Verified Solution • 2025–26 Session moderate QJEEM24Ma2b9d: JEE Main 2024 Mathematics - Class 12 JEEM24Ma2b9d Solution: Let A be a matrix and . If , then the remainder when is divided … Let A be a 3×33 \times 33×3 matrix and det(A)=2\operatorname{det}(A)=2det(A)=2 . If n=det(adj(adj(…..(adjA))⏟2024− times ))n=\operatorname{det}(\underbrace{\operatorname{adj}(\operatorname{adj}(\ldots . .(\operatorname{adj} A))}_{2024-\text { times }}))n=det(2024− times adj(adj(…..(adjA)))) , then the remainder when nnn is divided by 9 is equal to __________. Step-by-Step Verified Solution amp;∣A∣=2amp;adj(adj(adj…..(a)))⏟2024 times =∣A∣(n−1)2024amp;=∣A∣2024amp;=222024\begin{aligned}& |\mathrm{A}|=2 \\& \underbrace{\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \ldots . .(\mathrm{a})))}_{2024 \text { times }}=|\mathrm{A}|^{(\mathrm{n}-1)^{2024}} \\& =|\mathrm{A}|^{2024} \\& =2^{2^{2024}}\end{aligned}amp;∣A∣=2amp;2024 times adj(adj(adj…..(a)))=∣A∣(n−1)2024amp;=∣A∣2024amp;=222024 amp;22024=(22)22022=4(8)674=4(9−1)674amp;⇒22024≡4( mod 9)amp;⇒22024≡9 m+4, m← even amp;29 m+4≡16⋅(23)3 m≡16( mod 9)amp;≡7\begin{aligned}& 2^{2024}=\left(2^2\right) 2^{2022}=4(8)^{674}=4(9-1)^{674} \\& \Rightarrow 2^{2024} \equiv 4(\bmod 9) \\& \Rightarrow 2^{2024} \equiv 9 \mathrm{~m}+4, \mathrm{~m} \leftarrow \text { even } \\& 2^{9 \mathrm{~m}+4} \equiv 16 \cdot\left(2^3\right)^{3 \mathrm{~m}} \equiv 16(\bmod 9) \\& \quad \equiv 7\end{aligned}amp;22024=(22)22022=4(8)674=4(9−1)674amp;⇒22024≡4(mod9)amp;⇒22024≡9 m+4, m← even amp;29 m+4≡16⋅(23)3 m≡16(mod9)amp;≡7 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring