# Determinants and Areas

Let’s prove that the determinant of a $2\times 2$ matrix is the area of the parallelogram spanned by it’s column vectors! Here’s a picture of what that means.

Here are two vectors in $\mathbb{R}^2$, and a matrix with those vectors as columns. The determinant of this $2\times 2$ matrix is $ad-bc$. Next let’s look at the parallelogram associated with these two vectors.

To build this parallelogram, you put another copy of the vector $\begin{pmatrix}b\\d\end{pmatrix}$ that starts at the head of the vector $\begin{pmatrix}a\\c\end{pmatrix}$ and you put another copy of the vector $\begin{pmatrix}a\\c\end{pmatrix}$ that starts at the head of the vector $\begin{pmatrix}b\\d\end{pmatrix}$. Any two non-adjacent sides in this picture are parallel (even though my drawing might not make it seem like that).

Here’s that same picture with all of the important segments labelled.

To find the area of the parallelogram, let’s first find the area of the rectangle and then subtract the area of all of the other triangles.

The rectangle has side lengths $a+b$ and $c+d$, so its area is $(a+b)(c+d)$. Note that all of the triangles come in pairs. Let’s start by calculating the areas of all of the triangles, and then we’ll subtract this area from the area of the rectangle afterwards. The area of the right triangle with legs of length $b$ and $d$ is $\frac{1}{2}bd$, and there are two of them, so the total area to subtract away for those triangles is $bd$. There are 4 right triangles with leg lengths $b$ and $c$, so the total area that this accounts for is $4\cdot\frac{1}{2}bc=2bc$. Finally, there are two right triangles with leg lengths $a$ and $c$, so the total area that this accounts for is $2\cdot\frac{1}{2}ac=ac$.

So, to find the area of the parallelogram, we can compute $(a+b)(c+d)-bd-2bc-ac$. If we multiply out the first term, we get $ac+ad+bc+bd$.

Subtracting the other three terms, we get $ac+ad+bc+bd-bd-2bc-ac = ad-bc$. That’s what we were trying to get!

Another reason that this is cool is because we can actually think of $ad-bc$ geometrically in a SECOND way also! Not only is it the area of the parallelogram, but it’s ALSO the difference in area of two rectangles: the rectangle with side lengths $a$ and $d$ and the rectangle with side lengths $b$ and $c$, which are both in that picture! That’s not intuitive at all if you ask me!

How crazy is that?