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# If $B$ is a non-singular matrix and A is a square matrix then $det(B^{-1}AB)$ is equal to

$\begin{array}{1 1}(a)\;det(A^{-1})&(b)\;det(B^{-1})\\(c)\;det(A)&(d)\;det(B)\end{array}$

$det(B^{-1}AB)=det(B^{-1})det A \;det B$
$\Rightarrow det \;(B^{-1}).det\; B.det \;A=det (B^{-1}B).det \;A$
$\Rightarrow det(z).det A=1.det A$
$\Rightarrow det \;A$
Hence (c) is the correct answer.