A) \[l-A\]
B) \[A-l\]
C) A
D) \[A+l\]
Correct Answer: A
Solution :
If A is any square matrix, then \[A\,{{A}^{-1}}=I\] and \[{{A}^{-1}}I={{A}^{-1}}\]. Since, \[{{A}^{2}}-A+l=O\] \[\Rightarrow \] \[{{A}^{-1}}{{A}^{2}}-{{A}^{-1}}A+{{A}^{-1}}l=O\] (pre-multiply given relation by\[{{A}^{-1}})\] \[\Rightarrow \]\[({{A}^{-1}}A)A-({{A}^{-1}}A)+{{A}^{-1}}=O\] \[\Rightarrow \]\[A-l+{{A}^{-1}}=O\] \[\Rightarrow \]\[{{A}^{-1}}=l-A\]
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