Q. If $B$ is a non-singular matrix and $A$ is a square matrix such that $B^{- 1}AB$ exists, then det $(B^{- 1}AB)$ is equal to
A) $det(A^{- 1})$
B) $det(B^{- 1})$
C) $det(B)$
D) $det(A)$
Explanation
(d)
Given, $A = B^{- 1}\ AB$
$\Rightarrow BA = AB$
$\therefore\det\left( B^{- 1}AB \right) = \det\left( B^{- 1}BA \right) = det(A)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.