Let $$ A $$ be a square matrix of order 3 such that $$det(A)=-2 \text{ and }det(3adj(-6adj(3A)))=2^{m+n}\cdot3^{mn}$$, $$m>n. \text{ Then } 4m+2n\text{ is equal to } $$_______

$$A$$ is a $$3 \times 3$$ matrix with $$\det(A) = -2$$. We need $$\det(3 \cdot \text{adj}(-6 \cdot \text{adj}(3A))) = 2^{m+n} \cdot 3^{mn}$$ with $$m > n$$.

For an $$n \times n$$ matrix (here $$n = 3$$):

$$\det(kA) = k^n \det(A)$$, $$\det(\text{adj}(A)) = (\det(A))^{n-1}$$, $$\text{adj}(kA) = k^{n-1} \text{adj}(A)$$

$$\det(3A) = 3^3 \det(A) = 27 \times (-2) = -54$$

$$\det(\text{adj}(3A)) = (\det(3A))^{3-1} = (-54)^2 = 2916$$

$$\det(-6 \cdot \text{adj}(3A)) = (-6)^3 \cdot \det(\text{adj}(3A)) = -216 \times 2916$$

$$= -216 \times 2916 = -629856$$

$$= (\det(-6 \cdot \text{adj}(3A)))^2 = (-629856)^2 = 629856^2$$

$$= 3^3 \times 629856^2 = 27 \times 629856^2$$

Now factorise: $$629856 = 216 \times 2916 = 6^3 \times 54^2 = (2 \cdot 3)^3 \times (2 \cdot 3^3)^2 = 2^3 \cdot 3^3 \cdot 2^2 \cdot 3^6 = 2^5 \cdot 3^9$$

$$629856^2 = 2^{10} \cdot 3^{18}$$

$$27 \times 629856^2 = 3^3 \times 2^{10} \times 3^{18} = 2^{10} \times 3^{21}$$

So $$2^{m+n} \cdot 3^{mn} = 2^{10} \cdot 3^{21}$$ with $$m > n$$.

$$m + n = 10$$ and $$mn = 21$$.

Solving: $$m$$ and $$n$$ are roots of $$t^2 - 10t + 21 = 0$$, giving $$t = 7$$ or $$t = 3$$.

Since $$m > n$$: $$m = 7, n = 3$$.

$$4m + 2n = 28 + 6 = 34$$.

The answer is 34.