Let us find the solutions to the following system of equations, \begin{align*} x_1 -x_2 +2x_3 & =1\\ 2x_1+ x_2 + x_3 & =8\\ x_1 + x_2 & =5 \end{align*}
First, form the augmented matrix, \begin{align*} \begin{bmatrix} 1 & -1 & 2 & 1\\ 2 & 1 & 1 & 8\\ 1 & 1 & 0 & 5 \end{bmatrix} \end{align*} and work to reduced row-echelon form, first with $j=1$, \begin{align*} \xrightarrow{\rowopadd{-2}{1}{2}} & \begin{bmatrix} 1 & -1 & 2 & 1\\ 0 & 3 & -3 & 6\\ 1 & 1 & 0 & 5 \end{bmatrix} \xrightarrow{\rowopadd{-1}{1}{3}} \begin{bmatrix} \leading{1} & -1 & 2 & 1\\ 0 & 3 & -3 & 6\\ 0 & 2 & -2 & 4 \end{bmatrix} \end{align*} Now, with $j=2$, \begin{align*} \xrightarrow{\rowopmult{\frac{1}{3}}{2}} & \begin{bmatrix} \leading{1} & -1 & 2 & 1\\ 0 & 1 & -1 & 2\\ 0 & 2 & -2 & 4 \end{bmatrix} \xrightarrow{\rowopadd{1}{2}{1}} \begin{bmatrix} \leading{1} & 0 & 1 & 3\\ 0 & 1 & -1 & 2\\ 0 & 2 & -2 & 4 \end{bmatrix}\\ \xrightarrow{\rowopadd{-2}{2}{3}} & \begin{bmatrix} \leading{1} & 0 & 1 & 3\\ 0 & \leading{1} & -1 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{align*}
The system of equations represented by this augmented matrix needs to be considered a bit differently than that for Archetype B. First, the last row of the matrix is the equation $0=0$, which is always true, so it imposes no restrictions on our possible solutions and therefore we can safely ignore it as we analyze the other two equations. These equations are, \begin{align*} x_1+x_3&=3\\ x_2-x_3&=2. \end{align*}
While this system is fairly easy to solve, it also appears to have a multitude of solutions. For example, choose $x_3=1$ and see that then $x_1=2$ and $x_2=3$ will together form a solution. Or choose $x_3=0$, and then discover that $x_1=3$ and $x_2=2$ lead to a solution. Try it yourself: pick any value of $x_3$ you please, and figure out what $x_1$ and $x_2$ should be to make the first and second equations (respectively) true. We'll wait while you do that. Because of this behavior, we say that $x_3$ is a “free” or “independent” variable. But why do we vary $x_3$ and not some other variable? For now, notice that the third column of the augmented matrix is not a pivot column. With this idea, we can rearrange the two equations, solving each for the variable whose index is the same as the column index of a pivot column. \begin{align*} x_1&=3-x_3\\ x_2&=2+x_3 \end{align*}
To write the set of solution vectors in set notation, we have \begin{align*} S&=\setparts{\colvector{3-x_3\\2+x_3\\x_3}}{x_3\in\complex{\null}} \end{align*}
We will learn more in the next section about systems with infinitely many solutions and how to express their solution sets. Right now, you might look back at Example IS.