There are so many connections within and applications of linear algebra—I can only imagine that this will be a series of posts, to the extent that I continue writing about the subject. Here are a few connections that I’ve come across in my reading recently.

Compound Interest and Matrix Powers

We can multiply a matrix by itself \(\mathtt{n}\) times. The result is the matrix to the power \(\mathtt{n}\). We can use this when setting up a compound interest situation. For example, suppose we have three accounts, which each have a different interest rate compounded annually—say, 5%, 3%, and 2%. Without linear algebra, the amount in the first account can be modeled by the equation \[\mathtt{A(t)=p \cdot 1.05^t}\] where \(\mathtt{p}\) represents the starting amount in the account, and \(\mathtt{t}\) represents the time in years. With linear algebra, we can group all of the account interest rates into a matrix. The first year for each account would look like this: \[\mathtt{A(1)}=\begin{bmatrix}\mathtt{1.05}&\mathtt{0}&\mathtt{0}\\\mathtt{0}&\mathtt{1.03}&\mathtt{0}\\\mathtt{0}&\mathtt{0}&\mathtt{1.02}\end{bmatrix}^\mathtt{1}\begin{bmatrix}\mathtt{p_1}\\\mathtt{p_2}\\\mathtt{p_3}\end{bmatrix}=\begin{bmatrix}\mathtt{1.05p_1}\\\mathtt{1.03p_2}\\\mathtt{1.02p_3}\end{bmatrix}\]

For years beyond the first year, all we have to do is raise the matrix to the appropriate power. Since it’s diagonal, squaring it, cubing it, etc., will square, cube, etc., each entry. This computation can be a little more organized—and more straightforward for programming. A matrix has to be square (\(\mathtt{m \times m})\) in order to raise it to a power. Below we calculate the amount in each account after 100 years.

Centroids and Areas

We have seen that the determinant can be thought about as the area of the parallelogram formed by two vectors. We can use this fact to determine the area of a complex shape like the one shown below.

Since determinants are signed areas, as we move around the shape counterclockwise, calculating the determinant of each vector pair (and multiplying each determinant by one half so we just get each triangle), we get the total area of the shape.\[\frac{1}{2}\left(\begin{vmatrix}\mathtt{6}&\mathtt{6}\\\mathtt{0}&\mathtt{4}\end{vmatrix}+\begin{vmatrix}\mathtt{6}&\mathtt{3}\\\mathtt{4}&\mathtt{4}\end{vmatrix}+\begin{vmatrix}\mathtt{3}&\mathtt{3}\\\mathtt{4}&\mathtt{6}\end{vmatrix}+\begin{vmatrix}\mathtt{3}&\mathtt{-2}\\\mathtt{6}&\mathtt{\,\,\,\,6}\end{vmatrix}+\begin{vmatrix}\mathtt{-2}&\mathtt{-2}\\\mathtt{\,\,\,\,6}&\mathtt{\,\,\,\,3}\end{vmatrix}+\begin{vmatrix}\mathtt{-2}&\mathtt{0}\\\mathtt{\,\,\,\,3}&\mathtt{3}\end{vmatrix}\right)=\mathtt{36}\text{ units}\mathtt{^2}\]

That’s pretty hand-wavy, but it’s something that you can probably figure out with a little experimentation.

Another counterclockwise-moving calculation (though this one can be clockwise without changing the answer) is the calculation of the centroid of a closed shape. All that is required here is to calculate the sum of the position vectors of the vertices of the figure and then divide by the number of vertices.

\[\mathtt{\frac{1}{5}}\left(\begin{bmatrix}\mathtt{3}\\\mathtt{4}\end{bmatrix}+\begin{bmatrix}\mathtt{0}\\\mathtt{6}\end{bmatrix}+\begin{bmatrix}\mathtt{-3}\\\mathtt{\,\,\,\,4}\end{bmatrix}+\begin{bmatrix}\mathtt{-1}\\\mathtt{\,\,\,\,3}\end{bmatrix}\right)=\begin{bmatrix}\mathtt{-\frac{1}{5}}\\\mathtt{\,\,\,\,\frac{17}{5}}\end{bmatrix}\]