Archetype G and Archetype H are both systems of $m=5$ equations in $n=2$ variables. They have identical coefficient matrices, which we will denote here as the matrix $G$, \begin{equation*} G= \begin{bmatrix} 2 & 3\\ -1 & 4 \\ 3 & 10\\ 3 & -1\\ 6 & 9 \end{bmatrix} \end{equation*}
Adjoin the $5\times 5$ identity matrix, $I_5$, to form \begin{equation*} M= \begin{bmatrix} 2 & 3 & 1&0&0&0&0\\ -1 & 4 & 0&1&0&0&0\\ 3 & 10 & 0&0&1&0&0\\ 3 & -1 & 0&0&0&1&0\\ 6 & 9 & 0&0&0&0&1 \end{bmatrix} \end{equation*}
This row-reduces to \begin{equation*} N= \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 0 & \frac{3}{11} & \frac{1}{33}\\ 0 & \leading{1} & 0 & 0 & 0 & -\frac{2}{11} & \frac{1}{11}\\ 0 & 0 & \leading{1} & 0 & 0 & 0 & -\frac{1}{3}\\ 0 & 0 & 0 & \leading{1} & 0 & 1 & -\frac{1}{3}\\ 0 & 0 & 0 & 0 & \leading{1} & 1 & -1 \end{bmatrix} \end{equation*}
The first $n=2$ columns contain $r=2$ leading 1's, so we obtain $C$ as the $2\times 2$ identity matrix and extract $L$ from the final $m-r=3$ rows in the final $m=5$ columns. \begin{align*} C&= \begin{bmatrix} \leading{1} & 0\\ 0 & \leading{1} \end{bmatrix} & L&= \begin{bmatrix} \leading{1} & 0 & 0 & 0 & -\frac{1}{3}\\ 0 & \leading{1} & 0 & 1 & -\frac{1}{3}\\ 0 & 0 & \leading{1} & 1 & -1 \end{bmatrix} \end{align*}
Then we apply Theorem FS, \begin{align*} \nsp{G}=\nsp{C}&= \spn{\emptyset} =\set{\zerovector} &&\knowl{./knowls/theorem.BNS.knowl}{\text{Theorem BNS}}\\ \rsp{G}=\rsp{C}&= \spn{\set{ \colvector{1 \\ 0},\, \colvector{0 \\ 1} }} =\complex{2} &&\knowl{./knowls/theorem.BRS.knowl}{\text{Theorem BRS}}\\ \csp{G}=\nsp{L}&= \spn{\set{ \colvector{0\\-1\\-1\\1\\0},\, \colvector{\frac{1}{3}\\\frac{1}{3}\\1\\0\\1} }}&&\knowl{./knowls/theorem.BNS.knowl}{\text{Theorem BNS}}\\ &=\spn{\set{ \colvector{0\\-1\\-1\\1\\0},\, \colvector{1\\1\\3\\0\\3} }}\\ \lns{G}=\rsp{L}&= \spn{\set{ \colvector{1 \\ 0 \\ 0 \\ 0 \\ -\frac{1}{3}},\, \colvector{0 \\ 1 \\ 0 \\ 1 \\ -\frac{1}{3}},\, \colvector{0 \\ 0 \\ 1 \\ 1 \\ -1} }}&&\knowl{./knowls/theorem.BRS.knowl}{\text{Theorem BRS}}\\ &=\spn{\set{ \colvector{3 \\ 0 \\ 0 \\ 0 \\ -1},\, \colvector{0 \\ 3 \\ 0 \\ 3\ \\ -1},\, \colvector{0 \\ 0 \\ 1 \\ 1 \\ -1} }} \end{align*}
As mentioned earlier, Archetype G is consistent, while Archetype H is inconsistent. See if you can write the two different vectors of constants from these two archetypes as linear combinations of the two vectors that form the spanning set for $\csp{G}$. How about the two columns of $G$? Can you write each individually as a linear combination of the two vectors that form the spanning set for $\csp{G}$? They must be in the column space of $G$ also. Are your answers unique? Do you notice anything about the scalars that appear in the linear combinations you are forming?