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

Options

  1. A1
  2. B2
  3. C3
  4. D50

Correct answer

A. 1

Step-by-step solution

B 2 = 0 ⇒ B 4 = B 6 = B 8 = . . . . = B 50 = 0 and B 3 = B 2 B = 0 B = 0 ⇒  I + B 50 = I + C 1 50 B (higher powers are zero) ∴ d e t . I + B 50 - 50 B   = d e t .   I + 5 0 C 1 B - 5 0 B = d e t . I = 1

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