$A$ and $B$ are invertible matrices of the same order such that $\left|(A B)^{-1}\right|=8$ if $|A|=2$ then $|B|$ is

Given that $A$ and $B$ are invertible matrices of the same order, we have the property $|(AB)^{-1}| = \dfrac{1}{|AB|}$.

Since $|AB| = |A| |B|$, we can write $|(AB)^{-1}| = \dfrac{1}{|A| |B|}$.

Given $|(AB)^{-1}| = 8$ and $|A| = 2$, we substitute these values into the equation:

$8 = \dfrac{1}{2 \times |B|}$

Rearranging the equation to solve for $|B|$:

$|B| = \dfrac{1}{2 \times 8} = \dfrac{1}{16}$

Answer: $\dfrac{1}{16}$