Let B=left[begin{array}{ll}1 & 2 0 & 1end{ar | vedclass

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(D) Given $\left(A^T\right)^{-1}=A$, which implies $A A^T=A^T A=I$. Now, $X=A B A^T$. Then $X^{2021}=\left(A B A^T\right)^{2021} = (A B A^T)(A B A^T)

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$\left[\begin{array}{cc}1 & 4042 \ 0 & 1\end{array}ight]$

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(D) Given $\left(A^Tight)^{-1}=A$, which implies $A A^T=A^T A=I$. Now, $X=A B A^T$. Then $X^{2021}=\left(A B A^Tight)^{2021} = (A B A^T)(A B A^T) \dots (A B A^T)$. Using the associative property of matrix multiplication, we have $X^{2021}=A B (A^T A) B (A^T A) B \dots (A^T A) B A^T$. Since $A^T A=I$, this simplifies to $X^{2021}=A B I B I B \dots I B A^T = A B^{2021} A^T$. Now, we need to calculate $A^T X^{2021} A = A^T (A B^{2021} A^T) A = (A^T A) B^{2021} (A^T A) = I B^{2021} I = B^{2021}$. For matrix $B=\left[\begin{array}{ll}1 & 2 \ 0 & 1\end{array}ight]$, we observe: $B^2 = \left[\begin{array}{ll}1 & 2 \ 0 & 1\end{array}ight] \left[\begin{array}{ll}1 & 2 \ 0 & 1\end{array}ight] = \left[\begin{array}{ll}1 & 4 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}1 & 2 	imes 2 \ 0 & 1\end{array}ight]$. $B^3 = B^2 B = \left[\begin{array}{ll}1 & 4 \ 0 & 1\end{array}ight] \left[\begin{array}{ll}1 & 2 \ 0 & 1\end{array}ight] = \left[\begin{array}{ll}1 & 6 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}1 & 2 	imes 3 \ 0 & 1\end{array}ight]$. By induction, $B^n = \left[\begin{array}{cc}1 & 2n \ 0 & 1\end{array}ight]$. Therefore, $B^{2021} = \left[\begin{array}{cc}1 & 2 	imes 2021 \ 0 & 1\end{array}ight] = \left[\begin{array}{cc}1 & 4042 \ 0 & 1\end{array}ight]$.

Let $B=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$ and $A$ be a $2 \times 2$ matrix satisfying $\left(A^T\right)^{-1}=A$. If $X=A B A^T$, then $A^T X^{2021} A=$

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