Should the formula for the inverse of a 2x2 matrix be obvious? 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 $3731$?
 A: 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.
A: Mnemonic: make the product diagonals the determinant, then scale.
The off diagonals are zero because the area of a parallelogram with planar edge vectors $c_1,c_2$ is the length of the scaled projection $|c_1 \cdot i c_2| = |c_2 \cdot i c_1|$, and the mnemonic sets row $r_k$ in the inverse to $(ic_{3 - k})^T$.
A: It is probably too old question to answer, but I couldn't resist.
Consider $M_2\mathbb R$=$\{a+b\iota :a,b\in \mathbb C\}$. Additional relations are present: $\iota^2=1$, $\iota b=\bar b\iota$, which make enough to multiply $2\times 2$ matrices as split quaternions. For reader convenience $\iota=\begin{pmatrix} 1 & 0 \\\ 0 & -1 \end{pmatrix}$.
Now adjugate matrix to $a+b\iota $ is $\bar a-b\iota$. Let's calculate $(a+b\iota)(\bar a-b\iota)=a\bar a-b\bar b+(-ab+ba)\iota=a\bar a-b\bar b$, because complex numbers multiplication is commutative.
The determinant of matrix $a+b\iota$ is $a\bar a-b\bar b$.
A: Recall that the adjugate $\text{adj}(A)$ of a square matrix is a matrix that satisfies 
$$A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = \det(A).$$
Like the determinant, the adjugate is multiplicative. Categorically, the reason the determinant is multiplicative is that it comes from a functor (the exterior power), so one might expect that the adjugate also comes from a functor, and indeed it does (the same functor!). 
More precisely, let $T : V \to V$ be a linear transformation on a finite-dimensional vector space with basis $e_1, ... e_n$. Then the adjugate of the matrix of $T$ with respect to the basis $e_i$ is the matrix of $\Lambda^{n-1}(T) : \Lambda^{n-1}(V) \to \Lambda^{n-1}(V)$ with respect to an appropriate "dual basis"
$$(-1)^{i-1} \bigwedge_{j \neq i} e_j$$
of $\Lambda^{n-1}(V)$ (it becomes an actual dual basis if you identify $\Lambda^n(V)$ with the underlying field $k$ by sending $e_1 \wedge ... \wedge e_n$ to $1$). The exterior product $V \times \Lambda^{n-1}(V) \to \Lambda^n(V)$ can then be identified with the dual pairing $V \times V^{\ast} \to k$, and the action of the exterior product on endomorphisms of $V$ and $\Lambda^{n-1}(V)$ can be identified with the composition of endomorphisms of $V$ (remembering that $\text{End}(V)$ is canonically isomorphic to $\text{End}(V^{\ast})$). This categorifies the above statement.
When $n = 2$, the dual basis is $e_2, - e_1$ but $\Lambda^1$ is the identity functor, and the formula follows. The geometric intuition comes from thinking about the exterior product in terms of oriented areas of parallelograms in $\mathbb{R}^2$. 
A: 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.
A: I remember the inverse by looking at the corresponding linear fractional transformation. It sends $\frac{-d}{c}$ to $\infty$ and $\infty$ to $\frac{a}{c}$, so the inverse had better reverse this; it follows that the $c$ should stay put and the $a$ and $d$ should switch, and so the $b$ and $c$ get negated.
A: This is essentially the same as Tobias Hagge's answer and Jonny Evans's comment, but I thought that writing it up in this way would make things clearer.
Think about the product
$$
\begin{bmatrix}
a & b\\\\ c & d
\end{bmatrix}
\begin{bmatrix}
? & ?\\\\ ? & ?
\end{bmatrix}
=\begin{bmatrix}
ad-bc & 0\\\\ 0 & ad-bc
\end{bmatrix}.
$$
Focus on the zero in position $(2,1)$ in the RHS. In order to get it with the row-by-column rule, the first column of the unknown matrix must be $\begin{bmatrix}d\\\\-c\end{bmatrix}$. 
(Well, apart from the sign --- you could still get it wrong. But you can check that it is correct by computing the $(1,1)$ entry of the product.)
Now focus on the other zero entry in position (1,2) of the RHS, and you'll see that the second column must be $\begin{bmatrix}-b\\\\a\end{bmatrix}$. Again, if you're confused about the sign, just check the $(2,2)$ entry.
A: $\bullet$ The sign switch is familiar from complex numbers:
The regular representation of $\mathbb{C}$ over $\mathbb{R}$ is the embedding of $\mathbb{R}$-algebras $\mathbb{C} \to M_2(\mathbb{R})$ defined by $a+ib \mapsto \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$. The inverse of $a+ib$ is the conjugate $a-ib$ divided by the norm $a^2+b^2$, thus the inverse of $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ is the adjugate $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ divided by the determinant $a^2+b^2$.
$\bullet$ Both the sign switch and the swap of the diagonal entries can be illustrated with quaternions:
The regular representations of $\mathbb{H}$ over $\mathbb{C}$ is the embedding $\mathbb{H} \to M_2(\mathbb{C})$ mapping $u+jv \mapsto \begin{pmatrix} u & v \\ - \overline{v} & \overline{u} \end{pmatrix}$. The inverse of $u + jv$ is the conjugate $\overline{u} - j \overline{v}$ divided by the norm $|u|^2+|v|^2$. Thus, the inverse of $\begin{pmatrix} u & v \\ - \overline{v} & \overline{u} \end{pmatrix}$ is the adjugate $\begin{pmatrix} \overline{u} & -v \\ \overline{v} & u \end{pmatrix}$ divided by the determinant $|u|^2+|v|^2$.
A: My answer is not very highfaluting, but it is what I use to remember. Switch the diagonals, change the signs of the off-diagonals and divide by the determinant. Since the inverse of a diagonal matrix is easy, the switch should be easy to remember. On the other hand such mnemonics are dangerous. The critical points of a cubic $Ax^3+Bx^2+Cx+D$ are at $\frac{-B\pm\sqrt{B^2-3AC}}{3A}$, or so I remember.
A: Let $ A $ be an $ n \times n $ invertible matrix. We're looking for an $ M $ such that $ AM = I $. Let's write columns of $ A $ as $ A_1, \ldots, A_n $ ( since $ A $ is invertible, these form a basis of $ \mathbb{R}^n $ ), and those of $ M $ as $ M_1, \ldots, M_n $.
Focusing on $ M_j $, we get $ A_1 m_{1j} + \ldots + A_n m_{nj} = e_j $ ( where $ e_1, \ldots, e_n $ is the standard basis of $ \mathbb{R}^n $ ). Hence $ m_{ij} $ is that scalar $ t $ such that $ e_j - t A_i $ lies in span of $ \{ A_k : k \neq i \} $ [ This is the geometric part ].
So $ \det( A_1, \ldots, e_j - m_{ij} A_i , \ldots, A_n ) = 0 $, from which $ m_{ij} $ can be found [ Here determinants can be avoided when $ n = 2 $. For example $ e_1 - m_{11} A_1 $ is parallel to $ A_2 $, so equating slopes gives $ m_{11} $ ].
