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

Answer 1

The correct option is B $\left[\begin{array}{cc}1& 0\\ -2021i& 1\end{array}\right]$

$\because A\cdot A^T=\left[\begin{array}{cc}\frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\\ -\frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right]\left[\begin{array}{cc}\frac{1}{\sqrt{5}}& -\frac{2}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

So, $A\cdot Q^{2021}\cdot A^T=A(A^{T}BA{)}^{2021}\cdot A^T=B^{2021}$
$\because B^2=\left[\begin{array}{cc}1& 0\\ i& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ i& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2i& 1\end{array}\right]$

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

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

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