If A is an invertible matrix and B is an orthogonal matrix of the order same as that of `A` then `C = A^(-1) BA` is

Options

A. an orthogonal matrix
B. symmetric matrix
C. skew symmetric matrix
D. none of these

Answer

To solve the problem step by step, we need to analyze the given matrices and their properties.

### Step-by-Step Solution:

1. **Understanding the Given Matrices**:
- We have matrix \( A \) which is invertible.
- We have matrix \( B \) which is orthogonal. This means that \( B^T B = I \) (where \( I \) is the identity matrix).

2. **Define the Matrix \( C \)**:
- We define \( C \) as \( C = A^{-1} B A \).

3. **Finding the Transpose of \( C \)**:
- To determine whether \( C \) is symmetric or skew-symmetric, we first find \( C^T \):
\[ C^T = (A^{-1} B A)^T \]
- Using the property of transposes, we have:
\[ C^T = A^T B^T (A^{-1})^T \]

4. **Using the Inverse Property**:
- Since \( (A^{-1})^T = (A^T)^{-1} \), we can rewrite \( C^T \):
\[ C^T = A^T B^T (A^T)^{-1} \]

5. **Substituting the Orthogonality Condition**:
- Since \( B \) is orthogonal, we have \( B^T = B^{-1} \). Therefore:
\[ C^T = A^T B^{-1} (A^T)^{-1} \]

6. **Simplifying Further**:
- We can express \( (A^T)^{-1} \) as \( (A^{-1})^T \):
\[ C^T = A^T B^{-1} (A^{-1})^T \]

7. **Rearranging the Terms**:
- We can rearrange the expression:
\[ C^T = A^T (A^{-1})^T B^{-1} \]

8. **Using the Identity Property**:
- Since \( A A^{-1} = I \), we have:
\[ C^T = I B^{-1} = B^{-1} \]

9. **Conclusion**:
- Since \( C^T = B^{-1} \) and \( B \) is orthogonal, it follows that \( C \) must also be orthogonal. Thus, we conclude that:
\[ C \text{ is an orthogonal matrix.} \]

### Final Answer:
- Therefore, \( C \) is an orthogonal matrix.

---