Let A be an orthogonal non-singular matrix of order n, then `|A-I_n|` is equal to :

Options

A. `|I_n-A|`
B. `|A|`
C. `|A||I_n-A|`
D. `(-1)^n |A||I_n-A|`

Answer

To solve the problem, we need to find the determinant of the matrix \( A - I_n \), where \( A \) is an orthogonal non-singular matrix of order \( n \).

### Step-by-Step Solution:

1. **Understanding Orthogonal Matrices**:
An orthogonal matrix \( A \) satisfies the property \( A A^T = I_n \), where \( A^T \) is the transpose of \( A \) and \( I_n \) is the identity matrix of order \( n \). This means that the columns (and rows) of \( A \) are orthonormal vectors.

2. **Determinant of Orthogonal Matrices**:
The determinant of an orthogonal matrix is either \( 1 \) or \( -1 \). Thus, we can write:
\[ |A| = \pm 1 \]

3. **Finding the Determinant of \( A - I_n \)**:
We want to find \( |A - I_n| \). To do this, we can use the property of determinants:
\[ |A - I_n| = |A| \cdot |I_n - A^T| \]
Since \( A \) is orthogonal, we have \( |A| = \pm 1 \).

4. **Using the Property of Determinants**:
We know that:
\[ |I_n - A^T| = |I_n - A| \]
because the determinant of a matrix and its transpose are equal, i.e., \( |B| = |B^T| \).

5. **Final Expression**:
Thus, we can express the determinant as:
\[ |A - I_n| = |A| \cdot |I_n - A| = \pm |I_n - A| \]

6. **Conclusion**:
Therefore, the determinant \( |A - I_n| \) is equal to \( |I_n - A| \) multiplied by either \( 1 \) or \( -1 \). However, the exact value depends on the specific orthogonal matrix \( A \).

### Final Result:
\[ |A - I_n| = |I_n - A| \]