Let \(A\) be a matrix defined over a field that is in reduced row-echelon form (RREF). Then the solutions of \(Ax = b\) can be read off the augmented matrix \([A~b]\) immediately. What follows is a look at all the possible scenarios.

Suppose that the augmented matrix is
\(\begin{bmatrix} 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 3
\end{bmatrix}\), which is in RREF.
Columns \(1, 3,\) and \(4\) contain the leading ones. These columns are
called **pivot columns**. The variables that correspond to these
columns are called **pivot variables**. The remaining variables are
called **free variables**.

Suppose that the augmented matrix is
\(\begin{bmatrix} 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1
\end{bmatrix}\).
Notice that the third row has
\(0\)'s in the columns corresonding to the variables
but a nonzero in the right-most column, the column that corresponds to
the right-hand side values. This means that there is no solution because
the equation that the third row represents is “\(0 = 1\)”.
In general, if an augmented matrix in RREF has a row that contains all
\(0\)'s except the right-most entry, then the system has **no solution**.

If the augmented matrix does not have such a row, then there is at least one solution that can easily be obtained by setting each of the pivot variables to the corresponding right-hand side value and setting the free variables to \(0\). For example, if the augmented matrix is \(\begin{bmatrix} 1 & 2 & 0 & 0 & 5 \\ 0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 3 \end{bmatrix}\), a solution is given by setting the pivot variables \(x_1 = 5\), \(x_3 = 4\), \(x_4=3\) and the free variable \(x_2 = 0\).

In the case when the augmented matrix in RREF tells us that there is a solution, we can often say more about the solution set as we will see next.

Suppose that the augmented matrix does not have a row that contains all \(0\)'s except the right-most entry. If there is a free variable, then there will be infinitely many solutions unless the system is defined over a finite field. Consider the augmented matrix given by \(\begin{bmatrix} 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 3 \end{bmatrix}\). The second column is not a pivot column. We can set the variable corresponding to this column to any value and still obtain a solution for the system. In particular, setting \(x_2=s\), we obtain a solution with \(x_1 = -2s\), \(x_3 = 0\), \(x_4 = 3\). In other words, for every value of \(s\), \(x = \begin{bmatrix} -2s \\ s \\0 \\ 3\end{bmatrix}\) is a solution. For example, setting \(s = 0\) gives the solution \(x = \begin{bmatrix} 0 \\ 0 \\0 \\ 3\end{bmatrix}\) and setting \(s = -1\) gives the solution \(x = \begin{bmatrix} 2 \\ -1 \\0 \\ 3\end{bmatrix}\).

(To see how one obtain the solution above, one can write out the system in full: \begin{eqnarray} x_1 + 2x_2 = 0 \\ x_3 = 0 \\ x_4 = 3\end{eqnarray} Setting \(x_2 = s\) gives \begin{eqnarray} x_1 + 2s = 0 \\ x_3 = 0 \\ x_4 = 3\end{eqnarray} Now, solving for \(x_1\) gives the solution as shown above.)

If the augmented matrix does not tell us there is no solution and if there is no free variable (i.e. every column other than the right-most column is a pivot column), then the system has a unique solution. For example, if \(A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}\) and \(b = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}\), then there is a unique solution to the system \(A x = b\).

Each the following matrices defined over the rational numbers is an augmented matrix for some system of linear equations. Determine if the system has no solution, a unique solution, or infinitely many solutions.

\(\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1\end{bmatrix}\)

\(\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix}\)

\(\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 2 \end{bmatrix}\)