Solve this following

Let $D=\operatorname{diag}\left[d_{1}, d_{2}, d_{3}\right]$, where none of $d_{1}, d_{2}, d_{3}$ is 0 ; prove that $D^{-1}=\operatorname{diag}\left[d_{1}^{-1}, d_{2}^{-1}, d_{3}^{-1}\right]$.