Let’s consider what we’ve learned in this activity. Since we are interested in the question of whether a consistent linear system has a unique solution or infinitely many, we will only consider consistent systems. By the results of the previous section, this means that there is not a pivot in the rightmost column of the augmented matrix. Here are two possible examples:
In the second example, we have the equations
that we may rewrite in parametric form as
Here we see that
and
are basic variables that may be expressed in terms of the free variable
In this case, the presence of the free variable leads to infinitely many solutions.
Remember that every column of the coefficient matrix corresponds to a variable in our linear system. In the first example, we see that every column of the coefficient contains a pivot position, which means that every variable is uniquely determined. In the second example, the column of the coefficient matrix corresponding to
does not contain a pivot position, which results in
appearing as a free variable. This illustrates the following principle.
When a linear system has a unique solution, every column of the coefficient matrix has a pivot position. Since every row contains at most one pivot position, there must be at least as many rows as columns in the coefficient matrix. Therefore, the linear system has at least as many equations as variables, which is something we intuitively suspected in
Section 1.1.
As we’ll see in the future, we are more interested in the
number of free variables rather than in their choice. For convenience, we will adopt the convention that free variables correspond to columns without a pivot position, which allows us to quickly identify them. For example, the variables
and
appear as free variables in the following linear system: