When dealing with connections and cyclic \(D\)-modules, there are two possible ways you can write the connection as cyclic matrices,
\[ \Lambda' = \begin{pmatrix} & & & -a_{0} \\ 1 & & & -a_{1} \\ & \ddots& & \vdots \\ & & 1 & -a_{n-1} \\ \end{pmatrix} \quad \text{and} \quad \Lambda'' = \begin{pmatrix} & 1 & & \\ & & \ddots & \\ & & & 1 \\ -a_0 & -a_1 & \dots & -a_{n-1} \end{pmatrix}.\]
We will call the left hand side a left cyclic matrix and right hand side a right cyclic matrix.
For a cyclic \(D\)-module of the form \(M=D/DP\), where \(P = \partial^n + a_{n-1} \partial^{n-1} + \cdots + a_0\), if we take the basis \(e_0=1, e_1=\partial, \ldots, e_{n-1}=\partial^{n-1}\), then the matrix of left multiplication \(L_\partial: M\to M\) on \(M\) with respect to this basis is \(\Lambda'\).
The question is, what module structure does \(\Lambda''\) correspond to?
For a finite \(K\)-dimensional differential module \(M\) generated by \(e_1,\dots, e_n\), we can form the dual module \(M^* = \mathrm{Hom}_K(M,K)\) consisting of \(K\)-linear functions on \(M\). The module has a dual basis given by \(e_1^*,\dots, e_n^*\), where \(e_i^*(e_j) = \delta_{ij}\).
In general, for any two left \(D\)-modules \(M,N\), the module \(\mathrm{Hom}_O(M,N)\) has a natural left \(D\)-module structure given by, for any derivation \(\theta\in D\) and \(f\in \mathrm{Hom}_O(M,N)\),
\[ (\theta f)(m) := \theta (f(m)) - f(\theta m).\]
This means \(M^*=\mathrm{Hom}_K(M,K)\) has left \(D\)-module structure since \(K\) and \(M\) are left \(D\)-modules.
Let’s consider that \(M\) has a connection matrix \(\Lambda\), i.e. \(De_j = \sum_i \Lambda_{ij}e_i\) which is the matrix of certain chosen derivation \(\partial\) under the given basis.
Then the matrix element \(\Lambda_{ij}'\) of the connection on \(M^*\) is given by
\[\begin{align*} (De_j^*)(e_i) &= D(e_j^*e_i) - e_j^*(De_i) \\ &= D(\delta_{ij}) - e_j^*(\Lambda_{ji}e_j) \\ &= - \Lambda_{ji} \end{align*}\]
So actually the matrix of a dual connection is the negated transpose \(\Lambda' = -\Lambda^T\).