= Solution
Keep all four <relators> and all five generators, in the orders $(r_A,r_B,r_C,r_D)$ and $(d,e,a,b,c)$. Under <abelianization>, these generators become $(t,t,t^2,t^3,t^2)$. The <Fox calculus> rules are
$$
\frac{\partial(uv)}{\partial x}=\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial x},\qquad
\frac{\partial x^{-1}}{\partial x}=-x^{-1}.
$$
For example, $r_B=ad^{-1}cb^{-1}$ gives the nonzero <Fox derivative> entries $-ad^{-1}$, $1$, $-ad^{-1}cb^{-1}$, $ad^{-1}$ in columns $d,a,b,c$, respectively. After <abelianization> these are $-t,1,-1,t$. The full <Alexander matrix> is
$$
\boxed{A=\begin{pmatrix}
-1&-t&1&0&0\\
-t&0&1&-1&t\\
0&t&-t&1&-1\\
-t&-1&0&0&1
\end{pmatrix}.}
$$
The <Fox calculus> identity supplies a useful check:
$$
A\begin{pmatrix}t-1\\t-1\\t^2-1\\t^3-1\\t^2-1\end{pmatrix}=0.
$$
Deleting the $e$ column, adjacent to the unbounded region across the marked <meridian of a knot>, gives
$$
B=\begin{pmatrix}-1&1&0&0\\-t&1&-1&t\\0&-t&1&-1\\-t&0&0&1\end{pmatrix},\qquad
\det B=-t^2+3t-1.
$$
Thus $\det B$ is a representative of the <Alexander polynomial of a knot> of the <figure-eight knot>, with its customary ambiguity of multiplication by $\pm t^n$.
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