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General
Easy

Question

If A satisfies the equation x3 – 5x2 + 4x + k = 0, then A–1 exists if -

  1. k ≠ 1    
  2. k  ≠ 3    
  3. k  ≠ – 1    
  4. None    

The correct answer is: None


    A3 – 5A2 + 5A+ k.I = 0
    A–1 (A3 – 5A2 + 5A + kI) = A–1 × (0)
    A2 – 5A + 5I + kA–1 = 0
    A–1 = – fraction numerator 1 over denominator k end fraction (A2 – 5A + 5I)
    therefore A–1 exists if k  ≠ 0

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