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(C) Given $|A| = -2$ and order $n = 3$. We know $\operatorname{det}(k A) = k^n \operatorname{det}(A)$ and $\operatorname{det}(\operatorname{adj}(B)) =

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$34$

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(C) Given $|A| = -2$ and order $n = 3$. We know $\operatorname{det}(k A) = k^n \operatorname{det}(A)$ and $\operatorname{det}(\operatorname{adj}(B)) = (\operatorname{det}(B))^{n-1}$. First,$\operatorname{det}(3A) = 3^3 \operatorname{det}(A) = 27(-2) = -54$. Next,$\operatorname{det}(\operatorname{adj}(3A)) = (-54)^{3-1} = (-54)^2 = 54^2 = (2 \cdot 3^3)^2 = 2^2 \cdot 3^6$. Now,$\operatorname{det}(-6 \operatorname{adj}(3A)) = (-6)^3 \operatorname{det}(\operatorname{adj}(3A)) = (-2^3 \cdot 3^3) \cdot (2^2 \cdot 3^6) = -2^5 \cdot 3^9$. Then,$\operatorname{det}(\operatorname{adj}(-6 \operatorname{adj}(3A))) = (-2^5 \cdot 3^9)^{3-1} = (-2^5 \cdot 3^9)^2 = 2^{10} \cdot 3^{18}$. Finally,$\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3A))) = 3^3 \cdot \operatorname{det}(\operatorname{adj}(-6 \operatorname{adj}(3A))) = 3^3 \cdot 2^{10} \cdot 3^{18} = 2^{10} \cdot 3^{21}$. Comparing with $2^{m+n} \cdot 3^{mn}$,we get $m+n = 10$ and $mn = 21$. Since $m > n$,we have $m = 7$ and $n = 3$. Thus,$4m + 2n = 4(7) + 2(3) = 28 + 6 = 34$.

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

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