Proof that if a system of equations has a unique solution, then $ad - bc \neq 0$. I'm trying to understand a proof that if a system of equations
\begin{align*}
ax + by &= r \\ 
cx + dy &= s
\end{align*}
has a unique solution, then $ad - bc \neq 0$. The proof, paraphrased, looks like this:

Let $(x', y')$ be a solution to this system. Then
\begin{align*}
ax' + by' &= r \\ 
cx' + dy' &= s.
\end{align*}
So $(x' + 1, y')$ is not a solution, so either
\begin{align*}
a(x' + 1) + by' & = ax' + a + by' \neq r \\
& \text{ or } \\
c(x' + 1) + dy' &= cx' + c + dy \neq s.
\end{align*}
If $a = c = 0$, then these would be satisfy by the solution $(x' + 1, y)$, so we must have either $a \neq 0$ or $c \neq 0$. Without loss of generality, we can assume $a \neq 0$ by swapping the two equations and relabelling, if necessary. Then $(x' - b, y' + a)$ is not a solution either since $a \neq 0$ so $y' + a \neq y'$. But it satisfies the first equation:
$$
a(x' - b) + b(y' + a) = ax' - ab + by' + ab = ax' + by' = r.
$$
So we must have $c(x' - b) + d(y' + a) \neq s$. But
\begin{align*}
c(x' - b) + d(y' + a) &= cx' - bc + dy' + ad \\
& = (cx' + dy') + (ad - bc) \\
& = s + (ad - bc). 
\end{align*}
So $s + (ad - bc) \neq s$, so $ad - bc \neq 0$.

I'm able to follow every step of the proof, but I'm trying to understand how I would come up with it. I would appreciate any tips on the intuition or even alternative proofs that may be more intuitive.
 A: For a unique solution we need to have two distinct equations (distinct intersecting, parallel lines)for this to happen we require unequal slopes
$$\frac{a}{c} \ne \frac{b}{d} \implies ad-bc \ne 0.$$
Also by Cramer's rule or otherwise we have
$$x=\frac {ar-bs}{ad-bc}, y=\frac{as-cr}{ad-bc}$$ for $x,y$ to be finite we want $ad-bc\ne 0$.
A: For a geometric intuition, each equation represents a line and lines intersect exactly at one point (i.e. system has exactly one solution) if and only if slopes (i.e. angular coefficients) are different that is for $bd\neq 0$
$$m_1=-\frac{a}{b} \ne -\frac{c}{d}=m_2$$
which  implies the given expression. The special cases can be checked by inspection.


A: This method seems unnecessarily complex. Your equations imply:
\begin{align*}
adx + bdy &= rd \\ 
bcx + bdy &= bs
\end{align*}
Subtracting:
$$(ad-bc)x=rd-bs$$
If $ad-bc=0$ then we will have either no solutions for $x$ or infinitely many solutions. Therefore unique solutions imply $ad-bc \ne 0$. Similarly for $y$.
