Let A be a 3 × 3 matrix such that |adj (adj (adj A))| = 81. If S = {n ∈ Z : (|adj(adj A)|)^((n-1)^2/2) = |A|^(3n^2-5n-4)},

Question

Let A be a \(3 \times 3\) matrix such that \(|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \mathrm{A}))|=81.\) If \(\mathrm{S}=\left\{\mathrm{n} \in \mathbb{Z}:(|\operatorname{adj}(\operatorname{adj} A)|)^{\frac{(n-1)^{2}}{2}}=|A|^{\left(3 n^{2}-5 n-4\right)}\right\}\) then \(\sum\limits_{n \in S}\left|A^{\left(n^{2}+n\right)}\right|\) is equal to

(1) 866

(2) 750

(3) 820

(4) 732 

Answer 1

Correct option is: (4) 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