The matrix `A=[{:(1/sqrt2,1/sqrt2),((-1)/sqrt2,(-1)/sqrt2):}]` is

Options

A. unitery
B. orthogonal
C. nilpotent`
D. involutary

Answer

To determine the type 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 an orthogonal matrix, involutory matrix, or nilpotent matrix.

### Step 1: Check if \( A \) is an Orthogonal Matrix

An orthogonal matrix \( A \) satisfies the condition:
\[ A A^T = I \]
where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.

#### Calculate \( A^T \):
\[ A^T = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \]

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

Calculating the entries:
- First row, first column:
\[ \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} + \frac{1}{2} = 1 \]
- First row, second column:
\[ \frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} = -\frac{1}{2} - \frac{1}{2} = -1 \]
- Second row, first column:
\[ -\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = -\frac{1}{2} - \frac{1}{2} = -1 \]
- Second row, second column:
\[ -\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} = \frac{1}{2} + \frac{1}{2} = 1 \]

Thus,
\[ A A^T = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \neq I \]

### Conclusion for Orthogonal Matrix:
Since \( A A^T \neq I \), \( A \) is not an orthogonal matrix.

### Step 2: Check if \( A \) is an Involutory Matrix

An involutory matrix satisfies:
\[ A^2 = I \]

#### Calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \]

Calculating the entries:
- First row, first column:
\[ \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} = \frac{1}{2} - \frac{1}{2} = 0 \]
- First row, second column:
\[ \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} = \frac{1}{2} - \frac{1}{2} = 0 \]
- Second row, first column:
\[ -\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} = -\frac{1}{2} + \frac{1}{2} = 0 \]
- Second row, second column:
\[ -\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} = -\frac{1}{2} + \frac{1}{2} = 0 \]

Thus,
\[ A^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \neq I \]

### Conclusion for Involutory Matrix:
Since \( A^2 \neq I \), \( A \) is not an involutory matrix.

### Step 3: Check if \( A \) is a Nilpotent Matrix

A nilpotent matrix satisfies:
\[ A^m = 0 \quad \text{for some integer } m \]

From the previous calculation, we found that:
\[ A^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Conclusion for Nilpotent Matrix:
Since \( A^2 = 0 \), \( A \) is a nilpotent matrix.

### Final Answer:
The matrix \( A \) is a nilpotent matrix.