Let A be a 3 × 3 matrix such that aligned & |adj(adj(adj A))|=81 . If & S=\ n ∈ ℤ:(|adj(adj A)|)^ ((n-1)^2)/(2) =|A|^ (3 n^2-5 n-4) \ aligned , then ∑_ n ∈ S |A^ (n^2+n) | is equal to

Correct answer: D. 732 aligned & |adj(adj)(adjA)|=81 & |adjA|^4=81 & |adjA|=3 & |~A|^2=3 & |~A|= 3 & (|~A|^4)^ (n-1)^2 2 =|A|^ n^2-5 n-4 & 2(n-1)^2=3 n^2-5 n-4 & 2 n^2-4 n+2=3 n^2-5 n-4 & n^2-n-6=0 & (n-3)(n+2)=0 & n=3,-2 & ∑_ n ∈ s |A^ n^2+n | & =|A^2|+|A^ 12 | & =3+36=3+729=732 aligned

JEE Main 20257 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Matrices

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

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

Options

A866
B750
C820
D732

Correct answer

D. 732

Step-by-step 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}

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