As every MO user knows, and can easily prove, the inverse of the matrix $\begin{pmatrix} a & b \\\ c & d \end{pmatrix}$ is $\dfrac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. This can be proved, for example, by writing the inverse as $ \begin{pmatrix} r & s \\ t & u \end{pmatrix}$ and solving the resulting system of four equations in four variables.
As a grad student, when studying the theory of modular forms, I repeatedly forgot this formula (do you switch the $a$ and $d$ and invert the sign of $b$ and $c$ … or was it the other way around?) and continually had to rederive it. Much later, it occurred to me that it was better to remember the formula was obvious in a couple of special cases such as $\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$, and diagonal matrices, for which the geometric intuition is simple. One can also remember this as a special case of the adjugate matrix.
Is there some way to just write down $\dfrac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$, even in the case where $ad - bc = 1$, by pure thought—without having to compute? In particular, is there some geometric intuition, in terms of a linear transformation on a two-dimensional vector space, that renders this fact crystal clear?
Or may as well I be asking how to remember why $43 \times 87$ is equal to $3741$ and not $3731$?

Think about $\left({\phantom-d\phantom--b\atop-c\phantom{--}a}\right)$
as $tI - A$ where $t=a+d$ is the trace of $A$.
Since $A$ satisfies its own characteristic equation
(Cayley-Hamilton),
we have $A^2 - t A + \Delta \cdot I = 0$
where $\Delta = ad-bc$ is the determinant.  Thus $\Delta \cdot I = t A - A^2$.
Now divide both sides by $\Delta \cdot A$ to get $A^{-1} = \Delta^{-1}(tI-A)$,
QED.