Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If a2b2c2a+λ2b+λ2c+λ2a-λ2b-λ2c-λ2=kλa2b2c2abc111,λ0, then k is equal to:
MathematicsDeterminantsJEE Main
Options:
  • A 4λabc
  • B -4λ2
  • C 4λ2
  • D -4λabc
Solution:
1470 Upvotes Verified Answer
The correct answer is: 4λ2
R2R2-R1,R1R1-R3
λ2a-λλ2b-λλ2c-λ4aλ4bλ4cλa-λb-λ2c-λ2
R3R3+R1,R1R1-12R2
=-λ2-λ2-λ24aλ4bλ4cλa2b2c2
=4λ3111abca2b2c2
=4λ3a2b2c2abc111
K=4λ2

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.