Verified Solution • 2025–26 Session moderate

QJEEM24Ma613f: JEE Main 2024 Mathematics - Class 12 JEEM24Ma613f Solution: Let and . The sum of the prime factors of is equal to

Let A=[2amp;1amp;26amp;2amp;113amp;3amp;2]A=\left[\begin{array}{ccc}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{array}\right] and P=[1amp;2amp;05amp;0amp;27amp;1amp;5]P=\left[\begin{array}{lll}1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5\end{array}\right] . The sum of the prime factors of ∣P−1AP−2I∣\left|P^{-1} A P-2 I\right| is equal to

Multiple Choice Options

  • A 66
  • B 27
  • C 23
  • D 26

Step-by-Step Verified Solution

∣P−1AP−2I∣amp;=∣P−1AP−2P−1P∣amp;=∣P−1( A−2I)P∣amp;=∣P−1∣∣A−2I∣∣P∣amp;=∣A−2I∣amp;=∣0amp;1amp;26amp;0amp;113amp;3amp;0∣=69\begin{aligned}\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right| & =\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{P}^{-1} \mathrm{P}\right| \\& =\left|\mathrm{P}^{-1}(\mathrm{~A}-2 \mathrm{I}) \mathrm{P}\right| \\& =\left|\mathrm{P}^{-1}\right||\mathrm{A}-2 \mathrm{I}||\mathrm{P}| \\& =|\mathrm{A}-2 \mathrm{I}| \\& =\left|\begin{array}{ccc}0 & 1 & 2 \\6 & 0 & 11 \\3 & 3 & 0\end{array}\right|=69\end{aligned}

So, Prime factor of 69 is 3 & 23

So, sum = 26

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