Solution RREF.C19 Robert Beezer

We form the augmented matrix of the system, \begin{align*} &\begin{bmatrix} 2 & 1 & 6 \\ -1 & -1 & -2 \\ 3 & 4 & 4 \\ 3 & 5 & 2 \end{bmatrix} \end{align*} which row-reduces to \begin{align*} &\begin{bmatrix} \leading{1} & 0 & 4 \\ 0 & \leading{1} & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{align*} This augmented matrix represents the linear system $x_1=4$, $x_2=-2$, $0=0$, $0=0$, which clearly has only one possible solution. We can write this solution set then as \begin{align*} S&=\set{\colvector{4\\-2}} \end{align*}