Let A be a matrix of order 3 × 3 and |A|=5 . If |2 adj(3 ~A adj(2 ~A))|=2^α · 3^β · 5^γ α, β, γ ∈ N then α+β+γ is equal to

Correct answer: C. 27 aligned & |2 adj(3 ~A adj(2 ~A))| & 2^3 ·|3 ~A adj(2 ~A)|^2 & 2^3 ·(3^3)^2 ·|~A|^2 ·|adj(2 ~A)|^2 & 2^3 · 3^6 ·|~A|^2 ·(|2 ~A|^2)^2 & 2^3 · 3^6 ·|~A|^2[(2^3)^2 ·|~A|^2]^2 & 2^3 · 3^6 ·|~A|^2 · 2^ 12 ·|~A|^4 & 2^ 15 · 3^6 ·|~A|^6 & 2^ 15 · 3^6 · 5^6=2^α · 3^β · 5^γ & α=15, β=6, γ=6 & α+β+γ=27 aligned

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Matrices

Let A be a matrix of order 3 × 3 and |A|=5 . If |2 adj(3 ~A adj(2...

Let

A

be a matrix of order

3 \times 3

and

|A|=5

. If

|2 \operatorname{adj}(3 \mathrm{~A} \operatorname{adj}(2 \mathrm{~A}))|=2^\alpha \cdot 3^\beta \cdot 5^\gamma \alpha, \beta, \gamma \in \mathrm{N}

then

\alpha+\beta+\gamma

is equal to

Options

A $25$
B $26$
C $27$
D $28$

Correct answer

C. $27$

Step-by-step solution

\begin{aligned} & |2 \operatorname{adj}(3 \mathrm{~A} \operatorname{adj}(2 \mathrm{~A}))| \\ & 2^3 \cdot|3 \mathrm{~A} \operatorname{adj}(2 \mathrm{~A})|^2 \\ & 2^3 \cdot\left(3^3\right)^2 \cdot|\mathrm{~A}|^2 \cdot|\operatorname{adj}(2 \mathrm{~A})|^2 \\ & 2^3 \cdot 3^6 \cdot|\mathrm{~A}|^2 \cdot\left(|2 \mathrm{~A}|^2\right)^2 \\ & 2^3 \cdot 3^6 \cdot|\mathrm{~A}|^2\left[\left(2^3\right)^2 \cdot|\mathrm{~A}|^2\right]^2 \\ & 2^3 \cdot 3^6 \cdot|\mathrm{~A}|^2 \cdot 2^{12} \cdot|\mathrm{~A}|^4 \\ & 2^{15} \cdot 3^6 \cdot|\mathrm{~A}|^6 \\ & 2^{15} \cdot 3^6 \cdot 5^6=2^\alpha \cdot 3^\beta \cdot 5^\gamma \\ & \alpha=15, \quad \beta=6, \quad \gamma=6 \\ & \alpha+\beta+\gamma=27\end{aligned}

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