\[\begin{equation} \tag{40} \partial\left(\mathbf{X}^{-1}\right)=-\mathbf{X}^{-1}(\partial \mathbf{X}) \mathbf{X}^{-1} \end{equation}\]
Explanation: 1
\[\begin{equation} \underbrace{(I)^{\prime}}_{=0}=\left(\mathbf{X} \mathbf{X}^{-1}\right)^{\prime}=\mathbf{X}^{\prime} \mathbf{X}^{-1}+\mathbf{X}\left(\mathbf{X}^{-1}\right)^{\prime} \Rightarrow \end{equation}\]
\[\begin{equation} \mathbf{X}\left(\mathbf{X}^{-1}\right)^{\prime}=-\mathbf{X}^{\prime} \mathbf{X}^{-1} \quad \Rightarrow \end{equation}\]
\[\begin{equation} \left(\mathbf{X}^{-1}\right)^{\prime}=-\mathbf{X}^{-1} \mathbf{X}^{\prime} \mathbf{X}^{-1} \end{equation}\]
\[\begin{equation} \tag{41} \partial(\operatorname{det}(\mathbf{X}))=\operatorname{Tr}(\operatorname{adj}(\mathbf{X}) \partial \mathbf{X}) \end{equation}\]
Background
Adjugate Matrix
The adjugate of \(A\) is the transpose of the cofactor matrix \(C\) of \(X\), \[\begin{equation} \operatorname{adj}(\mathbf{X})=\mathbf{C}^{\top} \end{equation}\]
and, \[\begin{equation} \mathbf{X}^{-1}=\operatorname{det}(\mathbf{X})^{-1} \operatorname{adj}(\mathbf{X}) \quad \Rightarrow \end{equation}\]
\[\begin{equation} \operatorname{det}(\mathbf{X}) \mathbf{I} = \operatorname{adj}(\mathbf{X}) \mathbf{X} \end{equation}\]
Characteristic Polynomial
The characteristic polynomial of a square matrix is a polynomial which is invariant under matrix similarity and has the eigenvalues as roots. It has the determinant and the trace of the matrix as coefficients.
The characteristic polynomial of a sqaure matrix \(A\) is defined by \[\begin{equation} p_{A}(t)=\operatorname{det}(t I-A) \end{equation}\]
Proof 2
Via Matrix Computation
\[\begin{equation} \frac{\partial \operatorname{det}(\mathbf{X})}{\partial \mathbf{X}_{i j}}=\sum_{k} \operatorname{adj}^{\mathrm{T}}(\mathbf{X})_{i k} \delta_{j k}=\operatorname{adj}^{\mathrm{T}}(\mathbf{X})_{i j} \quad \Rightarrow \end{equation}\]
\[\begin{equation} d(\operatorname{det}(\mathbf{X}))=\sum_{i} \sum_{j} \operatorname{adj}^{\mathrm{T}}(\mathbf{X})_{i j} d \mathbf{X}_{i j} \quad \Rightarrow \end{equation}\]
\[\begin{equation} d(\operatorname{det}(\mathbf{X}))=\operatorname{tr}(\operatorname{adj}(\mathbf{X}) d \mathbf{X}) \end{equation}\]
Via Chain Rule
Lemma 1. \(\operatorname{det}^{\prime}(I)=\operatorname{tr}\), where \(\operatorname{det}^{\prime}\) is the differential of \(\operatorname{det}\).
Lemma 2. For an invertible matrix \(\mathbf{A}\), we have: \(\operatorname{det}^{\prime}(\mathbf{A})(\mathbf{T})=\operatorname{det} \mathbf{A} \operatorname{tr}(\mathbf{A}^{-1}\mathbf{T}))\)
\[\begin{equation} \tag{42} \partial(\operatorname{det}(\mathbf{X}))=\operatorname{det}(\mathbf{X}) \operatorname{Tr}\left(\mathbf{X}^{-1} \partial \mathbf{X}\right) \end{equation}\]