# Determinants and volumes, revisited

## Determinants and volumes, revisited

In an earlier video, we claimed that the determinant of a matrix $M$ is equal to the volume of the parallelopiped which is the image of the unit cube under the linear transformation associated with $M$ . We saw a proof of this for 2-by-2 matrices, but it wasn't clear how to generalise this to higher dimensions. In this video we will see another proof in 2-dimensions, one whose generalisation to higher dimensions is hopefully clearer.

The unit cube has vertices at $(0,0)$ , $(1,0)$ , $(0,1)$ and $(1,1)$ . If we apply the 2-by-2 matrix $M=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)$ then the vertices become $(0,0)$ , $(a,c)$ , $(b,d)$ and $(a+b,c+d)$ .

We will calculate the volume (area) of this parallelogram by rearrangement: cutting it into pieces and rearranging them using translations. The fact that the volume doesn't change when we do this is known as *Cavalieri's principle*.

We will cut the parallelogram into layers parallel to the edge connecting the origin to $(a,c)$ . Then we will slide these layers in the $(-a,-c)$ direction (i.e. along themselves) until they line up along the $y$ -axis.

The result of this is a new parallelogram such that two of its edges are vertical and the other two are still parallel to $(a,c)$ . Indeed, two vertices are still at the origin and at $(a,c)$ , but the vertex at $(b,d)$ has now moved a certain amount (let's say $-m$ ) parallel to the $(a,c)$ -direction, so that it is now at $(b-ma,d-mc)$ . What is $m$ ? It's whatever it needs to be to make the first coordinate of this point equal to zero, so that it sits on the $y$ -axis. In other words, this corner is at $(0,{d}^{\prime})$ where ${d}^{\prime}=d-mc$ .

In terms of the matrix $M$ , we have subtracted off $m$ times the first column from the second column to "clear" the top entry of the second column (make it equal to zero): $${C}_{2}\mapsto {C}_{2}-m{C}_{1}.$$

We do one more rearrangement (in higher dimensions we'd need to do more). Namely, we split the parallelogram into vertical slices and translate them vertically down until they line up along the $x$ -axis. The result is a rectangle base $a$ and height ${d}^{\prime}$ .

In terms of $M$ , we have already done a column operation to get $\left(\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill c\hfill & \hfill {d}^{\prime}\hfill \end{array}\right)$ and we now subtract off some multiple of column 2 from column 1 to get a diagonal matrix $\left(\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {d}^{\prime}\hfill \end{array}\right)$ . The volume (area) of the parallelogram we started with is the same as the area of this rectangle (because rearrangements don't change volumes) and this equals $a{d}^{\prime}$ , the determinant of this diagonal matrix. But this is equal to the determinant of $M$ because, just like row operations, column operations don't change the determinant.