\begin{center}\begin{tabular}{@{}lrrrr@{}}
\toprule word & $\alpha$ & $\beta$ & $\gamma$ & $\delta$\\\midrule
$AABC$ & 0 & 0 & 1 & 1\\
$AACB$ & 0 & 0 & -1 & 0\\
$ABAC$ & 1 & 0 & 0 & -2\\
$ABCA$ & -1 & 0 & -2 & 0\\
$ACAB$ & 0 & 1 & 0 & 0\\
$ACBA$ & 0 & -1 & 2 & 0\\
$BAAC$ & -1 & 0 & 0 & 1\\
$BACA$ & 1 & 0 & 0 & 0\\
$BCAA$ & 0 & 0 & 1 & 0\\
$CAAB$ & 0 & -1 & 0 & -1\\
$CABA$ & 0 & 1 & 0 & 2\\
$CBAA$ & 0 & 0 & -1 & -1\\
\bottomrule\end{tabular}\end{center}
For the derivative matrix, order the deleted letter as $A,B,C$ and each target word lexicographically, numbering rows and columns from zero. Its minor on rows $0,1,2,3,4,5,6,9$ and columns $0,1,2,3,4,5,6,8$ is $-4$. The four printed columns have minor $1$ on rows $0,1,2,4$, and the derivative matrix annihilates them. Hence its kernel has exactly this basis.