Let A be a square matrix of order 3 such that (A)=-2 and (3adj(-6adj(3A)))=2^m+n3^mn, mn.

Question

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

Answer: 34

Accepted Answer

Using $\det(\lambda M)=\lambda^n\det(M)$ for $n\times n$ matrices and $\det(\operatorname{adj}M)=\det(M)^{n-1}$:

$$\det(3\operatorname{adj}(-6\operatorname{adj}(3A)))=3^3\det(\operatorname{adj}(-6\operatorname{adj}(3A)))=3^3\big(-6\operatorname{adj}(3A)\big)^{2}\text{-type scaling}=3^3\cdot(-6)^6\cdot|3A|^4.$$

Since $|3A|=3^3\cdot(-2)=-54$, and combining powers of 2 and 3 carefully as in the source working, the result simplifies to
$$3^{21}\cdot2^{10}=2^{m+n}\cdot3^{mn}.$$

So $m+n=10$ and $mn=21$, giving (with $m>n$) $m=7,\ n=3$.

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

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