Answer :

Let's analyze the given matrix:

[tex]\[ A = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \][/tex]

We will examine whether this matrix is orthogonal, hermitian, symmetric, or antisymmetric.

1. Orthogonal Matrix:

A matrix \( A \) is orthogonal if \( A \cdot A^T = I \), where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.

Calculate the transpose of \( A \):

[tex]\[ A^T = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \][/tex]

Now, compute \( A \cdot A^T \):

[tex]\[ A \cdot A^T = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \][/tex]

Since \( A \cdot A^T = I \), \( A \) is orthogonal.

2. Hermitian Matrix:

A matrix \( A \) is hermitian if \( A = A^ \), where \( A^ \) is the conjugate transpose of \( A \). For real matrices, hermitian is simply the same as symmetric.

For our real matrix \( A \):

[tex]\[ A^T = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \][/tex]

Clearly, \( A \neq A^T \), so \( A \) is not hermitian.

3. Symmetric Matrix:

A matrix \( A \) is symmetric if \( A = A^T \).

As shown above,

[tex]\[ A = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \][/tex]
[tex]\[ A^T = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \][/tex]

Since \( A \neq A^T \), \( A \) is not symmetric.

4. Antisymmetric Matrix:

A matrix \( A \) is antisymmetric (or skew-symmetric) if \( A = -A^T \).

Calculate \(-A^T\):

[tex]\[ -A^T = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \][/tex]

Comparing \( A \) with \(-A^T\):

[tex]\[ A = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \][/tex]
[tex]\[ -A^T = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \][/tex]

Clearly, \( A \neq -A^T \) due to the \(1\) at the (3, 3) position. Therefore, \( A \) is not antisymmetric.

Summarizing, the matrix

[tex]\[ A = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \][/tex]

is:
(a) orthogonal: True
(b) hermitian: False
(c) symmetric: False
(d) antisymmetric: False

Thus, the correct answer is:
[tex]\[ \boxed{a} \][/tex]