Let us find the solutions to the following system of equations, \begin{align*} -7x_1 -6 x_2 - 12x_3 &=-33\\ 5x_1 + 5x_2 + 7x_3 &=24\\ x_1 +4x_3 &=5 \end{align*} First, form the augmented matrix, \begin{align*} \begin{bmatrix} -7&-6&- 12&-33\\ 5&5&7&24\\ 1&0&4&5 \end{bmatrix} \end{align*} and work to reduced row-echelon form, first with $j=1$, \begin{align*} \xrightarrow{\rowopswap{1}{3}} & \begin{bmatrix} 1&0&4&5\\ 5&5&7&24\\ -7&-6&-12&-33 \end{bmatrix} \xrightarrow{\rowopadd{-5}{1}{2}} \begin{bmatrix} 1&0&4&5\\ 0&5&-13&-1\\ -7&-6&-12&-33 \end{bmatrix}\\ \xrightarrow{\rowopadd{7}{1}{3}} & \begin{bmatrix} \leading{1}&0&4&5\\ 0&5&-13&-1\\ 0&-6&16&2 \end{bmatrix} \end{align*} Now, with $j=2$, \begin{align*} \xrightarrow{\rowopmult{\frac{1}{5}}{2}} & \begin{bmatrix} \leading{1}&0&4&5\\ 0&1&\frac{-13}{5}&\frac{-1}{5}\\ 0&-6&16&2 \end{bmatrix} \xrightarrow{\rowopadd{6}{2}{3}} \begin{bmatrix} \leading{1}&0&4&5\\ 0&\leading{1}&\frac{-13}{5}&\frac{-1}{5}\\ 0&0&\frac{2}{5}&\frac{4}{5} \end{bmatrix} \end{align*} And finally, with $j=3$, \begin{align*} \xrightarrow{\rowopmult{\frac{5}{2}}{3}} & \begin{bmatrix} \leading{1}&0&4&5\\ 0&\leading{1}&\frac{-13}{5}&\frac{-1}{5}\\ 0&0&1&2 \end{bmatrix} \xrightarrow{\rowopadd{\frac{13}{5}}{3}{2}} \begin{bmatrix} \leading{1}&0&4&5\\ 0&\leading{1}&0&5\\ 0&0&1&2 \end{bmatrix}\\ \xrightarrow{\rowopadd{-4}{3}{1}} & \begin{bmatrix} \leading{1}&0&0&-3\\ 0&\leading{1}&0&5\\ 0&0&\leading{1}&2 \end{bmatrix} \end{align*}
This is now the augmented matrix of a very simple system of equations, namely $x_1=-3$, $x_2=5$, $x_3=2$, which has an obvious solution. Furthermore, we can see that this is the only solution to this system, so we have determined the entire solution set, \begin{align*} S&=\set{\colvector{-3\\5\\2}} \end{align*}
You might compare this example with the procedure we used in Example US.