Question

If A and B are invertible matrices of order $3$. $\left|A\right|=2$ and $\left|{\left(AB\right)}^{-1}\right|=-\frac{1}{6}$. Find $\left|B\right|$

Answer 1

$\left|A\right|=2;\left|{\left(AB\right)}^{-1}\right|=-\frac{1}{6}\Rightarrow \left|B^{-1}\right|\left|A^{-1}\right|=-\frac{1}{6}....\left(1\right)$

$\left|A\right|\left|A^{-1}\right|=1\Rightarrow \left|A^{-1}\right|=\frac{1}{\left|A\right|}=\frac{1}{2}.....\left(2\right)$

from (1) and (2)

$\left|B^{-1}\right|\times \frac{1}{2}=-\frac{1}{6}\Rightarrow \left|B^{-1}\right|=-\frac{1}{3}....\left(3\right)$

$\left|B\right|\left|B^{-1}\right|=\mathrm{1...}\left(4\right)$

from (3) and (4)

$\left|B\right|(-\frac{1}{3})=1\Rightarrow \left|B\right|=-3$