Let $A$ be a square matrix of order $3$ such that $\det(A)=-2$ and $\det\bigl(3\operatorname{adj}(-6\operatorname{adj}(3A))\bigr)=2^{m+n}\cdot 3^{mn},\ m>n$. Then $4m+2n$ equals \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: Successively apply $|\operatorname{adj}(M)|=|M|^{n-1}$, $|kM|=k^{n}|M|$, $\operatorname{adj}(kM)=k^{n-1}\operatorname{adj}(M)$ for $3\times 3$ matrices. Compute the determinant entirely in scalar factors.
For $3\times 3$: $\operatorname{adj}(3A)=3^{2}\operatorname{adj}(A)=9\operatorname{adj}(A).$
$-6\operatorname{adj}(3A)=-54\operatorname{adj}(A).$
$\operatorname{adj}(-54\operatorname{adj}(A))=(-54)^{2}\operatorname{adj}(\operatorname{adj}(A))=2916\cdot|A|\,A=2916(-2)A=-5832\,A.$
$3\cdot\operatorname{adj}(-6\operatorname{adj}(3A))=-17496\,A.$
$\det(-17496\,A)=(-17496)^{3}\det(A)=-17496^{3}\cdot(-2)=2\cdot 17496^{3}.$
$17496=2^{3}\cdot 3^{7}\Rightarrow 17496^{3}=2^{9}\cdot 3^{21}.$
So $\det=2^{10}\cdot 3^{21}=2^{m+n}\cdot 3^{mn}\Rightarrow m+n=10,\ mn=21\Rightarrow (m,n)=(7,3).$
$4m+2n=28+6=34.$
Correct Answer: 34