Let A = [ 2 1 2 6 2 11 3 3 2 ] and P = [ 1 2 0 5 0 2 7 1 5 ] . The sum of the prime factors of | P - 1 A P - 2 I | is equal to [2024]

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Solution

Q.
Let $A = [\begin{matrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{matrix}]$ and $P = [\begin{matrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{matrix}] .$ The sum of the prime factors of $| P^{- 1} A P - 2 I |$ is equal to [2024]

1 66

2 27

3 26

4 23

Ans.
(3)

$P^{- 1} A P - 2 I = P^{- 1} (A - 2 I) P$

$\Rightarrow | P^{- 1} A P - 2 I | = | P^{- 1} | | A - 2 I | | P |$

$= | A - 2 I |$ $(∵ | P^{- 1} | \cdot | P | = 1)$

$∴$ $\det (A - 2 I) =$ $| \begin{matrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{matrix} |$

$= - 1 (- 33) + 2 (18) = 69 = 3 \times 23$

Sum of prime factors = 23 + 3 = 26

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