Let’s continue with the idea of reinterpreting some high school algebra concepts in the light of linear algebra. For example, we learn even before high school in some cases that a line on a coordinate plane can be defined by two points or it can be defined by a point and the slope of the line.

When we have two points, \(\mathtt{(x_1, y_1)}\) and \(\mathtt{(x_2, y_2)}\), we can determine the slope with \[\mathtt{\frac{y_2 – y_1}{x_2 – x_1}}\]

and then do some substitutions to work out the y-intercept.

The linear algebra way uses vectors, of course. And all we need is a point and a vector to define a line. Or, really, two vectors, since the point can be described as a position vector and the slope is also a vector.

We have the line here defined as a vector plus a scaled vector—scaled by k. (See here for adding vectors and here for scaling them.) \[\color{brown}{\begin{bmatrix}\mathtt{1}\\\mathtt{3}\end{bmatrix}} + \color{blue}{\begin{bmatrix}\mathtt{\,\,\,\,1}\\\mathtt{-1}\end{bmatrix}\mathtt{k}}\] That second, scaled, vector looks like it could do the job of defining the line all by itself, but free vectors like that don’t have a fixed location, so we need a position vector to “fix” that. In general terms, thinking about the free vector as extending from \(\mathtt{(x_1, y_1)}\) to \(\mathtt{(x_2, y_2)}\), we can write the equation for a line as \[\mathtt{l(k) = }\begin{bmatrix}\mathtt{x_1}\\\mathtt{y_1}\end{bmatrix} + \begin{bmatrix}\mathtt{x_2 – x_1}\\\mathtt{y_2 – y_1}\end{bmatrix}\mathtt{k}\] That form is called the parametric form of an equation and can be written as \(\mathtt{l(k) = p + kv}\), where p is a point (or position vector), v is the free vector, and k is a scalar value—the parameter that we change to get different points on the line.

Let’s put this into the context of a (reworded) word problem:

In 2014, County X had 783 miles of paved roads. Starting in 2015, the county has been building 8 miles of new paved roads each year. At this rate, if n is the number of years after 2014, what function gives the number of miles of paved road there will be in County X? (Assume that no paved roads go out of service.)

The equation we’re after is \(\mathtt{f(n) = 783 + 8n}\). As a vector function this can be written as \[\mathtt{f(n) = }\begin{bmatrix}\mathtt{0}\\\mathtt{783}\end{bmatrix} + \begin{bmatrix}\mathtt{1}\\\mathtt{8}\end{bmatrix}\mathtt{n}\] We can see here, perhaps a little more clearly with the vector representation, that our domain is restricted by the situation. Our parameter n is, at the very least, a positive real number, and really a positive integer.

It seems to me that here is at least one other example of a close relationship between linear algebra and current high school algebra instruction that would make absorbing linear algebra into some high school material feasible.

If you’d like to practice with some items related to this post, visit Linear Algebra Exercises I.