Should the formula for the inverse of a 2x2 matrix be obvious?

Question

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 $3731$?

Answer 1

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.

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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.

Answer 2

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.

Answer 3

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 $\mathbb{R}^2$.

Answer 4

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.

Answer 5

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.

Answer 6

$\bullet$ 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 $|u|^2+|v|^2$.

Answer 7

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.

Answer 8

For $\mathbf{A}$ near zero we have $$ (1-\mathbf{A})^{-1}\approx 1+\mathbf{A} $$ so it has to negate the off-diagonals.

(If you want to get all fancy about it you could notice that we use $\exp$ to map from the Lie algebra of invertible matrices to the Lie group itself and note that negation in the Lie algebra corresponds to the inverse in the group. But the zero element of the Lie algebra maps to $1$ in the group so the negation is only directly visible off the diagonal.)

Answer 9

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 $(ic_{3 - k})^T$.

Answer 10

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 $a\bar a-b\bar b$.

Answer 11

There's lots of great answers here, but they may be inaccessible to students first encountering this material, so here's my intuition for the 2x2 inverse at an undergraduate (maybe even high-school) level:

First, let's define two matrices $A$ and $B$, and our goal will be to find entries for $B$ such that $B= A^{-1}$.

$$A = \begin{bmatrix} \color{#d66}a & \color{#4b2}b \\ \color{#d66}c & \color{#4b2}d \end{bmatrix}, \space B = \begin{bmatrix} \color{#28d}e & \color{#28d}f \\ \color{#a6a}g & \color{#a6a}h \end{bmatrix} $$

and we'll give special names to the columns of $A$ and the rows of $B$:

$$\color{#d66}{A_x} = \begin{bmatrix} \color{#d66}a \\ \color{#d66}c \end{bmatrix}, \space \color{#4b2}{A_y} = \begin{bmatrix} \color{#4b2}b \\ \color{#4b2}d \end{bmatrix}, \space \color{#28d}{B_x} = \begin{bmatrix} \color{#28d}e & \color{#28d}f \end{bmatrix}, \space \color{#a6a}{B_y} = \begin{bmatrix} \color{#a6a}g & \color{#a6a}h \end{bmatrix}$$

and we'll think of these as just being vectors hanging out in 2-dimensional space.

Now, by the definition of the inverse we want $BA = A^{-1}A = I$. So what is $BA$?

$$BA = \begin{bmatrix} \color{#28d}e & \color{#28d}f \\ \color{#d6a}g & \color{#d6a}h \end{bmatrix} \begin{bmatrix} \color{#d66}a & \color{#4b2}b \\ \color{#d66}c & \color{#4b2}d \end{bmatrix} = \begin{bmatrix} \color{#28d}{B_x} \cdot \color{#d66}{A_x} & \color{#28d}{B_x} \cdot \color{#4b2}{A_y} \\ \color{#a6a}{B_y} \cdot \color{#d66}{A_x} & \color{#a6a}{B_y} \cdot \color{#4b2}{A_y}\end{bmatrix}$$

(I should note here that a corresponding expression for $AB$ can be found by imagining the columns of $B$ acting on the rows of $A$, but for the sake of simplicity I'll just be showing the geometric intuition for this version.)

Now, we've said that we want this expression to equal $I$ which gives us the following equation:

$$\begin{bmatrix} \color{#28d}{B_x} \cdot \color{#d66}{A_x} & \color{#28d}{B_x} \cdot \color{#4b2}{A_y} \\ \color{#a6a}{B_y} \cdot \color{#d66}{A_x} & \color{#a6a}{B_y} \cdot \color{#4b2}{A_y}\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Which is really just a set of four equations stating that the dot-product of a row from $B$ with a column from $A$ is equal to either $1$ or $0$. These four equations are the constraints that will constrain our answer for the four unknowns in the inverse matrix we are trying to find.

Each constraint has a simple geometric interpretation:
When the dot-product of two vectors equals $0$, their directions must be perpendicular. Given you know the direction of a vector, you must choose some length such that its dot-product with the other vector equals $1$. Thus we can use these four equations to constrain the direction and length of each of the rows of $B$.

For instance, in the top-right corner we have $\color{#28d}{B_x} \cdot \color{#4b2}{A_y} = 0$, which means that the first row of $B$ is perpendicular to the second column of $A$. We already know that we will need to rescale the result, so our goal right now is to merely find any vector which is perpendicular to $\color{#4b2}{A_y}$. The procedure to do so is simple enough: just swap the two components and negate one of them, giving us $\color{#28d}{B_x} = \begin{bmatrix} \color{#4b2}d & \color{#4b2}{-b} \end{bmatrix}$.

Going through a similar process for the bottom-left corner with $\color{#a6a}{B_y} \cdot \color{#d66}{A_x} = 0$, gives us $\color{#a6a}{B_y} = \begin{bmatrix} \color{#d66}{-c} & \color{#d66}{a} \end{bmatrix}$. Now there is some freedom as to which component gets negated, but it comes out in the wash when we choose the lengths so that $\color{#28d}{B_x} \cdot \color{#d66}{A_x}$ and $\color{#a6a}{B_y} \cdot \color{#4b2}{A_y}$ both equal $1$.

In this case, if you carry out the calculation, you'll see I've negated the components such that

$$\color{#28d}{B_x} \cdot \color{#d66}{A_x} = \color{#a6a}{B_y} \cdot \color{#4b2}{A_y} = \color{#d66}a \color{#4b2}d - \color{#4b2}b \color{#d66}c = \det(A)$$

In other words, with these row vectors for $B$, we have the the following equation:

$$\begin{bmatrix} \color{#28d}{B_x} \cdot \color{#d66}{A_x} & \color{#28d}{B_x} \cdot \color{#4b2}{A_y} \\ \color{#a6a}{B_y} \cdot \color{#d66}{A_x} & \color{#a6a}{B_y} \cdot \color{#4b2}{A_y}\end{bmatrix} = \begin{bmatrix} \det(A) & 0 \\ 0 & \det(A) \end{bmatrix}$$

(You'll find that regardless of which component you negate, you always end up with $\pm \det(A)$, and the negative sign will be undone by the next step)

So by dividing each row vector in $B$ by $\det(A)$ we will successfully rescale their lengths such that the dot-products along the diagonal both equal $1$. And thus we arrive at the final version of $B$:

$$ B = \frac{1}{\det(A)}\begin{bmatrix} \color{#4b2}d & \color{#4b2}{-b} \\ \color{#d66}{-c} & \color{#d66}{a} \end{bmatrix} = A^{-1} $$

Well, this may seem like a lot of words and math to explain a simple formula, but now that I've given the explanation, it's quite easy to remember the intuition:
Because the diagonal entries of $I$ are $1$, the length of a row vector in $A^{-1}$ is constrained by the corresponding column vector in $A$ such that their dot product is $1$. And because the off-diagonal entries of $I$ are $0$, the direction of any row vector in $A^{-1}$ is constrained to be perpendicular to the remaining column vectors of $A$. And notice by the way I've laid this out, this intuition works with any size matrix, although the formula is not so simple, obviously.

Answer 12

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} $ ].

Answer 13

$$( \operatorname{diag}(a,d))^{-1} = \operatorname{diag}\left( \frac1a, \frac1d \right) = \frac1{ad} \operatorname{diag}(d,a) $$ That answers at least part of the question.