Linear equations

A linear equation in a set of variables $x_1,x_2,x_3,\ldots,x_m$ is a equation of the form $$a_1 x_1 + a_2 x_2 + \cdots + a_m x_m = b$$ for constant numbers $a_1,\ldots,a_m, b$.

So, for example, $$-x_1 + 2x_2 - 3x_3 + 4x_4 = 5$$ would be a linear equation in 4 unknowns.

Most equations you have seen are not linear. For example, $$xy = 1.$$ or $$e^{x_1} + x_2 + x_3 = -1.$$

In general, linear equations are the simplest types of mathematical equations so we better be able to handle them quickly and efficiently.

Solving a single linear equation

Say we wanted to solve the equation $$-x_1 + 2x_2 - 3x_3 + 4x_4 = 5$$ How would we go about doing this?

First, we need to be a bit more precise about what we mean to “solve the equation”. What we want here is often called a *parametric” representation of the solution. We want to identify some free variables that can be used to determine all the solutions explicitly.

As you might have done or seen before, this can be accomplished by “solving out” a variable using the equation. We can rewrite the equation as $$x_1 = 2x_2 - 3x_3 + 4x_4 - 5$$

Now, given any possible values of $(x_2,x_3,x_4)$, we can use the new form of the equation to find a value for $x_1$ so that we solve the equation. For example, if we take $(1,1,1)$, then plug in to get $$x_1 = 2 \cdot 1 - 3 \cdot 1 + 4 \cdot 1 -5 = -2$$ and $(-2,1,1,1)$ is a solution to the equation.

Great! You might think we have mastered single linear equations. To find the solution to $$a_1 x_1 + \cdots + a_m x_m = b$$ in general, we pick some $a_i \neq 0$ and then we can solve for $x_i$ as $$x_i = b - \sum_{j \neq i} \frac{a_j}{a_i} x_j$$

We can try another example to test our understanding. How about solving $$x_2 - x_4 = 2$$

One possible point of confusion is: how many variables are we using? We only see 2 in the equation but the biggest index is a 4. So are we using 2 unknowns or 4 unknowns? We will often infer the number of unknowns in the linear equation (and system) from the context. The convention we will use will be this is an equation in 4 unknowns. But, do not be afraid to ask for clarification.

Back to the example. Let us pick $x_4$ to solve out $$x_4 = x_2 - 2$$ Then, our general (parametric) solution would look like $(x_1,x_2,x_3,x_2-2)$.

We might be satisfied with our understanding but there is actually one case that breaks our algorithm.

What happens if all $a_i = 0$ and $b \neq 0$, say $b = 1$? Then, we are looking at an immediate problem $$0 = 1$$ is not satisfiable for any values of our unknowns. In this case, there are no solutions.

This situation may seem contrived. Often in mathematics, a general understanding often hinges on properly understanding the edge cases of a problem. Linear algebra is no different. Encountering $$0 = 1$$ will be a natural possibility once we have developed enough tools.

Before we turn to systems of linear equations, let’s recap some important points:

Algorithm for solving a linear equation:

We can summarize the algorithm for solving $a_1x_1 + \cdots + a_mx_m = b$ described above.

Step 1

If we have an equation of the form $0 = b$ with $b \neq 0$, then output “No Solution”.

Step 2

Otherwise, identify some $x_i$ with $a_i \neq 0$ and solve for $x_i$ in terms of the other variables.

Dichotomy of solutions:

For a linear equation, $a_1x_1 + \cdots a_mx_m = b$, we have three possibilities

• there can be infinitely-many solutions (if $m > 1$ and either not all $a_i$ are $0$ or $b = 0$),
• there can be a unique solution (if $m = 1$ and either not all $a_i$ are $0$ or $b = 0$), or
• there can be no solutions at all (if all $a_i$ are $0$ and $b \neq 0$).

We will see this dichotomy continues to hold for systems of linear equations.