Let `A=[((1)/(sqrt(10)),(3)/(sqrt(10))),((sqrt(-3))/(sqrt(10)),(1)/(sqrt(10)))],` and B=Let `A=[((1,-i)),((0,1))],` where `i =sqrt(-1)`. If `M=A^(T)BA`,then the inverse of the matrix `AM^(2023)A^(T)` is

Options

A. `[(1,-2023i),(0,1)]`
B. `[(1,0),(2023i,1)]`
C. `[(1,2023i),(0,1)]`
D. `[(1,0),(2023i,1)]`

Answer

To solve the given problem, we need to follow these steps:

### Step 1: Define the matrices A and B
Given:
\[ A = \begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{pmatrix}, \quad B = \begin{pmatrix} 1 & -i \\ 0 & 1 \end{pmatrix} \]

### Step 2: Compute the transpose of matrix A
The transpose of matrix \( A \) is:
\[ A^T = \begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{pmatrix} \]

### Step 3: Compute the product \( A^T B A \)
We first compute the product \( B A \):
\[ B A = \begin{pmatrix} 1 & -i \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{10}} + 3i \cdot \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} - i \cdot \frac{3}{\sqrt{10}} \\ 0 & \frac{1}{\sqrt{10}} \end{pmatrix} \]
Calculating the entries:
\[ B A = \begin{pmatrix} \frac{1 + 3i}{\sqrt{10}} & \frac{3 - 3i}{\sqrt{10}} \\ 0 & \frac{1}{\sqrt{10}} \end{pmatrix} \]

Now, we compute \( A^T (B A) \):
\[ M = A^T B A = \begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} \frac{1 + 3i}{\sqrt{10}} & \frac{3 - 3i}{\sqrt{10}} \\ 0 & \frac{1}{\sqrt{10}} \end{pmatrix} \]

Calculating the entries of \( M \):
1. First row, first column:
\[ \frac{1}{\sqrt{10}} \cdot \frac{1 + 3i}{\sqrt{10}} + \frac{-3}{\sqrt{10}} \cdot 0 = \frac{1 + 3i}{10} \]
2. First row, second column:
\[ \frac{1}{\sqrt{10}} \cdot \frac{3 - 3i}{\sqrt{10}} + \frac{-3}{\sqrt{10}} \cdot \frac{1}{\sqrt{10}} = \frac{3 - 3i - 3}{10} = \frac{-3i}{10} \]
3. Second row, first column:
\[ \frac{3}{\sqrt{10}} \cdot \frac{1 + 3i}{\sqrt{10}} + \frac{1}{\sqrt{10}} \cdot 0 = \frac{3(1 + 3i)}{10} = \frac{3 + 9i}{10} \]
4. Second row, second column:
\[ \frac{3}{\sqrt{10}} \cdot \frac{3 - 3i}{\sqrt{10}} + \frac{1}{\sqrt{10}} \cdot \frac{1}{\sqrt{10}} = \frac{9 - 9i + 1}{10} = \frac{10 - 9i}{10} \]

Thus, we have:
\[ M = \begin{pmatrix} \frac{1 + 3i}{10} & \frac{-3i}{10} \\ \frac{3 + 9i}{10} & \frac{10 - 9i}{10} \end{pmatrix} \]

### Step 4: Compute \( M^{2023} \)
Since \( M \) can be expressed in terms of \( A^T \) and \( B \), we observe that:
\[ M^{2023} = (A^T B A)^{2023} = A^T B^{2023} A \]

### Step 5: Compute \( B^{2023} \)
From the pattern observed in the powers of \( B \):
\[ B^n = \begin{pmatrix} 1 & -ni \\ 0 & 1 \end{pmatrix} \]
Thus,
\[ B^{2023} = \begin{pmatrix} 1 & -2023i \\ 0 & 1 \end{pmatrix} \]

### Step 6: Compute the inverse of \( AM^{2023}A^T \)
We have:
\[ AM^{2023}A^T = A (A^T B^{2023} A) A^T = A A^T B^{2023} = I B^{2023} = B^{2023} \]
Thus, the inverse is:
\[ (AM^{2023}A^T)^{-1} = (B^{2023})^{-1} = \begin{pmatrix} 1 & 2023i \\ 0 & 1 \end{pmatrix} \]

### Final Answer
The inverse of the matrix \( AM^{2023}A^T \) is:
\[ \begin{pmatrix} 1 & 2023i \\ 0 & 1 \end{pmatrix} \]