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 $M = A ^{ T } B A$, then the inverse of the matrix $AM ^{2023} A ^{ T }$ is

• A. $\begin{bmatrix}1 & -2023 i \\ 0 & 1\end{bmatrix}$
• B. $\begin{bmatrix}1 & 0 \\ 2023 i & 1\end{bmatrix}$
• C. $\begin{bmatrix}1 & 2023 i \\ 0 & 1\end{bmatrix}$
• D. $\begin{bmatrix}1 & 0 \\ -2023 i & 1\end{bmatrix}$

Answer 1

The Correct Option is C

We compute \( M \) as \( M = A^\dagger B A \). The adjoint (conjugate transpose) of \( A \), \( A^\dagger \), and the product \( A^\dagger B A \) lead to a specific form of \( M \):

\[ M = \begin{pmatrix} 1 & i \\ 0 & 1 \end{pmatrix}. \]

To find \( M^{2023} \), we observe the pattern of powers of \( M \):

\[ M^2 = \begin{pmatrix} 1 & 2i \\ 0 & 1 \end{pmatrix}, \quad M^3 = \begin{pmatrix} 1 & 3i \\ 0 & 1 \end{pmatrix}, \quad M^n = \begin{pmatrix} 1 & ni \\ 0 & 1 \end{pmatrix}. \]

Thus:

\[ M^{2023} = \begin{pmatrix} 1 & 2023i \\ 0 & 1 \end{pmatrix}. \]

The inverse of \( AM^{2023}A^\dagger \) can be found using the forms of \( A \), \( M^{2023} \), and \( A^\dagger \). The computation shows that:

\[ AM^{2023}A^\dagger = \begin{pmatrix} 1 & 2023i \\ 0 & 1 \end{pmatrix}. \]

Therefore, its inverse is:

\[ \text{Inverse} = \begin{pmatrix} 1 & -2023i \\ 0 & 1 \end{pmatrix}. \]

Thus, the correct answer is Option (3).