Suppose now that we move only one particle at a time, meaning that \( \mathbf{r}^{\mathrm{new}} \) differs from \( \mathbf{r}^{\mathrm{old}} \) by the position of only one, say the i-th, particle . This means that \( \hat{D}(\mathbf{r}^{\mathrm{new}}) \) and \( \hat{D}(\mathbf{r}^{\mathrm{old}}) \) differ only by the entries of the i-th row. Recall also that the i-th row of a cofactor matrix \( \hat{C} \) is independent of the entries of the i-th row of its corresponding matrix \( \hat{D} \). In this particular case we therefore get that the i-th row of \( \hat{C}(\mathbf{r}^{\mathrm{new}}) \) and \( \hat{C}(\mathbf{r}^{\mathrm{old}}) \) must be equal. Explicitly, we have:

$$ \begin{equation*} C_{ij}(\mathbf{r}^{\mathrm{new}}) = C_{ij}(\mathbf{r}^{\mathrm{old}})\quad \forall\ j\in\{1,\dots,N\} \end{equation*} $$