If B is a non-singular matrix and A is a square matrix, then `det (B^(-1) AB)` is equal to (A) `det (A^(-1))` (B) `det (B^(-1))` (C) `det (A)` (D) `det (B)`

Options

A. `det(A^(-1))`
B. `det(B^(-1))`
C. `det(A)`
D. `det(B)`

Explanation

To solve the problem, we need to find the value of \( \det(B^{-1}AB) \) given that \( B \) is a non-singular matrix and \( A \) is a square matrix.

### Step-by-Step Solution:

1. **Understanding the Determinant Property**:
We know that for any two square matrices \( M \) and \( N \), the determinant of their product is the product of their determinants:
\[ \det(MN) = \det(M) \cdot \det(N) \]

2. **Applying the Property**:
We can apply this property to our expression \( \det(B^{-1}AB) \):
\[ \det(B^{-1}AB) = \det(B^{-1}) \cdot \det(A) \cdot \det(B) \]

3. **Using the Determinant of the Inverse**:
We also know that the determinant of the inverse of a matrix \( B \) is given by:
\[ \det(B^{-1}) = \frac{1}{\det(B)} \]

4. **Substituting the Inverse Determinant**:
Now, substituting \( \det(B^{-1}) \) in our expression:
\[ \det(B^{-1}AB) = \left(\frac{1}{\det(B)}\right) \cdot \det(A) \cdot \det(B) \]

5. **Simplifying the Expression**:
The \( \det(B) \) in the numerator and denominator cancels out:
\[ \det(B^{-1}AB) = \det(A) \]

6. **Conclusion**:
Thus, we find that:
\[ \det(B^{-1}AB) = \det(A) \]
Therefore, the answer is \( \det(A) \), which corresponds to option (C).