Verified Solution • 2025–26 Session moderate

QJEEM23Ma74eb: JEE Main 2023 Mathematics - Class 12 JEEM23Ma74eb Solution: Let . If , then is equal to :

Let A=[2amp;1amp;01amp;2amp;−10amp;−1amp;2]A=\left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right] . If ∣adj⁡(adj⁡(adj⁡2A))∣=(16)n|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))|=(16)^{n} , then nn is equal to :

Multiple Choice Options

  • A 9
  • B 8
  • C 10
  • D 12

Step-by-Step Verified Solution

We have,

amp;∣A∣=∣2amp;1amp;01amp;2amp;−10amp;−1amp;2∣=2(4−1)−1(2−0)+0amp;=6−2=4amp; So, ∣2 A∣=23∣ A∣=8×4=32amp; Now, ∣adj⁡(adj⁡(adj⁡2 A))∣=∣2 A∣(n−1)3amp;=(32)23=328amp;⇒16n=(32)8=28×168amp;⇒16n=162+8⇒n=10\begin{aligned}& |\mathrm{A}|=\left|\begin{array}{ccc}2 & 1 & 0 \\1 & 2 & -1 \\0 & -1 & 2\end{array}\right|=2(4-1)-1(2-0)+0 \\\\& =6-2=4 \\\\& \text { So, }|2 \mathrm{~A}|=2^3|\mathrm{~A}|=8 \times 4=32 \\\\& \text { Now, }|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 \mathrm{~A}))|=|2 \mathrm{~A}|^{(n-1)^3} \\\\& =(32)^{2^3}=32^8 \\\\& \Rightarrow 16^n=(32)^8=2^8 \times 16^8 \\\\& \Rightarrow 16^n=16^{2+8} \Rightarrow n=10\end{aligned}

Concepts :

(a) ∣k A∣=kn∣ A∣|k \mathrm{~A}|=k^n|\mathrm{~A}|

(b) ∣adj⁡A∣=∣A∣n−1|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{n-1}

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