Let   $\mathrm{A}=\left[\begin{array}{cc}\frac{1}{\sqrt{10}}& \frac{3}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}& \frac{1}{\sqrt{10}}\end{array}\right]\text{ and }\mathrm{B}=\left[\begin{array}{cc}1& -\mathrm{i}\\ 0& 1\end{array}\right]$ , where $\mathrm{i}=\sqrt{-1}\text{ If }\mathrm{M}={\mathrm{A}}^{\mathrm{T}}\mathrm{BA}$ then the inverse of the
matrix  ${\mathrm{AM}}^{2023}{\mathrm{A}}^{\mathrm{T}}$ is 

$\begin{array}{l}{\mathrm{AA}}^{\mathrm{T}}=\left[\begin{array}{c}\frac{1}{\sqrt{10}}\frac{3}{\sqrt{10}}\\ -\frac{3}{\sqrt{10}}\frac{1}{\sqrt{10}}\end{array}\right]\left[\begin{array}{c}\frac{1}{\sqrt{10}}\frac{3}{\sqrt{10}}\\ \frac{1}{\sqrt{10}}\frac{1}{\sqrt{10}}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{10}+\frac{9}{10}& \frac{-3}{10}+\frac{3}{10}\\ \frac{-3}{10}+\frac{3}{10}& \frac{9}{10}+\frac{1}{10}\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ \end{array}$

${\mathrm{M}}^2=\left({\mathrm{A}}^{\mathrm{T}}\mathrm{BA}\right)\left({\mathrm{A}}^{\mathrm{T}}\mathrm{BA}\right)={\mathrm{A}}^{\mathrm{T}}{\mathrm{B}}^2\mathrm{A}$

${\mathrm{M}}^3=\left({\mathrm{A}}^{\mathrm{T}}{\mathrm{B}}^2\mathrm{A}\right)\left({\mathrm{A}}^{\mathrm{T}}\mathrm{BA}\right)={\mathrm{A}}^{\mathrm{T}}{\mathrm{B}}^3\mathrm{A}$

continuing like this 

${\mathbf{M}}^{2023}={\mathrm{A}}^{\mathrm{T}}{\mathrm{B}}^{2023}\mathrm{A}$

${\mathrm{AM}}^{2023}{\mathrm{A}}^{\mathrm{T}}={\mathrm{AA}}^{\mathrm{T}}{\mathrm{B}}^{2023}{\mathrm{AA}}^{\mathrm{T}}={\mathrm{B}}^{2023}$

$\mathrm{B}=\left[\begin{array}{cc}1& -\mathrm{i}\\ 0& 1\end{array}\right]=\left(I+C\right)$ where 

$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],C=\left[\begin{array}{cc}0& -i\\ 0& 0\end{array}\right]\Rightarrow C^2=O$

${\mathrm{B}}^{2023}={\mathrm{I}}^{2023}{+}^{2023}{\mathrm{C}}_1{\mathrm{I}}^{2022}=\mathrm{I}+2023\mathrm{C}$

${\mathrm{B}}^{2023}=\left[\begin{array}{cc}1& -2023\mathrm{i}\\ 0& 1\end{array}\right]$

$\mathrm{Inverse} \mathrm{of} {\mathrm{AM}}^{2023}{\mathrm{A}}^{\mathrm{T}}=\mathrm{inverse} \mathrm{of} {\mathrm{B}}^{2023}=\left[\begin{array}{cc}1& 2023i\\ 0& 1\end{array}\right]$