not at the beginning.)
