Question

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. $\det \left( {{A^{ - 1}}} \right)$

Answer 1

- 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.