Some special square matrices are defined as follows. Nilpotent matrix: A square matrix. A is said to be nilpotent (of order 2)if, `A^(2)=O`. A square matrix is said to be nilpotent of order p, if p is the least positive integer such that `A^(p)=O`.
Idempotent matrix: A square matrix A is said to be idempotent if, `A^(2)=A`.
`e.g.[{:(,1,0),(,0,1):}]` is an idempotent matrix.
Involutory matrix: A square A is said to be involutary if `A^(2)=I`, `I `being the identity matrix.
`e.g..A=[{:(,1,0),(,0,1):}]` is an involutary matrix.
Orthogonal matrix: A square matrix A is said to be an orthogonal matrix if `A' A=I=A A'`
If `[{:(,0,2beta,gamma),(,alpha,beta,-gamma),(,alpha,-beta,gamma):}]` is orthogonal, then

Options

A. `alpha=pm(1)/(sqrt2)`
B. `beta=pm(1)/(sqrt6)`
C. `gamma=pm(1)/(sqrt3)`
D. all of those

Answer

To solve the problem, we need to determine the values of the parameters \( \alpha \), \( \beta \), and \( \gamma \) given that the matrix

\[ A = \begin{pmatrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{pmatrix} \]

is orthogonal. This means that \( A^T A = I \), where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.

### Step 1: Calculate \( A^T \)

The transpose of matrix \( A \) is given by:

\[ A^T = \begin{pmatrix} 0 & \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{pmatrix} \]

### Step 2: Compute \( A^T A \)

Now, we compute the product \( A^T A \):

\[ A^T A = \begin{pmatrix} 0 & \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{pmatrix} \begin{pmatrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{pmatrix} \]

Calculating the entries of \( A^T A \):

1. First row, first column:
\[ 0 \cdot 0 + \alpha \cdot \alpha + \alpha \cdot \alpha = 2\alpha^2 \]

2. First row, second column:
\[ 0 \cdot 2\beta + \alpha \cdot \beta + \alpha \cdot -\beta = 0 \]

3. First row, third column:
\[ 0 \cdot \gamma + \alpha \cdot -\gamma + \alpha \cdot \gamma = 0 \]

4. Second row, first column:
\[ 2\beta \cdot 0 + \beta \cdot \alpha + -\beta \cdot \alpha = 0 \]

5. Second row, second column:
\[ 2\beta \cdot 2\beta + \beta \cdot \beta + -\beta \cdot -\beta = 4\beta^2 + \beta^2 + \beta^2 = 6\beta^2 \]

6. Second row, third column:
\[ 2\beta \cdot \gamma + \beta \cdot -\gamma + -\beta \cdot \gamma = 0 \]

7. Third row, first column:
\[ \gamma \cdot 0 + -\gamma \cdot \alpha + \gamma \cdot \alpha = 0 \]

8. Third row, second column:
\[ \gamma \cdot 2\beta + -\gamma \cdot \beta + \gamma \cdot -\beta = 0 \]

9. Third row, third column:
\[ \gamma \cdot \gamma + -\gamma \cdot -\gamma + \gamma \cdot \gamma = 3\gamma^2 \]

Thus, we have:

\[ A^T A = \begin{pmatrix} 2\alpha^2 & 0 & 0 \\ 0 & 6\beta^2 & 0 \\ 0 & 0 & 3\gamma^2 \end{pmatrix} \]

### Step 3: Set \( A^T A = I \)

Setting \( A^T A = I \), we have:

\[ \begin{pmatrix} 2\alpha^2 & 0 & 0 \\ 0 & 6\beta^2 & 0 \\ 0 & 0 & 3\gamma^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

This gives us the following equations:

1. \( 2\alpha^2 = 1 \) → \( \alpha^2 = \frac{1}{2} \) → \( \alpha = \pm \frac{1}{\sqrt{2}} \)
2. \( 6\beta^2 = 1 \) → \( \beta^2 = \frac{1}{6} \) → \( \beta = \pm \frac{1}{\sqrt{6}} \)
3. \( 3\gamma^2 = 1 \) → \( \gamma^2 = \frac{1}{3} \) → \( \gamma = \pm \frac{1}{\sqrt{3}} \)

### Conclusion

The values of \( \alpha \), \( \beta \), and \( \gamma \) are:

- \( \alpha = \pm \frac{1}{\sqrt{2}} \)
- \( \beta = \pm \frac{1}{\sqrt{6}} \)
- \( \gamma = \pm \frac{1}{\sqrt{3}} \)

Thus, the correct option is D: All of the above.