Matrices Question 7
Question 7 - 2024 (29 Jan Shift 2)
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 $\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right|$ is equal to
(1) 26
(2) 27
(3) 66
(4) 23
Show Answer
Answer (1)
Solution
$$ \begin{aligned} \left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right| & =\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{P}^{-1} \mathrm{P}\right| \ & =\left|\mathrm{P}^{-1}(\mathrm{~A}-2 \mathrm{I}) \mathrm{P}\right| \ & =\left|\mathrm{P}^{-1}\right||\mathrm{A}-2 \mathrm{I}||\mathrm{P}| \ & =|\mathrm{A}-2 \mathrm{I}| \ & =\left|\begin{array}{ccc} 0 & 1 & 2 \ 6 & 0 & 11 \ 3 & 3 & 0 \end{array}\right|=69 \end{aligned} $$
So, Prime factor of 69 is $3 & 23$
So, sum $=26$
