Gauss-Jordan elimination on $$$\left[\begin{array}{ccccc|c}2 & 4 & 3 & 7 & -1 & -7\\1 & 2 & 2 & 5 & 3 & 6\\-3 & -6 & -1 & 0 & 1 & 2\end{array}\right]$$$

Perform the Gauss-Jordan elimination (reduce completely) on $$$\left[\begin{array}{cccccc}2 & 4 & 3 & 7 & -1 & -7\\1 & 2 & 2 & 5 & 3 & 6\\-3 & -6 & -1 & 0 & 1 & 2\end{array}\right]$$$.

Divide row $$$1$$$ by $$$2$$$: $$$R_{1} = \frac{R_{1}}{2}$$$.

$$$\left[\begin{array}{ccccc|c}1 & 2 & \frac{3}{2} & \frac{7}{2} & - \frac{1}{2} & - \frac{7}{2}\\1 & 2 & 2 & 5 & 3 & 6\\-3 & -6 & -1 & 0 & 1 & 2\end{array}\right]$$$

Subtract row $$$1$$$ from row $$$2$$$: $$$R_{2} = R_{2} - R_{1}$$$.

$$$\left[\begin{array}{ccccc|c}1 & 2 & \frac{3}{2} & \frac{7}{2} & - \frac{1}{2} & - \frac{7}{2}\\0 & 0 & \frac{1}{2} & \frac{3}{2} & \frac{7}{2} & \frac{19}{2}\\-3 & -6 & -1 & 0 & 1 & 2\end{array}\right]$$$

Add row $$$1$$$ multiplied by $$$3$$$ to row $$$3$$$: $$$R_{3} = R_{3} + 3 R_{1}$$$.

$$$\left[\begin{array}{ccccc|c}1 & 2 & \frac{3}{2} & \frac{7}{2} & - \frac{1}{2} & - \frac{7}{2}\\0 & 0 & \frac{1}{2} & \frac{3}{2} & \frac{7}{2} & \frac{19}{2}\\0 & 0 & \frac{7}{2} & \frac{21}{2} & - \frac{1}{2} & - \frac{17}{2}\end{array}\right]$$$

Since the element at row $$$2$$$ and column $$$2$$$ (pivot element) equals $$$0$$$, we need to swap the rows.

Find the first nonzero element in column $$$2$$$ under the pivot entry.

As can be seen, there are no such entries. Move to the next column.

Multiply row $$$2$$$ by $$$2$$$: $$$R_{2} = 2 R_{2}$$$.

$$$\left[\begin{array}{ccccc|c}1 & 2 & \frac{3}{2} & \frac{7}{2} & - \frac{1}{2} & - \frac{7}{2}\\0 & 0 & 1 & 3 & 7 & 19\\0 & 0 & \frac{7}{2} & \frac{21}{2} & - \frac{1}{2} & - \frac{17}{2}\end{array}\right]$$$

Subtract row $$$2$$$ multiplied by $$$\frac{3}{2}$$$ from row $$$1$$$: $$$R_{1} = R_{1} - \frac{3 R_{2}}{2}$$$.

$$$\left[\begin{array}{ccccc|c}1 & 2 & 0 & -1 & -11 & -32\\0 & 0 & 1 & 3 & 7 & 19\\0 & 0 & \frac{7}{2} & \frac{21}{2} & - \frac{1}{2} & - \frac{17}{2}\end{array}\right]$$$

Subtract row $$$2$$$ multiplied by $$$\frac{7}{2}$$$ from row $$$3$$$: $$$R_{3} = R_{3} - \frac{7 R_{2}}{2}$$$.

$$$\left[\begin{array}{ccccc|c}1 & 2 & 0 & -1 & -11 & -32\\0 & 0 & 1 & 3 & 7 & 19\\0 & 0 & 0 & 0 & -25 & -75\end{array}\right]$$$

Since the element at row $$$3$$$ and column $$$4$$$ (pivot element) equals $$$0$$$, we need to swap the rows.

Find the first nonzero element in column $$$4$$$ under the pivot entry.

As can be seen, there are no such entries. Move to the next column.

Divide row $$$3$$$ by $$$-25$$$: $$$R_{3} = - \frac{R_{3}}{25}$$$.

$$$\left[\begin{array}{ccccc|c}1 & 2 & 0 & -1 & -11 & -32\\0 & 0 & 1 & 3 & 7 & 19\\0 & 0 & 0 & 0 & 1 & 3\end{array}\right]$$$

Add row $$$3$$$ multiplied by $$$11$$$ to row $$$1$$$: $$$R_{1} = R_{1} + 11 R_{3}$$$.

$$$\left[\begin{array}{ccccc|c}1 & 2 & 0 & -1 & 0 & 1\\0 & 0 & 1 & 3 & 7 & 19\\0 & 0 & 0 & 0 & 1 & 3\end{array}\right]$$$

Subtract row $$$3$$$ multiplied by $$$7$$$ from row $$$2$$$: $$$R_{2} = R_{2} - 7 R_{3}$$$.

$$$\left[\begin{array}{ccccc|c}1 & 2 & 0 & -1 & 0 & 1\\0 & 0 & 1 & 3 & 0 & -2\\0 & 0 & 0 & 0 & 1 & 3\end{array}\right]$$$

The reduced matrix is $$$\left[\begin{array}{cccccc}1 & 2 & 0 & -1 & 0 & 1\\0 & 0 & 1 & 3 & 0 & -2\\0 & 0 & 0 & 0 & 1 & 3\end{array}\right]$$$A.