If A is a square matrix of order 3 such that det(A)=3 and det(adj(-4 adj(-3 adj(3 adj((2 ~A)^ -1 )))))=2^ m 3^ n , then m+2 n is equal to :

Correct answer: D. 4 aligned & |A|=3 & adj(-4 adj(-3 adj(3 adj((2 ~A)^ -1 ))) . & -4 adj(-.3 adj(3 adj(2 ~A)^ -1 )|^2. & 4^6|adj(-3 adj(3 adj(2 ~A)^ -1 ))|^2 & 2^ 12 · 3^ 12 |3 adj(2 ~A)^ -1 |^8 & 2^ 12 · 3^ 12 · 3^ 24 |adj(2 ~A)^ -1 |^8 aligned aligned & 2^ 12 · 3^ 36 |(2 ~A)^ -1 |^ 16 & 2^ 12 · 3^ 36 1 |2 ~A|^ 16 & 2^ 12 · 3^ 36 1 2^ 48 |~A|^ 16 & 2^ 12 · 3^ 36 1 2^ 48 · 3^ 16 & 3^ 20 2^ 36 =2^ -36 · 3^ 20 & ~m=-36 n=20 & ~m+2 n=4 aligned

JEE Main 20246 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Matrices

If A is a square matrix of order 3 such that det(A)=3 and...

A

is a square matrix of order 3 such that

\operatorname{det}(A)=3

and

\operatorname{det}\left(\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 \mathrm{~A})^{-1}\right)\right)\right)\right)\right)=2^{\mathrm{m}} 3^{\mathrm{n}}

, then

\mathrm{m}+2 \mathrm{n}

is equal to :

Options

Correct answer

D. 4

Step-by-step solution

\begin{aligned} & |\mathrm{A}|=3 \\ & \mid \operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 \mathrm{~A})^{-1}\right)\right)\right) \mid\right. \\ & \mid-4 \operatorname{adj}\left(-\left.3 \operatorname{adj}\left(3 \operatorname{adj}(2 \mathrm{~A})^{-1}\right)\right|^2\right. \\ & 4^6\left|\operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}(2 \mathrm{~A})^{-1}\right)\right)\right|^2 \\ & 2^{12} \cdot 3^{12}\left|3 \operatorname{adj}(2 \mathrm{~A})^{-1}\right|^8 \\ & 2^{12} \cdot 3^{12} \cdot 3^{24}\left|\operatorname{adj}(2 \mathrm{~A})^{-1}\right|^8\end{aligned}

\begin{aligned} & 2^{12} \cdot 3^{36}\left|(2 \mathrm{~A})^{-1}\right|^{16} \\ & 2^{12} \cdot 3^{36} \frac{1}{|2 \mathrm{~A}|^{16}} \\ & 2^{12} \cdot 3^{36} \frac{1}{2^{48}|\mathrm{~A}|^{16}} \\ & 2^{12} \cdot 3^{36} \frac{1}{2^{48} \cdot 3^{16}} \\ & \frac{3^{20}}{2^{36}}=2^{-36} \cdot 3^{20} \\ & \mathrm{~m}=-36 \quad \mathrm{n}=20 \\ & \mathrm{~m}+2 \mathrm{n}=4\end{aligned}

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