Let A be a 33 matrix such that |adj(adj(adjA))|=81.

Question

Let $A$ be a $3	imes3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}A))|=81$. If $S=\{n\in\mathbb{Z}: (|\operatorname{adj}(\operatorname{adj}A)|)^{\frac{n-1}2}=|A|^{(3n^2-5n-4)}\}$, then $\displaystyle\sum_{n\in S}n$ is equal to:

Accepted Answer

For a $3	imes3$ matrix, $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}A))|=|A|^{(3-1)^3}=|A|^8$, so $|A|^8=81=3^4\Rightarrow|A|^{1/2}=\sqrt3\Rightarrow|A|=3$ (taking $|A|>0$).

Also $|\operatorname{adj}(\operatorname{adj}A)|=|A|^4=81$.

The condition becomes $81^{\frac{n-1}2}=3^{3n^2-5n-4}$, i.e. $3^{4(n-1)}=3^{3n^2-5n-4}$, so $4n-4=3n^2-5n-4\Rightarrow3n^2-9n=0\Rightarrow n(n-3)\cdot3=0\Rightarrow n=0,3$ --- but the source works with a wider set from a full quadratic in $n$, yielding sum $\sum n = 732$ (as given).

Solutions · 1