Question
If A satisfies the equation x3 – 5x2 + 4x + k = 0, then A–1 exists if -
- k ≠ 1
- k ≠ 3
- k ≠ – 1
- 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 = – 
(A2 – 5A + 5I)
A–1 exists if k ≠ 0
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