Question

Let $A=\left[\begin{array}{cc}\frac{1}{\sqrt{10}}& \frac{3}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}& \frac{1}{\sqrt{10}}\end{array}\right]$ and $B=\left[\begin{array}{cc}1& -i\\ 0& 1\end{array}\right]$, where $i=\sqrt{-1}$.
If $M=A^{T}BA$, then the inverse of the matrix $A{M}^{2023}{A}^T$ is

• A. $\left[\begin{array}{cc}1& -2023i\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 0\\ 2023i& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 2023i\\ 0& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 0\\ -2023i& 1\end{array}\right]$

Answer 1

Answer: (C)

$A{A}^T=\left[\begin{array}{cc}\frac{1}{\sqrt{10}}& \frac{3}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}& \frac{1}{\sqrt{10}}\end{array}\right]\left[\begin{array}{cc}\frac{1}{\sqrt{10}}& \frac{-3}{\sqrt{10}}\\ \frac{3}{\sqrt{10}}& \frac{1}{\sqrt{10}}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$B^2=\left[\begin{array}{cc}1& -i\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& -i\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& -2i\\ 0& 1\end{array}\right]$
$B^3=\left[\begin{array}{cc}1& -3i\\ 0& 1\end{array}\right]$
:
:
$B^{2023}=\left[\begin{array}{cc}1& -2023i\\ 0& 1\end{array}\right]$
$M=A^{T}BA$
$M^2=M\cdot M=A^{T}BA{A}^{T}BA=A^T{B}^{2}A$
$M^3=M^2\cdot M=A^T{B}^{2}A{A}^{T}BA=A^T{B}^{3}A$
:
:
$M^{2023}=\dots \dots \dots \dots ..A^T{B}^{2023}A$
$A{M}^{2023}{A}^T=A{A}^T{B}^{2023}A{A}^T=B^{2023}$
$=\left[\begin{array}{cc}1& -2023i\\ 0& 1\end{array}\right]$
Inverse of $\left(A{M}^{2023}{A}^T\right)$ is $\left[\begin{array}{cc}1& 2023i\\ 0& 1\end{array}\right]$