If `{:A=alpha[(1,1+i),(1-i,-1)]:}a in R`, is a unitary matrix then `alpha^2` is

Options

A. `1/2`
B. `1/3`
C. `1/4`
D. `2/9`

Explanation

To solve the problem, we need to determine the value of \( \alpha^2 \) given that the matrix

\[ A = \alpha \begin{pmatrix} 1 & 1+i \\ 1-i & -1 \end{pmatrix} \]

is a unitary matrix. A matrix \( A \) is unitary if

\[ A A^* = I \]

where \( A^* \) is the conjugate transpose of \( A \) and \( I \) is the identity matrix.

### Step-by-Step Solution:

1. **Write down the matrix \( A \)**:
\[ A = \alpha \begin{pmatrix} 1 & 1+i \\ 1-i & -1 \end{pmatrix} \]

2. **Find the conjugate transpose \( A^* \)**:
To find \( A^* \), we first take the transpose of \( A \) and then take the complex conjugate of each element.

- The transpose of \( A \) is:
\[ A^T = \begin{pmatrix} 1 & 1-i \\ 1+i & -1 \end{pmatrix} \]

- The complex conjugate of \( A^T \) is:
\[ A^* = \begin{pmatrix} 1 & 1-i \\ 1+i & -1 \end{pmatrix} \]

Therefore,
\[ A^* = \alpha \begin{pmatrix} 1 & 1-i \\ 1+i & -1 \end{pmatrix} \]

3. **Set up the equation for unitarity**:
Since \( A \) is unitary, we have:
\[ A A^* = I \]

Substituting \( A \) and \( A^* \):
\[ \left( \alpha \begin{pmatrix} 1 & 1+i \\ 1-i & -1 \end{pmatrix} \right) \left( \alpha \begin{pmatrix} 1 & 1-i \\ 1+i & -1 \end{pmatrix} \right) = I \]

4. **Multiply the matrices**:
\[ A A^* = \alpha^2 \begin{pmatrix} 1 & 1+i \\ 1-i & -1 \end{pmatrix} \begin{pmatrix} 1 & 1-i \\ 1+i & -1 \end{pmatrix} \]

Now, we will perform the multiplication:
- First row, first column:
\[ 1 \cdot 1 + (1+i)(1+i) = 1 + (1 + 2i - 1) = 2 \]
- First row, second column:
\[ 1 \cdot (1-i) + (1+i)(-1) = (1-i) - (1+i) = -2i \]
- Second row, first column:
\[ (1-i) \cdot 1 + (-1)(1+i) = (1-i) - (1+i) = -2i \]
- Second row, second column:
\[ (1-i)(1-i) + (-1)(-1) = (1 - 2i + i^2) + 1 = (1 - 2i - 1) + 1 = 2 \]

Therefore, we have:
\[ A A^* = \alpha^2 \begin{pmatrix} 2 & -2i \\ -2i & 2 \end{pmatrix} \]

5. **Set the product equal to the identity matrix**:
\[ \alpha^2 \begin{pmatrix} 2 & -2i \\ -2i & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

This gives us the equations:
- \( 2\alpha^2 = 1 \)
- \( -2i\alpha^2 = 0 \)

From the first equation:
\[ \alpha^2 = \frac{1}{2} \]

6. **Final answer**:
Thus, the value of \( \alpha^2 \) is:
\[ \alpha^2 = \frac{1}{2} \]