Question:
If $B$ is a non-singular matrix and $A$ is a square matrix, then $\operatorname{det}\left(B^{-1} A B\right)$ is equal to
(a) Det $\left(A^{-1}\right)$
(b) Det $\left(B^{-1}\right)$
(c) Det (A)
(d) Det (B)
Solution:
(c) $\operatorname{Det}(A)$
$B$ is non-singular.
This implies that $|B| \neq 0$, that $B$ is invertible and that $B^{-1}$ exists.
Here, $B$ is invertible.
$\therefore\left|B^{-1}\right|=|B|^{-1}=\frac{1}{|B|}$
$\Rightarrow\left|B^{-1} A B\right|=\left|B^{-1}\right||A B|$
$\Rightarrow\left|B^{-1} A B\right|=|B|^{-1}|A||B|$
$\Rightarrow\left|B^{-1} A B\right|=\frac{1}{|B|}|A||B|$
$\Rightarrow\left|B^{-1} A B\right|=|A|$