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QJEEM24Ma884a: JEE Main 2024 Mathematics - Class 12 JEEM24Ma884a Solution: If is a square matrix of order 3 such that and , then is equal…

If AA is a square matrix of order 3 such that det⁡(A)=3\operatorname{det}(A)=3 and det⁡(adj⁡(−4adj⁡(−3adj⁡(3adj⁡((2 A)−1)))))=2m3n\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 m+2n\mathrm{m}+2 \mathrm{n} is equal to :

Multiple Choice Options

  • A 2
  • B 4
  • C 3
  • D 6

Step-by-Step Verified Solution

amp;∣A∣=3amp;∣adj⁡(−4adj⁡(−3adj⁡(3adj⁡(2A)−1)))∣amp;=∣−4adj⁡(−3adj⁡(3adj⁡((2A)−1)))∣2\begin{aligned}& |A|=3 \\& \left|\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}(2 A)^{-1}\right)\right)\right)\right| \\& =\left|-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 A)^{-1}\right)\right)\right)\right|^2\end{aligned}

amp;=46∣−3adj⁡(aadj⁡((2A)−1))∣4amp;=46⋅312∣aadj⁡((2A)−1)∣8amp;=46⋅312⋅324∣(2A)−1∣16amp;=46⋅3362−48∣A−1∣16amp;=2−36336316=2−36320amp;m=−36amp;n=20amp;m+2n=4\begin{aligned}& =4^6\left|-3 \operatorname{adj}\left(\operatorname{aadj}\left((2 A)^{-1}\right)\right)\right|^4 \\& =4^6 \cdot 3^{12}\left|\operatorname{aadj}\left((2 A)^{-1}\right)\right|^8 \\& =4^6 \cdot 3^{12} \cdot 3^{24}\left|(2 A)^{-1}\right|^{16} \\& =4^6 \cdot 3^{36} 2^{-48}\left|A^{-1}\right|^{16} \\& =\frac{2^{-36} 3^{36}}{3^{16}}=2^{-36} 3^{20} \\& m=-36 \\& n=20 \\& m+2 n=4\end{aligned}

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