Let A be a 3 × 3 matrix such that adj ( adj ( adj A)) = 81. If S= \ n in mathbb Z :( adj ( adj A) ) frac (n-1) 2 2 = A (3 n 2 -5 n-4 ) \ , then sum n in S A (n 2 +n ) is equal to

Mathematics · Matrices

JEE Main 2025 — 7 April, Morning Shift — Question 27

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

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

equal to

Option A:
750
Option B:
866
Option C:
732
Correct
Option D:
820

Answer: C

Step-by-step solution

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

\begin{aligned} & =|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 \quad n=-2,3 \\& \sum_{x \leftarrow-S}\left|A^{n^{2}+n}\right|=\left|A^{2}\right|+\left|A^{12}\right| \\& =3+3^{6}=732 \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Mathematics
Chapter: Matrices
Topic: Adjoint of a Square Matrix