Exercise: Prove that if $ad-bc \neq 0$, then the system
$$
ax + by = j \\
cx + dy = k
$$
has an unique solution.
This is from the very first subsection of Hefferon's Linear Algebra, and also the very first time I'm getting into rigorous mathematics. Now I am convinced that this could be shown from the defining property of the determinants, but that's way too far from my level.
My reasoning is as follows:
Solving the first equation for $x$ yields $x=(j-by)/a$. Substituting this expression in the second equation we get $c(j-by)/a+dy=k$. Solving this equation for $y$ gives $(ad-bc)y=ak-cj$. Since $ad-bc \neq 0$, it is safe to say that $y=(ak-cj)/(ad-bc)$. Back-substituting $y$ into the expression for $x$, we also get an unique value for $x$. Therefore $y=(ak-cj)/(ad-bc), x=(j-by)/a$ is the unique solution. QED?
However, I think my proof breaks down at some point because if $a=0$ then we cannot take the first step to solve the equation for $x$. The only part I asserted that we are not dividing by zero, is the part that I assumed $ad-bc \neq 0$.
I would like to take hints to make my proof watertight. I have peeked into the solutions manual and have noted that the author seperates the proof into seperate cases. I would like to know which cases I should consider and why. I would also like to know if it is provable without doing so.
Best Answer
In your first step, you assumed that $a$ in non zero, that's a problem. But your idea is not very far from the atual way. Let's multiply the first euation by $d$ and the second by $b$ to get $adx+dby=dj$ and $cbx+dby=bk$. Now we get $dj-adx=bk-cbx$ as both are equal to $dby$ so $dj-bk=(ad-bc)x$, and now you can divide by $ad-bc$.