If A = [ cases 1 5 & 2 5 - 2 5 & 1 5 cases ], B = [ cases 1 & 0 i &...

{AA}^{⊤} = \left[\begin{cases}\dfrac{1}{\sqrt{5}} & \dfrac{2}{\sqrt{5}} \\ \dfrac{- 2}{\sqrt{5}} & \dfrac{1}{\sqrt{5}}\end{cases}\right]\left[\begin{cases}\dfrac{1}{\sqrt{5}} & \dfrac{- 2}{\sqrt{5}} \\ \dfrac{2}{\sqrt{5}} & \dfrac{1}{\sqrt{5}}\end{cases}\right] = \left[\begin{cases}1 & 0 \\ 0 & 1\end{cases}\right] = I

$Q = {A}^{⊤}BA$

$⇒ {Q}^{2} = \left({A}^{⊤}BA\right)\left({A}^{⊤}BA\right)$

$= {A}^{⊤}B\left(A{A}^{⊤}\right)BA$

$= {A}^{⊤}B\left(I\right)BA$

$= {A}^{⊤}{B}^{2}A$

$⇒ {Q}^{3} = {A}^{⊤}{B}^{3}A$

$⇒ {Q}^{2021} = {A}^{T}{B}^{2021}A$

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

$P = A\left({A}^{⊤}{B}^{2021}A\right){A}^{⊤}$

given $A{A}^{⊤} = I$

$P = {B}^{2021}$

{B}^{2} = \left[\begin{cases}1 & 0 \\ i & 1\end{cases}\right]\left[\begin{cases}1 & 0 \\ i & 1\end{cases}\right] = \left[\begin{cases}1 & 0 \\ 2i & 1\end{cases}\right]

{B}^{3} = \left[\begin{cases}1 & 0 \\ 2 & 1\end{cases}\right]\left[\begin{cases}1 & 0 \\ i & 1\end{cases}\right] = \left[\begin{cases}1 & 0 \\ 3i & 1\end{cases}\right]

{B}^{2021} = \left[\begin{cases}1 & 0 \\ 2021i & 1\end{cases}\right]

inverse of P = \left({P}^{- 1}\right) = {\left({B}^{2021}\right)}^{- 1} = \left[\begin{cases}1 & 0 \\ - 2021i & 1\end{cases}\right]