Question

If A and B are invertible matrices of order 2, |A| =2 and $|(AB{)}^{-1}|=-\frac{1}{6}$, find |B|.

Answer 1

As $|P^{-1}|=\frac{1}{|P|}\therefore |(AB{)}^{-1}|=\frac{1}{|AB|}=\frac{1}{|A||B|}=\frac{1}{2\times |B|}$

Also since $|(AB{)}^{-1}|=-\frac{1}{6}$ so, $\frac{1}{2\times |B|}=-\frac{1}{6}\therefore |B|=-3$.