CUET Preparation Today
If B is a non-singular matrix and A is a square matrix, then $det (B^{-1}AB)$ = |
$det(A^{-1})$ $det(B^{-1})$ $det(A)$ $det(B)$ |
$det(A)$ |
The correct answer is Option (3) → $det(A)$ $|B^{-1}AB|=|B^{-1}||A||B|$ $[|AB|=|A||B|]$ $=\frac{1}{|B|}|A||B|$ $[|B^{-1}|=\frac{1}{|B|},|B|≠0]$ $=|A|$ |
