The coefficient matrix in Archetype I is \begin{equation*} I=\begin{bmatrix} 1 & 4 & 0 & -1 & 0 & 7 & -9 \\ 2 & 8 & -1 & 3 & 9 & -13 & 7\\ 0 & 0 & 2 & -3 & -4 & 12 & -8\\ -1 & -4 & 2 & 4 & 8 & -31 & 37 \end{bmatrix} \end{equation*}
To build the row space, we transpose the matrix, \begin{equation*} \transpose{I}= \begin{bmatrix} 1 & 2 & 0 & -1\\ 4 & 8 & 0 & -4\\ 0 & -1 & 2 & 2\\ -1 & 3 & -3 & 4\\ 0 & 9 & -4 & 8\\ 7 & -13 & 12 & -31\\ -9 & 7 & -8 & 37 \end{bmatrix} \end{equation*}
Then the columns of this matrix are used in a span to build the row space, \begin{equation*} \rsp{I}=\csp{\transpose{I}}= \spn{\set{ \colvector{1\\4\\0\\-1\\0\\7\\-9},\, \colvector{2\\8\\-1\\3\\9\\-13\\7},\, \colvector{0\\0\\2\\-3\\-4\\12\\-8},\, \colvector{-1\\-4\\2\\4\\8\\-31\\37} }}. \end{equation*}
However, we can use Theorem BCS to get a slightly better description. First, row-reduce $\transpose{I}$, \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 0 & -\frac{31}{7}\\ 0 & \leading{1} & 0 & \frac{12}{7}\\ 0 & 0 & \leading{1} & \frac{13}{7}\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}. \end{equation*}
Since the pivot columns have indices $D=\set{1,\,2,\,3}$, the column space of $\transpose{I}$ can be spanned by just the first three columns of $\transpose{I}$, \begin{equation*} \rsp{I}=\csp{\transpose{I}}= \spn{\set{ \colvector{1\\4\\0\\-1\\0\\7\\-9},\, \colvector{2\\8\\-1\\3\\9\\-13\\7},\, \colvector{0\\0\\2\\-3\\-4\\12\\-8} }}. \end{equation*}