Question

If $A =\begin{pmatrix}\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{pmatrix}, B =\begin{pmatrix}1 & 0 \\ i & 1\end{pmatrix}, i=\sqrt{-1}$, and $Q = A ^{ T } B A$, then the inverse of the matrix $A Q^{2021} A^{ T }$ is equal to :

• A. $\begin{pmatrix}\frac{1}{\sqrt{5}} & -2021 \\ 2021 & \frac{1}{\sqrt{5}}\end{pmatrix}$
• B. $\begin{pmatrix}1 & 0 \\ - 2021 i & 1\end{pmatrix}$
• C. $\begin{pmatrix}1 & 0 \\ 2021 i & 1\end{pmatrix}$
• D. $\begin{pmatrix}1 & -2020 i \\ 0 & 1\end{pmatrix}$

Answer 1

Answer: (B)

$AA ^{ T }=\begin{pmatrix}\frac{1}{5} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{5}} & \frac{-2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{pmatrix}$
$AA ^{ T }=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}= I$
$Q ^{2}= A ^{ T } BA A ^{ T } B A = A ^{ T } BIB A$
$\Rightarrow Q ^{2}= A ^{ T } B ^{2} A$
$Q ^{3}= A ^{ T } B ^{2} AA ^{ T } B A $
$\Rightarrow Q ^{3}= A ^{ T } B ^{3} A$
Similarly : $Q ^{2021}= A ^{ T } B ^{2021} A \ldots \ldots .$ (1)
Now $B ^{2}=\begin{pmatrix}1 & 0 \\ i & 1\end{pmatrix} \begin{pmatrix}1 & 0 \\ i & 1\end{pmatrix}=\begin{pmatrix}1 & 0 \\ 2 i & 1\end{pmatrix}$
$ B ^{3}=\begin{pmatrix} 1 & 0 \\ 2 i & 1 \end{pmatrix}\begin{pmatrix}1 & 0 \\ i & 1
\end{pmatrix}$
$ \Rightarrow B ^{3}=\begin{pmatrix}1 & 0 \\ 3 i & 1 \end{pmatrix}$
Similarly $B ^{2021}=\left(\begin{array}{cc}1 & 0 \\ 2021 i & 1\end{array}\right)$
$\therefore AQ ^{2021} A ^{ T }= AA ^{ T } B ^{2021} AA ^{ T }= IB ^{2021} I$
$\Rightarrow AQ ^{2021} A ^{ T }= B ^{2021}=\begin{pmatrix}1 & 0 \\ 2021 i & 1\end{pmatrix}$
$\therefore \left( AQ ^{2021} A ^{ T }\right)^{-1}=\begin{pmatrix}1 & 0 \\ 2021 i & 1\end{pmatrix}^{-1}=\begin{pmatrix}1 & 0 \\ -2021 i & 1\end{pmatrix}$