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 $$Q = A^T B A$$, then the inverse of the matrix $$AQ^{2021}A^T$$ is equal to:

We have the three given matrices

$$ A=\frac1{\sqrt5}\begin{bmatrix}1&2\\-2&1\end{bmatrix},\qquad B=\begin{bmatrix}1&0\\ i&1\end{bmatrix},\qquad i=\sqrt{-1}. $$

First, notice that

$$ A^T A=\frac1{5}\begin{bmatrix}1&-2\\2&1\end{bmatrix} \begin{bmatrix}1&2\\-2&1\end{bmatrix} =\frac1{5}\begin{bmatrix}5&0\\0&5\end{bmatrix} =I, $$

so $$A^T=A^{-1}$$ and $$AA^T=I$$. This orthogonality will greatly simplify our work.

The matrix in the statement is

$$ Q=A^TBA. $$

Because a similarity transform does not change powers, the well-known formula

$$ ( A^{-1} B A )^{n}=A^{-1} B^{\,n} A $$

gives at once

$$ Q^{2021}=A^T B^{2021} A. $$

We are asked to study

$$ AQ^{2021}A^T =A\bigl(A^T B^{2021} A\bigr)A^T. $$

Using associativity and the identities $$AA^T=I$$ and $$A^T A=I$$ we obtain

$$ AQ^{2021}A^T =(AA^T)\,B^{2021}\,(AA^T) =I\,B^{2021}\,I =B^{2021}. $$

Thus the matrix whose inverse is required is simply $$B^{2021}$$. Therefore

$$ \bigl(AQ^{2021}A^T\bigr)^{-1} =\bigl(B^{2021}\bigr)^{-1} =B^{-2021} =(B^{-1})^{2021}. $$

Next we find $$B^{-1}$$. For a $$2\times2$$ matrix $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ with $$ad-bc\neq0$$, the inverse is $$\frac1{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$. Here $$\det B=1\cdot1-0\cdot i=1,$$ so

$$ B^{-1}=\begin{bmatrix}1&0\\-i&1\end{bmatrix}. $$

An even easier way is to see $$B=I+N$$ with $$N=\begin{bmatrix}0&0\\ i&0\end{bmatrix},\qquad N^{2}=0,$$ so $$(I+N)^{-1}=I-N$$, yielding the same inverse.

Write $$B^{-1}=I+N',\; N'=\begin{bmatrix}0&0\\-i&0\end{bmatrix}.$$ Again $$\bigl(N'\bigr)^2=0,$$ so the binomial expansion truncates after the linear term:

$$ (B^{-1})^{2021}=(I+N')^{2021}=I+2021\,N' =\begin{bmatrix} 1 & 0\\ -2021i & 1 \end{bmatrix}. $$

Therefore

$$ \bigl(AQ^{2021}A^T\bigr)^{-1} =\begin{bmatrix} 1 & 0\\ -2021i & 1 \end{bmatrix}. $$

Comparing with the provided options, this matches Option B.

Hence, the correct answer is Option B.