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Question
If B is a non-singular matrix and A is a square matrix, then det (B−1 AB) is equal to ___________ .
Options
Det (A−1)
Det (B−1)
Det (A)
Det (B)
Solution
Det (A)
B is non - singular.
\[\text{ This implies that }\left| B \right| \neq 0,\text{ that B is invertible and that }B^{- 1}\text{ exists }. \]
Here, B is invertible .
\[ \therefore \left| B^{- 1} \right| = \left| B \right|^{- 1} = \frac{1}{\left| B \right|}\]
\[ \Rightarrow \left| B^{- 1} AB \right| = \left| B^{- 1} \right|\left| AB \right|\]
\[ \Rightarrow \left| B^{- 1} AB \right| = \left| B \right|^{- 1} \left| A \right|\left| B \right| \]
\[ \Rightarrow \left| B^{- 1} AB \right| = \frac{1}{\left| B \right|}\left| A \right|\left| B \right| \]
\[ \Rightarrow \left| B^{- 1} AB \right| = \left| A \right|\]
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