Let A = begin{bmatrix} frac{1}{sqrt{10}} & fr | vedclass

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(D) First,observe that $A$ is an orthogonal matrix,meaning $AA^{T} = A^{T}A = I$,where $I$ is the identity matrix.Next,calculate the powers of $B$:$B^

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$\begin{bmatrix} 1 & 2023i \ 0 & 1 \end{bmatrix}$

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(D) First,observe that $A$ is an orthogonal matrix,meaning $AA^{T} = A^{T}A = I$,where $I$ is the identity matrix.Next,calculate the powers of $B$:$B^2 = \begin{bmatrix} 1 & -i \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -i \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2i \ 0 & 1 \end{bmatrix}$$B^3 = \begin{bmatrix} 1 & -2i \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -i \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -3i \ 0 & 1 \end{bmatrix}$By induction,$B^{n} = \begin{bmatrix} 1 & -ni \ 0 & 1 \end{bmatrix}$,so $B^{2023} = \begin{bmatrix} 1 & -2023i \ 0 & 1 \end{bmatrix}$.Given $M = A^{T}BA$,we find $M^{n} = (A^{T}BA)(A^{T}BA)...(A^{T}BA) = A^{T}B^{n}A$.Thus,$M^{2023} = A^{T}B^{2023}A$.Now,calculate $AM^{2023}A^{T}$:$AM^{2023}A^{T} = A(A^{T}B^{2023}A)A^{T} = (AA^{T})B^{2023}(AA^{T}) = I \cdot B^{2023} \cdot I = B^{2023} = \begin{bmatrix} 1 & -2023i \ 0 & 1 \end{bmatrix}$.The inverse of $\begin{bmatrix} 1 & k \ 0 & 1 \end{bmatrix}$ is $\begin{bmatrix} 1 & -k \ 0 & 1 \end{bmatrix}$.Therefore,the inverse of $B^{2023}$ is $\begin{bmatrix} 1 & 2023i \ 0 & 1 \end{bmatrix}$.

Let $A = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}$,where $i = \sqrt{-1}$. If $M = A^{T}BA$,then the inverse of the matrix $AM^{2023}A^{T}$ is $.........$

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