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
Select the correct option:
A
26
B
27
C
66
D
23
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$
\begin{aligned}
\left|P^{-1} A P-2 I\right|=\left|P^{-1} A P-2 P^{-1} P\right| & \\
& =\left|P^{-1}(A-2 I) P\right| \\
& =\left|P^{-1}\|A-2 I\| P\right| \\
& =|A-2 I| \\
& =\left|\begin{array}{ccc}
0 & 1 & 2 \\
6 & 0 & 11 \\
3 & 3 & 0
\end{array}\right|=69
\end{aligned}
$$
So, Prime factor of 69 is 3\&23
So, sum $=26$
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