JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(A=\left[\begin{array}{llc}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{array}\right]\) 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
- A \(26\)
- B \(27\)
- C \(66\)
- D \(23\)
Answer & Solution
Correct Answer
(A) \(26\)
Step-by-step Solution
Detailed explanation
\(\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right|\) \( =\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{P}^{-1} \mathrm{P}\right|\) \( =\left|\mathrm{P}^{-1}(\mathrm{~A}-2 \mathrm{I}) \mathrm{P}\right|\) \( =\left|\mathrm{P}^{-1}\right||\mathrm{A}-2 \mathrm{I}||\mathrm{P}| \)…
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