If A and B are invertible matrices of order 3, |A| = 2 and ≤ft|(A B)^…

$|A|=2$

$\left|(A B)^{-1}\right|=-\frac{1}{6}$

so $\left|(A B)^{-1}\right|=-\frac{1}{6}$

$=\left|B^{-1} A^{-1}\right|=\frac{-1}{6}$    (as (AB)^-1 = B^-1 A^-1)

We know that

$A A^{-1}=I$

so  $\left|A A^{-1}\right|=|I|$

so  $|A|=\frac{1}{|A^{-1}|}$

or  $|A^{-1}|=\frac{1}{|A|}$

$\Rightarrow \left|B^{-1}\right||A-1|=-\frac{1}{6}$

$\Rightarrow \frac{1}{|A||B|}=-\frac{1}{6}$

So  $\frac{1}{2|B|}=-\frac{1}{6}$

So  |B| = -3