# Solve this

Question:

If $\mathrm{A}=\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right), \mathrm{B}=\left(\begin{array}{ll}1 & 0 \\ i & 1\end{array}\right), i=\sqrt{-1}$, and $Q=A^{\mathrm{T}} B A$, then the inverse of the matrix $\mathrm{A} \mathrm{Q}^{2021} \mathrm{~A}^{\mathrm{T}}$ is equal to :

1. $\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & -2021 \\ 2021 & \frac{1}{\sqrt{5}}\end{array}\right)$

2. $\left(\begin{array}{cc}1 & 0 \\ -2021 i & 1\end{array}\right)$

3. $\left(\begin{array}{cc}1 & 0 \\ 2021 i & 1\end{array}\right)$

4. $\left(\begin{array}{cc}1 & -2021 i \\ 0 & 1\end{array}\right)$

Correct Option: , 2

Solution:

$\mathrm{AA}^{\mathrm{T}}=\left(\begin{array}{cc}\frac{1}{5} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & \frac{-2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right)$

$\mathrm{AA}^{\mathrm{T}}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)=\mathrm{I}$

$\mathrm{Q}^{2}=\mathrm{A}^{\mathrm{T}} \mathrm{B} \mathrm{A} \mathrm{A}^{\mathrm{T}} \mathrm{B} \mathrm{A}=\mathrm{A}^{\mathrm{T}} \mathrm{BIB} \mathrm{A}$

$\Rightarrow \mathrm{Q}^{2}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^{2} \mathrm{~A}$

$\mathrm{Q}^{3}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^{2} \mathrm{~A} \mathrm{~A}^{\mathrm{T}} \mathrm{B} \mathrm{A} \Rightarrow \mathrm{Q}^{3}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^{3} \mathrm{~A}$

Similarly : $\mathrm{Q}^{2021}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^{2021} \mathrm{~A} \ldots \ldots . .$

Now $\mathrm{B}^{2}=\left(\begin{array}{ll}1 & 0 \\ \mathrm{i} & 1\end{array}\right)\left(\begin{array}{ll}1 & 0 \\ \mathrm{i} & 1\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 2 \mathrm{i} & 1\end{array}\right)$

$\mathrm{B}^{3}=\left(\begin{array}{cc}1 & 0 \\ 2 \mathrm{i} & 1\end{array}\right)\left(\begin{array}{cc}1 & 0 \\ \mathrm{i} & 1\end{array}\right) \Rightarrow \mathrm{B}^{3}=\left(\begin{array}{cc}1 & 0 \\ 3 \mathrm{i} & 1\end{array}\right)$

Similarly $\mathrm{B}^{2021}=\left(\begin{array}{cc}1 & 0 \\ 2021 \mathrm{i} & 1\end{array}\right)$

$\therefore \mathrm{AQ}^{2021} \mathrm{~A}^{\mathrm{T}}=\mathrm{AA}^{\mathrm{T}} \mathrm{B}^{2021} \mathrm{AA}^{\mathrm{T}}=\mathrm{IB}^{2021} \mathrm{I}$

$\Rightarrow \mathrm{AQ}^{2021} \mathrm{~A}^{\mathrm{T}}=\mathrm{B}^{2021}=\left(\begin{array}{cc}1 & 0 \\ 2021 \mathrm{i} & 1\end{array}\right)$

$\therefore\left(\mathrm{AQ}^{2021} \mathrm{~A}^{\mathrm{T}}\right)^{-1}=\left(\begin{array}{cc}1 & 0 \\ 2021 \mathrm{i} & 1\end{array}\right)^{-1}=\left(\begin{array}{cc}1 & 0 \\ -2021 \mathrm{i} & 1\end{array}\right)$