Archetype D is a linear system of 3 equations in 4 variables. Row-reducing the augmented matrix yields \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 3 & -2 & 4 \\ 0 & \leading{1} & 1 & -3 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*} and we see $r=2$ pivot columns. Also, $D=\set{1,\,2}$ so the dependent variables are then $x_1$ and $x_2$. $F=\set{3,\,4,\,5}$ so the two free variables are $x_3$ and $x_4$. We will express a generic solution for the system by two slightly different methods, though both arrive at the same conclusion.
First, we will decompose (Proof Technique DC) a solution vector. Rearranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality, \begin{align*} \colvector{x_1\\x_2\\x_3\\x_4}&= \colvector{4-3x_3+2x_4\\ -x_3+3x_4\\x_3\\x_4} \end{align*} Now we will use the definitions of column vector addition and scalar multiplication to express this vector as a linear combination, \begin{align*} &=\colvector{4\\0\\0\\0}+ \colvector{-3x_3\\-x_3\\x_3\\0}+ \colvector{2x_4\\3x_4\\0\\x_4}&&\knowl{./knowls/definition.CVA.knowl}{\text{Definition CVA}}\\ &=\colvector{4\\0\\0\\0}+ x_3\colvector{-3\\-1\\1\\0}+ x_4\colvector{2\\3\\0\\1}&&\knowl{./knowls/definition.CVSM.knowl}{\text{Definition CVSM}}\\ \end{align*}
We will develop the same linear combination a bit quicker, using three steps. While the method above is instructive, the method below will be our preferred approach.
Step 1. Write the vector of variables as a fixed vector, plus a linear combination of $n-r$ vectors, using the free variables as the scalars. \begin{equation*} \vect{x}=\colvector{x_1\\x_2\\x_3\\x_4}= \colvector{\ \\\ \\\ \\\ }+x_3\colvector{\ \\\ \\\ \\\ }+x_4\colvector{\ \\\ \\\ \\\ } \end{equation*}
Step 2. Use 0's and 1's to ensure equality for the entries of the vectors with indices in $F$ (corresponding to the free variables). \begin{equation*} \vect{x}=\colvector{x_1\\x_2\\x_3\\x_4}= \colvector{\ \\\ \\0\\0}+x_3\colvector{\ \\\ \\1\\0}+x_4\colvector{\ \\\ \\0\\1} \end{equation*}
Step 3. For each dependent variable, use the augmented matrix to formulate an equation expressing the dependent variable as a constant plus multiples of the free variables. Convert this equation into entries of the vectors that ensure equality for each dependent variable, one at a time. \begin{align*} x_1=4-3x_3+2x_4&&\Rightarrow&& \vect{x}=\colvector{x_1\\x_2\\x_3\\x_4}= \colvector{4\\\ \\0\\0}+x_3\colvector{-3\\\ \\1\\0}+x_4\colvector{2\\\ \\0\\1}\\ x_2=0-1x_3+3x_4&&\Rightarrow&& \vect{x}=\colvector{x_1\\x_2\\x_3\\x_4}= \colvector{4\\0\\0\\0}+x_3\colvector{-3\\-1\\1\\0}+x_4\colvector{2\\3\\0\\1} \end{align*}
This final form of a typical solution is especially pleasing and useful. For example, we can build solutions quickly by choosing values for our free variables, and then compute a linear combination. Such as \begin{align*} x_3=2,\,x_4=-5&&\Rightarrow&& \vect{x}=\colvector{x_1\\x_2\\x_3\\x_4}= \colvector{4\\0\\0\\0}+(2)\colvector{-3\\-1\\1\\0}+(-5)\colvector{2\\3\\0\\1} =\colvector{-12\\-17\\2\\-5} \end{align*} or, \begin{align*} x_3=1,\,x_4=3&&\Rightarrow&& \vect{x}=\colvector{x_1\\x_2\\x_3\\x_4}= \colvector{4\\0\\0\\0}+(1)\colvector{-3\\-1\\1\\0}+(3)\colvector{2\\3\\0\\1} =\colvector{7\\8\\1\\3} \end{align*}
You will find the second solution listed in the write-up for Archetype D, and you might check the first solution by substituting it back into the original equations.
While this form is useful for quickly creating solutions, it is even better because it tells us exactly what every solution looks like. We know the solution set is infinite, which is pretty big, but now we can say that a solution is some multiple of $\colvector{-3\\-1\\1\\0}$ plus a multiple of $\colvector{2\\3\\0\\1}$ plus the fixed vector $\colvector{4\\0\\0\\0}$. Period. So it only takes us three vectors to describe the entire infinite solution set, provided we also agree on how to combine the three vectors into a linear combination.