If A=[ … ],i=√-1, and Q=A^TBA, then the inverse of the matrix AQ^2021 A^T is equal to:

$${\mathrm{AA}}^{⊤}=[\begin{matrix}\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{matrix}][\begin{matrix}\frac{1}{\sqrt{5}} & \frac{-2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{matrix}]=[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]=I$$AA⊤=[5​1​5​−2​​5​2​5​1​​][5​1​5​2​​5​−2​5​1​​]=[10​01​]=I

$$Q={A}^{⊤}BA$$Q=A⊤BA

$$\Rightarrow {Q}^{2}=({A}^{⊤}BA)({A}^{⊤}BA)$$⇒Q2=(A⊤BA)(A⊤BA)

$$={A}^{⊤}B(A{A}^{⊤})BA$$=A⊤B(AA⊤)BA

$$={A}^{⊤}B(I)BA$$=A⊤B(I)BA

$$={A}^{⊤}{B}^{2}A$$=A⊤B2A

$$\Rightarrow {Q}^{3}={A}^{⊤}{B}^{3}A$$⇒Q3=A⊤B3A

$$\Rightarrow {Q}^{2021}={A}^{T}{B}^{2021}A$$⇒Q2021=ATB2021A

Now let $P={\mathrm{AQ}}^{2021}{A}^{T}$P=AQ2021AT

$$P=A({A}^{⊤}{B}^{2021}A){A}^{⊤}$$P=A(A⊤B2021A)A⊤

given $A{A}^{⊤}=I$AA⊤=I

$$P={B}^{2021}$$P=B2021

$${B}^{2}=[\begin{matrix}1 & 0 \\ i & 1\end{matrix}][\begin{matrix}1 & 0 \\ i & 1\end{matrix}]=[\begin{matrix}1 & 0 \\ 2i & 1\end{matrix}]$$B2=[1i​01​][1i​01​]=[12i​01​]

$${B}^{3}=[\begin{matrix}1 & 0 \\ 2 & 1\end{matrix}][\begin{matrix}1 & 0 \\ i & 1\end{matrix}]=[\begin{matrix}1 & 0 \\ 3i & 1\end{matrix}]$$B3=[12​01​][1i​01​]=[13i​01​]

$${B}^{2021}=[\begin{matrix}1 & 0 \\ 2021i & 1\end{matrix}]$$B2021=[12021i​01​]

inverse of $P=({P}^{-1})={({B}^{2021})}^{-1}=[\begin{matrix}1 & 0 \\ -2021i & 1\end{matrix}]$P=(P−1)=(B2021)−1=[1−2021i​01​]