We solve the following system using augmented matrices and row operations. This is the same system of equations solved in Example US using equation operations. \begin{align*} x_1+2x_2+2x_3&=4\\ x_1+3x_2+3x_3&=5\\ 2x_1+6x_2+5x_3&=6 \end{align*} Form the augmented matrix, \begin{align*} A=\begin{bmatrix} 1&2&2&4\\ 1&3&3&5\\ 2&6&5&6 \end{bmatrix} \end{align*} and apply row operations, \begin{align*} \xrightarrow{\rowopadd{-1}{1}{2}} & \begin{bmatrix} 1&2&2&4\\ 0&1&1&1\\ 2&6&5&6 \end{bmatrix} \xrightarrow{\rowopadd{-2}{1}{3}} \begin{bmatrix} 1&2&2&4\\ 0&1&1&1\\ 0&2&1&-2 \end{bmatrix}\\ \xrightarrow{\rowopadd{-2}{2}{3}} & \begin{bmatrix} 1&2&2&4\\ 0&1&1&1\\ 0&0&-1&-4 \end{bmatrix} \xrightarrow{\rowopmult{-1}{3}} \begin{bmatrix} 1&2&2&4\\ 0&1& 1&1\\ 0&0&1&4 \end{bmatrix} \end{align*}
So the matrix \begin{equation*} B=\begin{bmatrix} 1&2&2&4\\ 0&1& 1&1\\ 0&0&1&4 \end{bmatrix} \end{equation*} is row equivalent to $A$ and by Theorem REMES the system of equations below has the same solution set as the original system of equations. \begin{align*} x_1+2x_2+2x_3&=4\\ x_2+ x_3&=1\\ x_3&=4 \end{align*}
Solving this “simpler” system is straightforward and is identical to the process in Example US.