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 \( \hat{D}^{-1} \):

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