If B is a non-singular matrix and A is a square matrix, then det (B−1 AB) is equal to ___________ .

Question

KollegeApply · Accepted Answer

Det (A) B is non - singular. 	ext{ This implies that }\left| B ight| 
eq 0,	ext{ that B is invertible and that }B^{- 1}	ext{ exists }. Here, B is invertible . 	herefore \left| B^{- 1} ight| = \left| B ight|^{- 1} = \frac{1}{\left| B ight|} \Rightarrow \left| B^{- 1} AB ight| = \left| B^{- 1} ight|\left| AB ight| \Rightarrow \left| B^{- 1} AB ight| = \left| B ight|^{- 1} \left| A ight|\left| B ight| \Rightarrow \left| B^{- 1} AB ight| = \frac{1}{\left| B ight|}\left| A ight|\left| B ight| \Rightarrow \left| B^{- 1} AB ight| = \left| A ight|

AChoose Your Answer