I want to get to moving stuff around using vectors and matrices, but I’ll stop for a second and touch on the determinant, since linear algebra seems to think it’s important. And, to be honest, it is kind of interesting.

The determinant is the area of the parallelogram created by two vectors. Two vectors will always create a parallelogram like the one shown below, unless they are just scaled versions of each other—but we’ll get to that.

The two vectors shown here are \(\color{blue}{\mathtt{u} = \begin{bmatrix}\mathtt{u_1}\\\mathtt{u_2}\end{bmatrix}}\) and \(\color{red}{\mathtt{v} = \begin{bmatrix}\mathtt{v_1}\\\mathtt{v_2}\end{bmatrix}}\).

We can determine the area of the parallelogram by first determining the area of the large rectangle and then subtracting the triangle areas. Note, by the way, that there are two pairs of two congruent triangles.

So, the area of the large rectangle is \(\mathtt{(u_1 + -v_1)(u_2 + v_2)}\). The negative is interesting. We need it because we want to use positive values when calculating the area of the rectangle. If you play around with different pairs of vectors and different rectangles, you will notice that one of the vector components will always have to be negative in the area calculation, if a parallelogram is formed.

The two large congruent right triangles have a combined area of \(\mathtt{u_{1}u_{2}}\). And the two smaller congruent right triangles have a combined area of \(\mathtt{-v_{1}v_{2}}\). Thus, distributing and subtracting, we get \[\mathtt{u_{1}u_{2} + u_{1}v_{2} – v_{1}u_{2} – v_{1}v_{2} – u_{1}u_{2} – (-v_{1}v_{2})}\]

Then, after simplifying, we have \(\mathtt{u_{1}v_{2} – u_{2}v_{1}}\). If the two vectors u and v represented a linear transformation and were written as column vectors in a matrix, then we could say that there is a determinant of the matrix and show the determinant of the matrix in the way it is usually presented: \[\begin{vmatrix}\mathtt{u_1} & \mathtt{v_1}\\\mathtt{u_2} & \mathtt{v_2}\end{vmatrix} = \mathtt{u_{1}v_{2} – u_{2}v_{1}}\]

One thing to note is that this is a signed area. The sign records a change in orientation that we won’t go into at the moment. Also, if we have vectors that are simply scaled versions of one another—the components of one vector are scaled versions of the other—then the determinant will be zero, which is pretty much what we want, since the area will be zero. Let’s use lambda (\(\mathtt{\lambda}\)) as our scalar to be cool. \[\,\,\,\,\,\,\quad\,\,\,\,\,\begin{vmatrix}\mathtt{u_1} & \mathtt{\lambda u_1}\\\mathtt{u_2} & \mathtt{\lambda u_2}\end{vmatrix} = \mathtt{\lambda u_{1}u_{2} – \lambda u_{1}u_{2} = 0}\]

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