If A = 1 5 2 5 - 2 5 1 5 , B = 1 0 i 1 , i = - 1 , and Q = A T B A , then the inverse of the matrix A Q 2021…

${\mathrm{AA}}^{\top }=\left[\begin{array}{ll}\frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\\ \frac{-2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right]\left[\begin{array}{ll}\frac{1}{\sqrt{5}}& \frac{-2}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=\mathrm{I}$

$Q=A^{\top }BA$

$\Rightarrow Q^2=\left(A^{\top }BA\right)\left(A^{\top }BA\right)$

$=A^{\top }B\left(A{A}^{\top }\right)BA$

$=A^{\top }B\left(I\right)BA$

$=A^{\top }{B}^{2}A$

$\Rightarrow Q^3=A^{\top }{B}^{3}A$

$\Rightarrow {\mathrm{Q}}^{2021}={\mathrm{A}}^{\mathrm{T}}{\mathrm{B}}^{2021} \mathrm{A}$

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

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

given $A{A}^{\top }=I$

$\mathrm{P}={\mathrm{B}}^{2021}$

$B^2=\left[\begin{array}{ll}1& 0\\ i& 1\end{array}\right]\left[\begin{array}{ll}1& 0\\ i& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2i& 1\end{array}\right]$

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

${\mathrm{B}}^{2021}=\left[\begin{array}{cc}1& 0\\ 2021\mathrm{i}& 1\end{array}\right]$

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