JEE 2024 — Mathematics PYQ
JEE | Mathematics | 2024LetA=263amp;1amp;2amp;3amp;2amp;11amp;2 and P=157amp;2amp;0amp;1amp;0amp;2amp;5. The sum of the prime factors of 1|P–1AP – 2I| is equal to
Choose the correct answer:
- A.
26
(Correct Answer) - B.
27
- C.
66
- D.
23
26
Explanation
LetA=263amp;1amp;2amp;3amp;2amp;11amp;2 and P=157amp;2amp;0amp;1amp;0amp;2amp;5
letB=P−1AP−2I
PostmultiplybyP⇒PB=AP−2P{∵PP−1=I & PI=P}
PostmultiplybyP−1⇒PBP−1=A−2I
Now∣PBP−1∣=∣A−2I∣
∣P∣.∣B∣.∣P−1∣=2−263amp;1amp;2−2amp;3amp;2amp;11amp;2−2
PB=AP−2P{∵PP−1=I & PI=P}
PBP−1=A−2I⇒∣PBP−1∣=∣A−2I∣
∣P∣.∣B∣.∣P∣1=063amp;1amp;0amp;3amp;2amp;11amp;0
∣B∣=−1(−33)+2(18)⇒∣P−1AP−2I∣=33+36=69
Primefactorsof69is3 & 23⇒sum=3+23=26
Explanation
LetA=263amp;1amp;2amp;3amp;2amp;11amp;2 and P=157amp;2amp;0amp;1amp;0amp;2amp;5
letB=P−1AP−2I
PostmultiplybyP⇒PB=AP−2P{∵PP−1=I & PI=P}
PostmultiplybyP−1⇒PBP−1=A−2I
Now∣PBP−1∣=∣A−2I∣
∣P∣.∣B∣.∣P−1∣=2−263amp;1amp;2−2amp;3amp;2amp;11amp;2−2
PB=AP−2P{∵PP−1=I & PI=P}
PBP−1=A−2I⇒∣PBP−1∣=∣A−2I∣
∣P∣.∣B∣.∣P∣1=063amp;1amp;0amp;3amp;2amp;11amp;0
∣B∣=−1(−33)+2(18)⇒∣P−1AP−2I∣=33+36=69
Primefactorsof69is3 & 23⇒sum=3+23=26

