![\text { Let } D=\operatorname{diag}\left[d_{1}, d_{2}, d_{3}\right], \text { where none of } d_{1}, d_{2}, d_{3} \text { is } 0 ; \text { prove that } D^{-1}=\operatorname{diag}\left[d_{1}^{-1}, d_{2}^{-1}, d_{3}^{-1}\right] \text {. }](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7eeafb5b41cf009a58fb9da00117a772_l3.png "Rendered by QuickLaTeX.com")
Solution:
Given matrix as ![D=\operatorname{diag}\left[d_{1}, d_{2}, d_{3}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3d22720a95ea803d009549d7edee2c30_l3.png "Rendered by QuickLaTeX.com")
Therefore matrix 
Now let’s calculate the determinant of the given matrix first.
So, the given matrix has inverse.
Find: the adjoint of the given matrix.
Step: 1 Find the minor matrix of 
Step: 2 Find the co-factor matrix of
Step: 3 By transpose of 
Finally the inverse of the matrix is
![\begin{array}{c} D^{-1}=\frac{1}{|D|} \text { adj } D=\frac{1}{d_{1} d_{2} d_{3}}\left(\begin{array}{cccc} d_{2} d_{3} & 0 & 0 \\ 0 & d_{1} d_{3} & 0 \\ 0 & 0 & d_{1} d_{2} \end{array}\right) \\ =\left(\begin{array}{ccc} d_{1}^{-1} & 0 & 0 \\ 0 & d_{2}^{-1} & 0 \\ 0 & 0 & d_{3}^{-1} \end{array}\right)=\operatorname{diag}\left[d_{1}^{-1}, d_{2}^{-1}, d_{3}^{-1}\right] \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cd9fe555f6084a5e9e0942d87a60d42d_l3.png "Rendered by QuickLaTeX.com")
As a result ![D^{-1}=\operatorname{diag}\left[d_{1}^{-1} \cdot d_{2}^{-1}, d_{3}^{-1}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-151e34692d0f5bd5155bb178a83acbc0_l3.png "Rendered by QuickLaTeX.com")
