If \[B\] is a nonsingular matrix and \[A\] is a square matrix then \[\det \left( {{B^{ - 1}}AB} \right)\] is equal to
A. \[\det \left( A \right)\]
B. \[\det \left( B \right)\]
C. \[\det \left( {{B^{ - 1}}} \right)\]
D. \[\det \left( {{A^{ - 1}}} \right)\]

- Hint: First of all, split the matrices inside the determinants by using the multiplicative properties of determinants. Then make the product of two matrices equal to a unit matrix to obtain the required answer.

Complete step-by-step solution -

Given \[B\] is a nonsingular matrix and \[A\] is a square matrix.
Now, consider \[\det \left( {{B^{ - 1}}AB} \right)\]
We know that \[\det \left( {ABC} \right) = \det \left( A \right)\det \left( B \right)\det \left( C \right)\], so we have
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( {{B^{ - 1}}} \right)\det \left( A \right)\det \left( B \right)\]
As determinants obeys commutative property of multiplication, we have
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( {{B^{ - 1}}} \right)\det \left( B \right)\det \left( A \right)\]
We know that, \[\det \left( {AB} \right) = \det \left( A \right)\det \left( B \right)\]
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( {{B^{ - 1}}B} \right)\det \left( A \right)\]
As \[{B^{ - 1}}B = I\], we have
\[\det \left( {{B^{ - 1}}AB} \right) = \det \left( I \right)\det \left( A \right)\]
We know that, \[\det \left( I \right) = 1\]
\[\therefore \det \left( {{B^{ - 1}}AB} \right) = \det \left( A \right)\]
Thus, the correct option is A. \[\det \left( A \right)\]
Note: The multiplicative property of determinants tells us that the det of product of the given matrices is equal to their product of individual det of matrices. The product of a matrix and its inverse matrix gives us a unit matrix.