Let A be a matrix of order 3, such that `A^(T)A=I`. Then find the value of det. `(A^(2)-I)`.

Explanation

To solve the problem, we need to find the value of \( \text{det}(A^2 - I) \) given that \( A \) is a \( 3 \times 3 \) matrix such that \( A^T A = I \). This means that \( A \) is an orthogonal matrix.

### Step-by-Step Solution:

1. **Understanding the properties of orthogonal matrices**:
Since \( A^T A = I \), we know that \( A \) is an orthogonal matrix. One important property of orthogonal matrices is that their determinant can only be \( 1 \) or \( -1 \). Therefore, we have:
\[ \text{det}(A) = \pm 1 \]

2. **Express \( A^2 - I \)**:
We want to find \( \text{det}(A^2 - I) \). We can rewrite \( A^2 - I \) as:
\[ A^2 - I = A^2 - A^T A \]
This is because \( A^T A = I \).

3. **Factor the expression**:
We can factor \( A^2 - I \) as follows:
\[ A^2 - I = (A - I)(A + I) \]

4. **Use the property of determinants**:
The determinant of a product of matrices is the product of their determinants. Therefore:
\[ \text{det}(A^2 - I) = \text{det}((A - I)(A + I)) = \text{det}(A - I) \cdot \text{det}(A + I) \]

5. **Analyzing the determinants**:
Since \( A \) is orthogonal, we can analyze the eigenvalues of \( A \). The eigenvalues of an orthogonal matrix are of the form \( e^{i\theta} \) (complex numbers on the unit circle) or \( \pm 1 \). For a \( 3 \times 3 \) orthogonal matrix, the eigenvalues could be \( 1, e^{i\theta}, e^{-i\theta} \) or \( -1, e^{i\theta}, e^{-i\theta} \).

6. **Eigenvalues of \( A^2 \)**:
The eigenvalues of \( A^2 \) will be the squares of the eigenvalues of \( A \). Therefore, if \( \lambda \) is an eigenvalue of \( A \), then \( \lambda^2 \) is an eigenvalue of \( A^2 \). The eigenvalues of \( A^2 \) will be \( 1, e^{2i\theta}, e^{-2i\theta} \) or \( 1, -1, e^{2i\theta} \).

7. **Finding \( \text{det}(A^2 - I) \)**:
The eigenvalues of \( A^2 - I \) will be \( 0, e^{2i\theta} - 1, e^{-2i\theta} - 1 \). Since one of the eigenvalues is \( 0 \), we have:
\[ \text{det}(A^2 - I) = 0 \cdot (e^{2i\theta} - 1)(e^{-2i\theta} - 1) = 0 \]

### Conclusion:
Thus, the value of \( \text{det}(A^2 - I) \) is:
\[ \text{det}(A^2 - I) = 0 \]