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Question

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

The correct answer is

732

\(|\operatorname{adj}(\operatorname{adj})(\operatorname{adj} A)|=81\)

\(\Rightarrow|\operatorname{adj} \mathrm{A}|^{4}=81\)

\(\Rightarrow|\operatorname{adj} A|=3\)

\(\Rightarrow|A|^{2}=3\)

\(\Rightarrow|\mathrm{A}|=\sqrt{3}\)

\(\left(|A|^{4}\right)^{\frac{(\mathrm{n}-1)^{2}}{2}}=|\mathrm{A}|^{3 n^{2}-5 n-4}\)

\(\Rightarrow 2(\mathrm{n}-1)^{2}=3 \mathrm{n}^{2}-5 \mathrm{n}-4\)

\(\Rightarrow 2 \mathrm{n}^{2}-4 \mathrm{n}+2=3 \mathrm{n}^{2}-5 \mathrm{n}-4\)

\(\Rightarrow \mathrm{n}^{2}-\mathrm{n}-6=0\)

\(\Rightarrow(\mathrm{n}-3)(\mathrm{n}+2)=0\)

\(\Rightarrow \mathrm{n}=3,-2\)

\(\sum\limits_{\mathrm{n} \in \mathrm{S}}\left|\mathrm{A}^{\mathrm{n}^{2}+\mathrm{n}}\right|\)

\(=\left|\mathrm{A}^{2}\right|+\left|\mathrm{A}^{12}\right|\)

= 3 + 36 = 3 + 729 = 732  

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