Let A be a 3 3 matrix such that |adj(adj(adj A))|=81.

Question

Let $A$ be a $3 	imes 3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \mathrm{A}))|=81$.

If $S=\left\{n \in \mathbb{Z}:(|\operatorname{adj}(\operatorname{adj} A)|)^{\frac{(n-1)^2}{2}}=|A|^{\left(3 n^2-5 n-4ight)}ight\}$, then $\sum_\limits{n \in S}\left|A^{\left(n^2+night)}ight|$ is equal to :

Accepted Answer

$$\begin{aligned}
& |\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))|=81 \
& =|A|^{(n-1)^3}=(3)^4 \Rightarrow|A|^8=3^4 \Rightarrow|A|=3^{1 / 2} \
& |\operatorname{adj}(\operatorname{adj} A)|^{\frac{(n-1)^2}{2}}=|A|^{\left(3 n^2-5 n-4\right)} \
& {\left[|A|^{(n-1)^2}\right]^{\frac{(n-1)^2}{2}}=|A|^{3 n^2-5 n-4}} \
& |A|^{2(n-1)^2}=|A|^{3 n^2-5 n-4} \
& \Rightarrow 2(n-1)^2=3 n^2-5 n-4 \
& \quad n^2-n-6=0 \
& \Rightarrow n=-2,3 \
& \sum_{x \leftarrow 5}\left|A^{n^2+n}\right|=\left|A^2\right|+\left|A^{12}\right| \
& =3+3^6=732
\end{aligned}$$

Solutions · 1