Let A be a 3 × 3 matrix and det (A)=2 . If n = det (underbrace adj ( adj (ldots ldots .( adj A) 2024- times ))) Then the remainder when n is divided by 9 is equal to

Mathematics · Matrices

JEE Main 2024 — 31 January, Shift 2 — Question 23

Let $A$ be a $3 \times 3$ matrix and $\operatorname{det}(A)=2$ .

If $\mathrm{n}=\operatorname{det}(\underbrace{\operatorname{adj}(\operatorname{adj}(\ldots \ldots .(\operatorname{adj} A)}_{2024-\text { times }})))$ Then the remainder when n is divided by 9 is equal to

Answer: 7

Numerical answer — enter this value.

Step-by-step solution

$|\mathrm{A}|=2$$

$\underbrace{\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \ldots..(\mathrm{a})))}=|\mathrm{A}|^{(\mathrm{n}1)^{2024}}$$2024times$

$\begin{aligned}&=|\mathrm{A}|^{2^{2024}}&=2^{2^{2024}}\end{aligned}$$

$2^{2024}=\left(2^{2}\right)2^{2022}=4(8)^{674}=4(91)^{674}$

$\Rightarrow2^{2024}\equiv4(\bmod9)$

$\Rightarrow2^{2024}\equiv9\mathrm{~m}+4,\mathrm{~m}\leftarrow$ even $2^{9\mathrm{~m}+4}\equiv16\cdot\left(2^{3}\right)^{3\mathrm{~m}} \equiv 16(\bmod 9)$

$\equiv 7$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Mathematics
Chapter: Matrices
Topic: Adjoint of a Square Matrix

Let A be a 3 × 3 real matrix such that $A\begin{pmatrix}1\\0\\1\end{pmatrix} =2\begin{pmatrix}0\\1\\1\end{pmatrix},\quad…

Answer: 7

Step-by-step solution

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