Archetype B has a coefficient matrix given as \begin{align*} B=&\begin{bmatrix} -7&-6&-12\\ 5&5&7\\ 1&0&4 \end{bmatrix} \\ \end{align*} Exercising Theorem CINM we set \begin{align*} M=& \begin{bmatrix} -7&-6&-12&1&0&0\\ 5&5&7&0&1&0\\ 1&0&4&0&0&1 \end{bmatrix}.\\ \end{align*} which row reduces to \begin{align*} N=& \begin{bmatrix} 1&0&0&-10 & -12 & -9\\ 0&1&0&\frac{13}{2} & 8 & \frac{11}{2}\\ 0&0&1&\frac{5}{2} & 3 & \frac{5}{2} \end{bmatrix}.\\ \end{align*} So \begin{align*} \inverse{B}=&\begin{bmatrix} -10 & -12 & -9\\ \frac{13}{2} & 8 & \frac{11}{2}\\ \frac{5}{2} & 3 & \frac{5}{2} \end{bmatrix} \end{align*} once we check that $\inverse{B}B=I_3$ (the product in the opposite order is a consequence of the theorem).