Given,

$\mathrm{A}=\left[\begin{array}{cc}\frac{1}{\sqrt{10}}& \frac{3}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}& \frac{1}{\sqrt{10}}\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}1& -i\\ 0& 1\end{array}\right]$

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

${\mathrm{B}}^2=\left[\begin{array}{cc}1& -\mathrm{i}\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& -\mathrm{i}\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& -2\mathrm{i}\\ 0& 1\end{array}\right]$

${\mathrm{B}}^3=\left[\begin{array}{cc}1& -2\mathrm{i}\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& -i\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& -3i\\ 0& 1\end{array}\right]$

So, we can write

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

$\mathrm{M}={\mathrm{A}}^{\mathrm{T}}\mathrm{BA}$        (given)

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

${\mathrm{M}}^2=\mathrm{M}·\mathrm{M}={\mathrm{A}}^{\mathrm{T}}{\mathrm{BAA}}^{\mathrm{T}}\mathrm{BA}=A{\mathrm{A}}^{\mathrm{T}}.{\mathrm{AA}}^{\mathrm{T}}.{\mathrm{B}}^2$

$={\mathrm{AA}}^{\mathrm{T}}{\mathrm{B}}^2$

${\mathrm{M}}^3={\mathrm{AA}}^{\mathrm{T}}{\mathrm{B}}^3$

${\mathrm{M}}^{2023}={\mathrm{AA}}^{\mathrm{T}}{\mathrm{B}}^{2023}={\mathrm{B}}^{2023}$                ($\because A{A}^T$ is an identity matrix)

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

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