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]
[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]
