Let [A=[[arrayccc 2 & 1 & 2 6 & 2 & 11 3 & 3 & 2 array]]] and [P=[[arraylll 1 & 2 & 0 5 & …

Question

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

Accepted Answer

$$\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}\end{aligned}$$\right|=69
\]

So, Prime factor of 69 is 3 \& 23

So, sum = 26

Solutions · 1