Let $A$ be a $3 \times 3$ matrix such that $\begin{aligned} &…

Let $A$ be a $3 \times 3$ matrix such that
$\begin{aligned}
& |\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \mathrm{A}))|=81 . \text { If } \\ & \mathrm{S}=\left\{\mathrm{n} \in \mathbb{Z}:(|\operatorname{adj}(\operatorname{adj} A)|)^{\frac{(n-1)^2}{2}}=|A|^{\left(3 n^2-5 n-4\right)}\right\}
\end{aligned}$
, then $\sum_{n \in S}\left|A^{\left(n^2+n\right)}\right|$ is equal to
  1. 866
  2. 750
  3. 820
  4. 732

Solution

$\begin{aligned} & |\operatorname{adj}(\operatorname{adj})(\operatorname{adjA})|=81 \\ & \Rightarrow|\operatorname{adjA}|^4=81 \\ & \Rightarrow|\operatorname{adjA}|=3 \\ & \Rightarrow|\mathrm{~A}|^2=3 \\ & \Rightarrow|\mathrm{~A}|=\sqrt{3} \\ & \left(|\mathrm{~A}|^4\right)^{\frac{(\mathrm{n}-1)^2}{2}}=|\mathrm{A}|^{\mathrm{n}^2-5 \mathrm{n}-4} \\ & \Rightarrow 2(\mathrm{n}-1)^2=3 \mathrm{n}^2-5 \mathrm{n}-4 \\ & \Rightarrow 2 \mathrm{n}^2-4 \mathrm{n}+2=3 \mathrm{n}^2-5 \mathrm{n}-4 \\ & \Rightarrow \mathrm{n}^2-\mathrm{n}-6=0 \\ & \Rightarrow(\mathrm{n}-3)(\mathrm{n}+2)=0 \\ & \Rightarrow \mathrm{n}=3,-2 \\ & \sum_{\mathrm{n} \in \mathrm{s}}\left|\mathrm{A}^{\mathrm{n}^2+\mathrm{n}}\right| \\ & =\left|\mathrm{A}^2\right|+\left|\mathrm{A}^{12}\right| \\ & =3+36=3+729=732\end{aligned}$

Asked in: JEE Main 2025 (07 Apr Shift 1)

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