Mathematics · MatricesJEE Main 2025 — 3 April, Morning Shift — Question 20Let A be a matrix of order 3×33 \times 33×3 and ∣A∣=5|\mathrm{A}|=5∣A∣=5. If ∣2adj(3 Aadj(2 A))∣=2α.3β.5γα,β,γ∈N|2 \operatorname{adj}(3 \mathrm{~A} \operatorname{adj}(2 \mathrm{~A}))|=2^{\alpha} .3^{\beta} .5^{\gamma} \alpha, \beta, \gamma \in \mathrm{N}∣2adj(3 Aadj(2 A))∣=2α.3β.5γα,β,γ∈N then α+β+γ\alpha+\beta+\gammaα+β+γ is equal toAOption A: 25BOption B: 26COption C: 27CorrectDOption D: 28Answer: CStep-by-step solution∣2\mid 2∣2 adj (3 Aadj(2 A))∣(3 \mathrm{~A} \operatorname{adj}(2 \mathrm{~A})) \mid(3 Aadj(2 A))∣ 23.∣3 Aadj(2 A)∣22^{3} .|3 \mathrm{~A} \operatorname{adj}(2 \mathrm{~A})|^{2}23.∣3 Aadj(2 A)∣2 23⋅(33)2⋅∣ A∣2⋅∣adj(2 A)∣22^{3} \cdot\left(3^{3}\right)^{2} \cdot|\mathrm{~A}|^{2} \cdot|\operatorname{adj}(2 \mathrm{~A})|^{2}23⋅(33)2⋅∣ A∣2⋅∣adj(2 A)∣2 23⋅36⋅∣ A∣2⋅(∣2 A∣2)22^{3} \cdot 3^{6} \cdot|\mathrm{~A}|^{2} \cdot\left(|2 \mathrm{~A}|^{2}\right)^{2}23⋅36⋅∣ A∣2⋅(∣2 A∣2)2 23⋅36⋅∣ A∣2[(23)2⋅∣ A∣2]22^{3} \cdot 3^{6} \cdot|\mathrm{~A}|^{2}\left[\left(2^{3}\right)^{2} \cdot|\mathrm{~A}|^{2}\right]^{2}23⋅36⋅∣ A∣2[(23)2⋅∣ A∣2]2 23⋅36⋅∣ A∣2⋅212⋅∣ A∣42^{3} \cdot 3^{6} \cdot|\mathrm{~A}|^{2} \cdot 2^{12} \cdot|\mathrm{~A}|^{4}23⋅36⋅∣ A∣2⋅212⋅∣ A∣4 215.36.∣A∣62^{15} .3^{6} .|\mathrm{A}|^{6}215.36.∣A∣6 215⋅36⋅56=2α⋅3β⋅5γ2^{15} \cdot 3^{6} \cdot 5^{6}=2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma}215⋅36⋅56=2α⋅3β⋅5γ α=15,β=6,γ=6\alpha=15, \quad \beta=6, \quad \gamma=6α=15,β=6,γ=6 α+β+γ=27\alpha+\beta+\gamma=27α+β+γ=27Answer key and solution verified before publishing.Practise MatricesStart with this question, then two more from the same chapter — with a tutor that explains every step. Free.Solve a similar one free→ExamJEE Main 2025Paper3 April, Morning ShiftSubjectMathematicsChapterMatricesTopicAdjoint of a Square Matrix← Question 19Consider the following reactions A + NaCl + H 2 SO 4 xrightarrow Little amount CrO 2 Cl 2 + Side Products CrO 2 Cl 2( vapour) + NaOH arrow…Question 21 →Let a line passing through the point (4,1,0) intersect the line L 1 ; x-1/2=y-2/3=z-3/4 at the point A (alpha, beta, gamma) and the line L…