Example

Consider a linear system $A {\mathbf x}= {\mathbf b}$ which after the application of the Gauss-Jordan method reduces to a matrix $[C \; \; {\mathbf d}]$ with

$$[C \;\; {\mathbf d}] = \begin{bmatrix}1 & 0 & 2 & -1 & 0 & 0 & 2 ... ...\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}.$$

For this particular matrix $[C \;\; {\mathbf d}],$ we want to see the set of solutions. We start with some observations.
Observations:

1. The number of non-zero rows in $C$ is $4.$ This number is also equal to the number of non-zero rows in $[C \;\; {\mathbf d}].$
2. The first non-zero entry in the non-zero rows appear in columns $1, 2, 5$ and $6.$
3. Thus, the respective variables $x_1, x_2, x_5$ and $x_6$ are the basic variables.
4. The remaining variables, $x_3, x_4$ and $x_7$ are free variables.
5. We assign arbitrary constants $k_1, k_2$ and $k_3$ to the free variables $x_3, x_4$ and $x_7,$ respectively.

Hence, we have the set of solutions as

$$\begin{bmatrix}x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\\ x_7 \end{bmatrix} = \begin{bmatrix}8 - 2 k_1 + k_2 - 2 k_3\\ 1 - k_1 - 3 k_2 - 5 k_3 \\ k_1\\ k_2\\ 2 + k_3 \\ 4 - k_3\\ k_3\end{bmatrix}$$

$$= \begin{bmatrix}8\\ 1\\ 0\\ 0\\ 2\\ 4\\ 0 \end{bmatrix} + k_1 \be... ... \end{bmatrix} + k_3 \begin{bmatrix}-2\\ -5\\ 0\\ 0\\ 1\\ -1\\ 1 \end{bmatrix},$$

where $k_1, k_2$ and $k_3$ are arbitrary.
Let ${\mathbf u}_0=\begin{bmatrix}8\\ 1\\ 0\\ 0\\ 2\\ 4\\ 0 \end{bmatrix}, \; {\mat... ...rix}, \; {\mathbf u}_2= \begin{bmatrix}1\\ -3\\ 0\\ 1\\ 0\\ 0\\ 0 \end{bmatrix}$ and ${\mathbf u}_3 = \begin{bmatrix}-2\\ -5\\ 0\\ 0\\ 1\\ -1\\ 1 \end{bmatrix}.$
Then it can easily be verified that $C {\mathbf u}_0 = {\mathbf d},$ and for $1 \leq i \leq 3,\;$ $C {\mathbf u}_i = {\mathbf 0}.$

A similar idea is used in the proof of the next theorem and is omitted. The interested readers can read the proof in Appendix 14.1.