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.

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