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:

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