If $\hat{D}$ is invertible, then we must obviously have $\hat{D}^{-1}\hat{D}= \mathbf{1}$, or explicitly in terms of the individual elements of $\hat{D}$ and $\hat{D}^{-1}$:

$$\begin{equation} \sum_{k=1}^N d_{ik}^{\phantom X}d_{kj}^{-1} = \delta_{ij}^{\phantom X} \tag{2} \end{equation}$$