If A=begin{bmatrix} frac{1}{sqrt{5}} & frac{2 | vedclass

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(B) First,observe that $A$ is an orthogonal matrix because $AA^{T} = I$.$AA^{T} = \begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \ \frac{-2}

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$\begin{bmatrix} 1 & 0 \ -2021i & 1 \end{bmatrix}$

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(B) First,observe that $A$ is an orthogonal matrix because $AA^{T} = I$.$AA^{T} = \begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{-2}{\sqrt{5}} \ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.Given $Q = A^{T}BA$,we find $Q^{n} = (A^{T}BA)(A^{T}BA)...(A^{T}BA) = A^{T}B^{n}A$.Thus,$Q^{2021} = A^{T}B^{2021}A$.Now,calculate powers of $B = \begin{bmatrix} 1 & 0 \ i & 1 \end{bmatrix}$.$B^{2} = \begin{bmatrix} 1 & 0 \ i & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ i & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 2i & 1 \end{bmatrix}$.By induction,$B^{n} = \begin{bmatrix} 1 & 0 \ ni & 1 \end{bmatrix}$,so $B^{2021} = \begin{bmatrix} 1 & 0 \ 2021i & 1 \end{bmatrix}$.Now,$AQ^{2021}A^{T} = A(A^{T}B^{2021}A)A^{T} = (AA^{T})B^{2021}(AA^{T}) = I B^{2021} I = B^{2021}$.Therefore,$AQ^{2021}A^{T} = \begin{bmatrix} 1 & 0 \ 2021i & 1 \end{bmatrix}$.The inverse of a matrix $\begin{bmatrix} 1 & 0 \ k & 1 \end{bmatrix}$ is $\begin{bmatrix} 1 & 0 \ -k & 1 \end{bmatrix}$.Thus,$(AQ^{2021}A^{T})^{-1} = \begin{bmatrix} 1 & 0 \ -2021i & 1 \end{bmatrix}$.

If $A=\begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{bmatrix}$,$B=\begin{bmatrix} 1 & 0 \\ i & 1 \end{bmatrix}$,$i=\sqrt{-1}$,and $Q=A^{T}BA$,then the inverse of the matrix $AQ^{2021}A^{T}$ is equal to:

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