thrm{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] = I \\\\\ & \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 & -3 \mathrm{i} \\ 0 & 1 \end{array}\right] \end{aligned} $$

Similarly,

$$ \begin{aligned} & \mathrm{B}^{2023}=\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \\\\ & \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \\\\ & \mathrm{M}^2=\mathrm{M} \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \mathrm{A}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{~A} \\\\ & \mathrm{M}^3=\mathrm{M}^2 \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{AA}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^3 \mathrm{~A} \end{aligned} $$

Similarly,

$$ \begin{aligned} & \mathrm{M}^{2023}= \mathrm{A}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{~A} \\\\ & \mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}=\mathrm{AA}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{AA}^{\mathrm{T}} = \mathrm{I}.\mathrm{B}^{2023}.\mathrm{I}=\mathrm{B}^{2023} \\\\ & =\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \end{aligned} $$

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

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