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 $3731$?
Best Answer
EDIT (8/14/2020): A couple people have suggested that this answer should come with a warning -- this is a pretty fancy approach to an elementary question, motivated by the fact that I know the OP's interests. Some of the other answers below are probably better if you just want to invert some matrices :). I've also fixed a couple of minor typos.
My favorite way to remember this is to think of $SL_2(\mathbb{R})$ as a circle bundle over the upper half-plane, where $SL_2(\mathbb{R})$ acts on the upper half-plane via fractional linear transformations; then the map sends an element of $SL_2(\mathbb{R})$ to the image of $i$ under the corresponding fractional linear transformation. The fiber over a point is the corresponding coset of the stabilizer of $i$.
This naturally gives the Iwasawa decomposition of $SL_2(\mathbb{R})$ as $$SL_2(\mathbb{R})=NAK$$ where
$$K=\left\{\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} , ~0\leq\theta<2\pi \right\}$$
$$A=\left\{\begin{pmatrix} r & 0\\ 0 &1/r\end{pmatrix},~ r\in \mathbb{R}\setminus\{0\}\right\}$$
$$N=\left\{\begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix},~ x\in \mathbb{R}\right\}$$
Here $K$ is the stabilizer of $i$ in the upper half-plane picture; viewed as acting on the plane via the usual action of $SL_2(\mathbb{R})$ on $\mathbb{R}^2$ it is just rotation by $\theta$ (and likewise if we view the upper half plane as the unit disk, sending $i$ to $0$ via a fractional linear transformation). $A$ is just scaling by $r^2$, in the upper half-plane picture, and is stretching in the $\mathbb{R}^2$ picture. $N$ is translation by $x$ in the upper half-plane picture, and is a skew transformation in the $\mathbb{R}^2$ picture.
In each case, the inverse is geometrically obvious: for $K$, replace $\theta$ with $-\theta$; for $A$ replace $r$ with $1/r$, and for $N$, replace $x$ with $-x$. Since $$SL_2(\mathbb{R})=NAK$$ this lets us invert every $2\times 2$ matrix by "pure thought", at least if you remember the Iwasawa decomposition (which is easy from the geometric picture, I think). Of course this easily extends to $GL_2$; if $A$ has determinant $d$, then $A^{-1}$ had better have determinant $d^{-1}$.
If you'd like to derive the formula you've written down by "pure thought" it suffices to look at any one of these cases if you remember the general form of the inverse; or you can simply put them all together to give a rigorous derivation.