If a square matrix A is such that `"AA"^(T)=l=A^(T)A`, then `|A|` is equal to

Options

A. 0
B. `+-1`
C. `+-2`
D. None of these

Explanation

To solve the problem, we need to find the determinant of the square matrix \( A \) given that \( AA^T = I \) and \( A^TA = I \), where \( I \) is the identity matrix.

### Step-by-Step Solution:

1. **Understanding the Given Condition:**
We are given that \( AA^T = I \). This means that the matrix \( A \) is orthogonal. An orthogonal matrix has the property that its transpose is also its inverse.

2. **Taking the Determinant of Both Sides:**
We can take the determinant of both sides of the equation \( AA^T = I \):
\[ \det(AA^T) = \det(I) \]
Since the determinant of the identity matrix \( I \) is 1, we have:
\[ \det(AA^T) = 1 \]

3. **Using the Property of Determinants:**
The property of determinants states that \( \det(AB) = \det(A) \cdot \det(B) \). Therefore:
\[ \det(AA^T) = \det(A) \cdot \det(A^T) \]
Since \( \det(A^T) = \det(A) \), we can rewrite this as:
\[ \det(AA^T) = \det(A) \cdot \det(A) = (\det(A))^2 \]

4. **Setting Up the Equation:**
From the previous steps, we have:
\[ (\det(A))^2 = 1 \]

5. **Solving for the Determinant:**
Taking the square root of both sides gives us:
\[ \det(A) = \pm 1 \]

### Conclusion:
Thus, the determinant of the matrix \( A \) can either be \( 1 \) or \( -1 \).

### Final Answer:
\[ |A| = \pm 1 \]