If B is a non-singular matrix and A is a square matrix,


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)


(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.


$\Rightarrow\left|B^{-1} A B\right|=\left|B^{-1}\right||A B|$

$\Rightarrow\left|B^{-1} A B\right|=|B|^{-1}|A||B|$

$\Rightarrow\left|B^{-1} A B\right|=\frac{1}{|B|}|A||B|$

$\Rightarrow\left|B^{-1} A B\right|=|A|$

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