Question

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

• A. $\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& -2021\\ 2021& \frac{1}{\sqrt{5}}\end{array}\right)$
• B. $\left(\begin{array}{cc}1& 0\\ -2021i& 1\end{array}\right)$
• C. $\left(\begin{array}{cc}1& 0\\ 2021i& 1\end{array}\right)$
• D. $\left(\begin{array}{cc}1& -2020i\\ 0& 1\end{array}\right)$

Answer 1

Answer: (B)

$A{A}^T=\left(\begin{array}{cc}\frac{1}{5}& \frac{2}{\sqrt{5}}\\ \frac{-2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& \frac{-2}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right)$
$A{A}^T=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$
$Q^2=A^{T}BA{A}^{T}BA=A^{T}BIBA$
$\Rightarrow Q^2=A^T{B}^{2}A$
$Q^3=A^T{B}^{2}A{A}^{T}BA$
$\Rightarrow Q^3=A^T{B}^{3}A$
Similarly : $Q^{2021}=A^T{B}^{2021}A\dots \dots .$ (1)
Now $B^2=\left(\begin{array}{cc}1& 0\\ i& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ i& 1\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 2i& 1\end{array}\right)$
$B^3=\left(\begin{array}{cc}1& 0\\ 2i& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ i& 1<br/>\end{array}\right)$
$\Rightarrow B^3=\left(\begin{array}{cc}1& 0\\ 3i& 1\end{array}\right)$
Similarly $B^{2021}=\left(\begin{array}{cc}1& 0\\ 2021i& 1\end{array}\right)$
$\therefore A{Q}^{2021}{A}^T=A{A}^T{B}^{2021}A{A}^T=I{B}^{2021}I$
$\Rightarrow A{Q}^{2021}{A}^T=B^{2021}=\left(\begin{array}{cc}1& 0\\ 2021i& 1\end{array}\right)$
$\therefore {\left(A{Q}^{2021}{A}^T\right)}^{-1}={\left(\begin{array}{cc}1& 0\\ 2021i& 1\end{array}\right)}^{-1}=\left(\begin{array}{cc}1& 0\\ -2021i& 1\end{array}\right)$