CUET Preparation Today
If B is a non-singular matrix and A is a square matrix, then $det (B^{-1}AB)$ is equal to |
$det (A^{-1})$ $det (B^{-1})$ det (A) det (B) |
det (A) |
We have, $|B^{-1} AB|=|B^{-1}||A||B|$ $⇒|B^{-1} AB|=\frac{1}{|B|}|A||B|$ $\left[∵|B^{-1}|=\frac{1}{|B|}\right]$ $⇒|B^{-1} AB|=|A|$ |
