If `alpha, beta, gamma` are three real numbers and `A=[(1,cos (alpha-beta),cos(alpha-gamma)),(cos (beta-alpha),1,cos (beta-gamma)),(cos (gamma-alpha),cos (gamma-beta),1)]`
then which of following is/are true ?

Options

A. A is singular
B. A is symmetric
C. A is orthogonal
D. A is not invertible

Explanation

To solve the problem, we need to analyze the given matrix \( A \) and determine its properties. The matrix is defined as:

\[ A = \begin{pmatrix} 1 & \cos(\alpha - \beta) & \cos(\alpha - \gamma) \\ \cos(\beta - \alpha) & 1 & \cos(\beta - \gamma) \\ \cos(\gamma - \alpha) & \cos(\gamma - \beta) & 1 \end{pmatrix} \]

### Step 1: Check if the matrix is symmetric
A matrix is symmetric if \( A = A^T \). This means that the element at position \( (i, j) \) should be equal to the element at position \( (j, i) \).

- For \( A \):
- The element at \( (1, 2) \) is \( \cos(\alpha - \beta) \) and the element at \( (2, 1) \) is \( \cos(\beta - \alpha) \).
- Since \( \cos(\beta - \alpha) = \cos(\alpha - \beta) \), the first condition holds.

- Similarly, we check the other elements:
- \( (1, 3) \) and \( (3, 1) \): \( \cos(\alpha - \gamma) = \cos(\gamma - \alpha) \)
- \( (2, 3) \) and \( (3, 2) \): \( \cos(\beta - \gamma) = \cos(\gamma - \beta) \)

Since all corresponding elements are equal, we conclude that \( A \) is symmetric.

### Step 2: Determine the determinant of the matrix
To find the determinant of the matrix \( A \), we can use the property of determinants for symmetric matrices or calculate directly.

Using the determinant formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

Substituting the values from matrix \( A \):

\[ \text{det}(A) = 1 \cdot (1 \cdot 1 - \cos(\beta - \gamma) \cdot \cos(\gamma - \beta)) - \cos(\alpha - \beta) \cdot (\cos(\beta - \alpha) \cdot \cos(\gamma - \beta) - \cos(\alpha - \gamma) \cdot 1) + \cos(\alpha - \gamma) \cdot (\cos(\beta - \alpha) \cdot \cos(\gamma - \alpha) - 1 \cdot \cos(\beta - \gamma)) \]

However, we can also note that the determinant of a symmetric matrix can be zero if the rows (or columns) are linearly dependent.

### Conclusion
1. **The matrix \( A \) is symmetric.**
2. **The determinant of \( A \) is zero, indicating that the rows are linearly dependent.**