The matrix `A={:[((1)/(sqrt(2)),(1)/(sqrt(2))),((-1)/(sqrt(2)),(-1)/(sqrt(2)))]:}` is

Options

A. Unitary
B. Orthogonal
C. Nilpotent
D. Involutary

Answer

To determine the properties of the matrix \( A = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \), we will check if it is orthogonal, nilpotent, or involutory.

### Step 1: Check if the matrix is Orthogonal
A matrix \( A \) is orthogonal if \( A A^T = I \), where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.

1. **Find the transpose of \( A \)**:
\[ A^T = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \]

2. **Multiply \( A \) by \( A^T \)**:
\[ A A^T = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \]

Performing the multiplication:
\[ = \begin{pmatrix} \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\right) & \left(\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}}\right) \\ \left(-\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\right) & \left(-\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}}\right) \end{pmatrix} \]
\[ = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \]

3. **Check if \( A A^T = I \)**:
The identity matrix \( I \) for a 2x2 matrix is \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \). Since \( A A^T \neq I \), the matrix is not orthogonal.

### Step 2: Check if the matrix is Nilpotent
A matrix \( A \) is nilpotent if there exists some integer \( m \) such that \( A^m = 0 \).

1. **Calculate \( A^2 \)**:
\[ A^2 = A \cdot A = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \]
Performing the multiplication:
\[ = \begin{pmatrix} \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}}\right) & \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}}\right) \\ \left(-\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}}\right) & \left(-\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}}\right) \end{pmatrix} \]
\[ = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

2. **Conclusion**:
Since \( A^2 = 0 \), the matrix \( A \) is nilpotent.

### Step 3: Check if the matrix is Involutory
A matrix \( A \) is involutory if \( A^2 = I \).

1. **Since we already calculated \( A^2 \)**:
We found \( A^2 = 0 \), which is not equal to the identity matrix \( I \).

### Final Conclusion
The matrix \( A \) is nilpotent.