If A and B are square matrices of order 3 such that `det. (A) = -2` and `det.(B)= 1`, then `det.(A^(-1)adjB^(-1).adj(2A^(-1))` is equal to

Explanation

det. `(A^(-1)" adj. "(B^(-1))**"adj. "(2A^(-1)))`
`=" det."(A^(-1). B/(|B|). (2^(2)" adj. "(A^(-1))))" "["as adj. "(kA)=k^(n-1) ("adj. A")]`
`=" det."(A^(-1))." det. "(B/(|B|))." det. " ((4A)/(|A|))`
`=(1)/("det. (A)") . 1/(|B|^(3))." det. "(B)." det. "(-2A)`
`=1/(-2) . (-2)^(3)" det. "(A)`
`=1/(-2) . (-2)^(3) (-2) = -8`