JEE Main 2024 (Online) 6th April Evening ShiftQuestion 1
If $A$ is a square matrix of order $3$ such that $\det(A)=3$ and $\det\left(\operatorname{adj}\left(-4\operatorname{adj}\left(-3\operatorname{adj}\left(3\operatorname{adj}\left((2A)^{-1}\right)\right)\right)\right)\right)=2^{m}3^{n}$, then $m+2n$ is equal to :
Determinant
Solution
Order 3: $|\operatorname{adj}B|=|B|^2$, $|kB|=k^3|B|$. $|(2A)^{-1}|=\frac1{24}$. Then $\operatorname{adj}$: $\frac1{24^2}$; times $3$: $27\cdot24^{-2}$; adj: $27^2\cdot24^{-4}$; times $-3$: $-27\cdot27^2\cdot24^{-4}$; adj: $27^6\cdot24^{-8}$; times $-4$: $-64\cdot27^6\cdot24^{-8}$; adj: $2^{12}\,27^{12}\,24^{-16}$. $=2^{12}3^{36}\cdot2^{-48}3^{-16}=2^{-36}3^{20}$. So $m=-36,\ n=20$, $m+2n=4$. Option B.