Elementary Linear Algebra: For University of Lethbridge Math 1410

So far, whenever we have solved a system of linear equations, we have always found exactly one solution. This is not always the case; we will find in this section that some systems do not have a solution, and others have more than one.

There are obviously infinitely many solutions to this system; as long as \(x=y\text{,}\) we have a solution. We can picture all of these solutions by thinking of the graph of the equation \(y=x\) on the traditional \(x,y\) coordinate plane.

Each of these equations can be viewed as lines in the coordinate plane, and since their slopes are different, we know they will intersect somewhere (see Figure 3.4.1.(a)). In this example, they intersect at the point \((1,1)\) — that is, when \(x=1\) and \(y=1\text{,}\) both equations are satisfied and we have a solution to our linear system. Since this is the only place the two lines intersect, this is the only solution.

It is clear that while we have two equations, they are essentially the same equation; the second is just a multiple of the first. Therefore, when we graph the two equations, we are graphing the same line twice (see Figure 3.4.1.(b); the thicker line is used to represent drawing the line twice). In this case, we have an infinite solution set, just as if we only had the one equation \(x+y=1\text{.}\) We often write the solution as \(x=1-y\) to demonstrate that \(y\) can be any real number, and \(x\) is determined once we pick a value for \(y\text{.}\)

We should immediately spot a problem with this system; if the sum of \(x\) and \(y\) is 1, how can it also be 2? There is no solution to such a problem; this linear system has no solution. We can visualize this situation in Figure 3.4.1.(c); the two lines are parallel and never intersect.

If we were to consider a linear system with three equations and two unknowns, we could visualize the solution by graphing the corresponding three lines. We can picture that perhaps all three lines would meet at one point, giving exactly 1 solution; perhaps all three equations describe the same line, giving an infinite number of solutions; perhaps we have different lines, but they do not all meet at the same point, giving no solution. We further visualize similar situations with, say, 20 equations with two variables.

From here, we can go on to consider three or more equations in three variables; there might be no solution, or the planes might intersect in a single point, or along a common line, or they might even all describe the same plane.

While it becomes harder to visualize when we add variables, no matter how many equations and variables we have, solutions to linear equations always come in one of three forms: exactly one solution, infinitely many solutions, or no solution. This is a fact that we will not prove here, but it deserves to be stated.

This leads us to a definition. Here we don’t differentiate between having one solution and infinitely many solutions, but rather just whether or not a solution exists.

How can we tell what kind of solution (if one exists) a given system of linear equations has? The answer to this question lies with properly understanding the reduced row echelon form of a matrix. To discover what the solution is to a linear system, we first put the matrix into reduced row echelon form and then interpret that form properly.

Before we start with a simple example, let us make a note about finding the reduced row echelon form of a matrix.

As a general rule, when we are learning a new technique, it is best to not use technology to aid us. This helps us learn not only the technique but some of its “inner workings.” We can then use technology once we have mastered the technique and are now learning how to use it to solve problems.

From here on out, in our examples, when we need the reduced row echelon form of a matrix, we will not show the steps involved. Rather, we will give the initial matrix, then immediately give the reduced row echelon form of the matrix. We trust that the reader can verify the accuracy of this form by both performing the necessary steps by hand or utilizing some technology to do it for them.

Our first example explores officially a quick example used in the introduction of this section.

Let’s try another example, one that uses more variables.

In the two previous examples we have used the word “free” to describe certain variables. What exactly is a free variable? How do we recognize which variables are free and which are not?

Look back to the reduced matrix in Example 3.4.4. Notice that there is only one leading 1 in that matrix, and that leading 1 corresponded to the \(x_1\) variable. That told us that \(x_1\) was not a free variable; since \(x_2\) did not correspond to a leading 1, it was a free variable.

Look also at the reduced matrix in Example 3.4.5. There were two leading 1s in that matrix; one corresponded to \(x_1\) and the other to \(x_2\text{.}\) This meant that \(x_1\) and \(x_2\) were not free variables; since there was not a leading 1 that corresponded to \(x_3\text{,}\) it was a free variable.

We formally define this and a few other terms in this following definition.

These definitions help us understand when a consistent system of linear equations will have infinitely many solutions. If there are no free variables, then there is exactly one solution; if there are any free variables, there are infinitely many solutions.

Note: Key Idea 3.4.7 applies only to consistent systems. If a system is inconsistent, then no solution exists and talking about free and basic variables is meaningless.

When a consistent system has only one solution, each equation that comes from the reduced row echelon form of the corresponding augmented matrix will contain exactly one variable. If the consistent system has infinitely many solutions, then there will be at least one equation coming from the reduced row echelon form that contains more than one variable. The “first” variable will be the basic (or dependent) variable; all others will be free variables.

We have now seen examples of consistent systems with exactly one solution and others with infinitely many solutions. How will we recognize that a system is inconsistent? Let’s find out through an example.

In previous sections we have only encountered linear systems with unique solutions (exactly one solution). Now we have seen three more examples with different solution types. The first two examples in this section had infinitely many solutions, and the third had no solution. How can we tell if a system is inconsistent?

A linear system will be inconsistent only when it implies that 0 equals 1. We can tell if a linear system implies this by putting its corresponding augmented matrix into reduced row echelon form. If we have any row where all entries are 0 except for the entry in the last column, then the system implies \(0=1\text{.}\) More succinctly, if we have a leading 1 in the last column of an augmented matrix, then the linear system has no solution.

Let’s summarize what we have learned up to this point. Consider the reduced row echelon form of the augmented matrix of a system of linear equations. (That sure seems like a mouthful in and of itself. However, it boils down to “look at the reduced form of the usual matrix.”) If there is a leading 1 in the last column, the system has no solution. Otherwise, if there is a leading 1 for each variable, then there is exactly one solution; otherwise there are infinitely many solutions. (i.e., there are free variables.)

Systems with exactly one solution or no solution are the easiest to deal with; systems with infinitely many solutions are a bit harder to deal with. Therefore, we’ll do a little more practice. First, a definition: if there are infinitely many solutions, what do we call one of those infinitely many solutions?

The easiest way to find a particular solution is to pick values for the free variables which then determines the values of the dependent variables. Again, more practice is called for.

The constants and coefficients of a matrix work together to determine whether a given system of linear equations has one, infinitely many, or no solution. The concept will be fleshed out more in later chapters, but in short, the coefficients determine whether a matrix will have exactly one solution or not. In the “or not” case, the constants determine whether or not infinitely many solutions or no solution exists. (So if a given linear system has exactly one solution, it will always have exactly one solution even if the constants are changed.) Let’s look at an example to get an idea of how the values of constants and coefficients work together to determine the solution type.

As in the previous example, if \(k\neq6\text{,}\) we can make the second row, second column entry a leading one and hence we have one solution. However, if \(k=6\text{,}\) then our last row is \([0\ 0\ 1]\text{,}\) meaning we have no solution.

We have been studying the solutions to linear systems mostly in an “academic” setting; we have been solving systems for the sake of solving systems. In the next section, we’ll look at situations which create linear systems that need solving (i.e., “word problems”).