If A is a square matrix of order 3 such that det(A)=3 and det ( adj (-4 adj (-3 adj (3 adj ((2 ~A)^-1 ) ) ) ) )=2^ m 3^ n, then m+2 n is equal to :

$\begin{aligned} & |\mathrm{A}|=3 \\ & \mid \operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 \mathrm{~A})^{-1}\right)\right)\right) \mid\right. \\ & \mid-4 \operatorname{adj}\left(-\left.3 \operatorname{adj}\left(3 \operatorname{adj}(2 \mathrm{~A})^{-1}\right)\right|^2\right. \\ & 4^6\left|\operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}(2 \mathrm{~A})^{-1}\right)\right)\right|^2 \\ & 2^{12} \cdot 3^{12}\left|3 \operatorname{adj}(2 \mathrm{~A})^{-1}\right|^8 \\ & 2^{12} \cdot 3^{12} \cdot 3^{24}\left|\operatorname{adj}(2 \mathrm{~A})^{-1}\right|^8\end{aligned}$​∣A∣=3∣adj(−4adj(−3adj(3adj((2 A)−1)))∣∣−4adj(−3adj(3adj(2 A)−1)​246​adj(−3adj(3adj(2 A)−1))​2212⋅312​3adj(2 A)−1​8212⋅312⋅324​adj(2 A)−1​8​ $\begin{aligned} & 2^{12} \cdot 3^{36}\left|(2 \mathrm{~A})^{-1}\right|^{16} \\ & 2^{12} \cdot 3^{36} \frac{1}{|2 \mathrm{~A}|^{16}} \\ & 2^{12} \cdot 3^{36} \frac{1}{2^{48}|\mathrm{~A}|^{16}} \\ & 2^{12} \cdot 3^{36} \frac{1}{2^{48} \cdot 3^{16}} \\ & \frac{3^{20}}{2^{36}}=2^{-36} \cdot 3^{20} \\ & \mathrm{~m}=-36 \quad \mathrm{n}=20 \\ & \mathrm{~m}+2 \mathrm{n}=4\end{aligned}$​212⋅336​(2 A)−1​16212⋅336∣2 A∣161​212⋅336248∣ A∣161​212⋅336248⋅3161​236320​=2−36⋅320 m=−36n=20 m+2n=4​