JEE MAIN - Mathematics (2024 - 29th January Evening Shift - No. 20)
Let $$A=\left[\begin{array}{ccc}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{array}\right]$$ and $$P=\left[\begin{array}{lll}1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5\end{array}\right]$$. The sum of the prime factors of $$\left|P^{-1} A P-2 I\right|$$ is equal to
66
27
23
26
Explanation
$$\begin{aligned} \left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right| & =\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{P}^{-1} \mathrm{P}\right| \\ & =\left|\mathrm{P}^{-1}(\mathrm{~A}-2 \mathrm{I}) \mathrm{P}\right| \\ & =\left|\mathrm{P}^{-1}\right||\mathrm{A}-2 \mathrm{I}||\mathrm{P}| \\ & =|\mathrm{A}-2 \mathrm{I}| \\ & =\left|\begin{array}{ccc} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{array}\right|=69 \end{aligned}$$
So, Prime factor of 69 is 3 & 23
So, sum = 26
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