A square matrix `B` is said to be an orthogonal matrix of order `n` if `BB^(T)=I_(n)` if `n^(th)` order square matrix `A` is orthogonal, then |adj(adjA)| is

Options

A. may be `-1` if `n` is even
B. greater than `1` in all cases
C. always `1`
D. always 1 if `n` is odd

Explanation

To solve the problem, we need to find the value of \(|\text{adj}(\text{adj} A)|\) for an orthogonal matrix \(A\) of order \(n\).

### Step-by-Step Solution:

1. **Understanding Orthogonal Matrices**:
An \(n \times n\) matrix \(A\) is orthogonal if \(AA^T = I_n\), where \(A^T\) is the transpose of \(A\) and \(I_n\) is the identity matrix of order \(n\).

**Hint**: Recall that the definition of an orthogonal matrix involves its transpose and the identity matrix.

2. **Determinant of Orthogonal Matrices**:
For any orthogonal matrix \(A\), we have:
\[ \det(A) \cdot \det(A^T) = \det(I_n) = 1 \]
Since \(\det(A^T) = \det(A)\), we can write:
\[ \det(A)^2 = 1 \]
This implies that:
\[ \det(A) = \pm 1 \]

**Hint**: Use the property of determinants that relates the determinant of a matrix and its transpose.

3. **Adjoint of a Matrix**:
The adjoint of a matrix \(A\), denoted as \(\text{adj}(A)\), is related to the determinant by the formula:
\[ \text{adj}(A) = \det(A) A^{-1} \]
For the adjoint of the adjoint, we have:
\[ \text{adj}(\text{adj}(A)) = \det(A)^{n-1} A \]
Therefore, the determinant of the adjoint of the adjoint is given by:
\[ |\text{adj}(\text{adj}(A))| = |\det(A)^{n-1} A| = |\det(A)|^{n-1} \cdot |A| \]

**Hint**: Remember that the determinant of the adjoint can be expressed in terms of the determinant of the original matrix.

4. **Finding \(|\text{adj}(\text{adj} A)|\)**:
Since \(|A| = \det(A)\) and we know \(\det(A) = \pm 1\), we can substitute this into our equation:
\[ |\text{adj}(\text{adj}(A))| = |\det(A)|^{n-1} \cdot |\det(A)| = |\det(A)|^n \]
Thus, we have:
\[ |\text{adj}(\text{adj}(A))| = (\pm 1)^n \]

**Hint**: Consider the cases for \(n\) being even or odd.

5. **Conclusion**:
- If \(n\) is even, \((\pm 1)^n = 1\).
- If \(n\) is odd, \((\pm 1)^n = \pm 1\).

Therefore, the final result is:
\[ |\text{adj}(\text{adj}(A))| = \begin{cases} 1 & \text{if } n \text{ is even} \\ \pm 1 & \text{if } n \text{ is odd} \end{cases} \]