If B is a 3 × 3 matrix such that B 2 = 0 , then d e t .   I + B 50 - 50 B is equal to :

If B is a 3×3  matrix such that B2=0, then det. I+B50-50B is equal to :

  1. 1
  2. 2
  3. 3
  4. 50

Solution

B2=0B4=B6=B8=....=B50=0  and B3=B2B=0B=0

⇒ I+B50=I+C150B (higher powers are zero)

 det.I+B50-50B 

  =det. I+50C1B-50B

  =det.I

  =1

Asked in: JEE Main 2014 (09 Apr Online)

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