A) \[nA-(n-1)I\]
B) \[nA-I\]
C) \[{{2}^{n-1}}A-(n-1)I\]
D) \[{{2}^{n-1}}A-I\]
Correct Answer: A
Solution :
We have, \[{{A}^{2}}=2A-I\] \[\therefore \] \[{{A}^{2}}A=2AA-IA\] \[=2{{A}^{2}}-A=2(2A-I)-A\] \[\Rightarrow \]\[{{A}^{3}}=3A-2I\] \[\Rightarrow \]\[{{A}^{3}}.A=3A\,A-2IA\] \[=3{{A}^{2}}-2A=3(2A-I)-2A\] \[\Rightarrow \]\[{{A}^{4}}-4A-3I\] Similarly, \[{{A}^{n}}-nA-(n-1)I\]You need to login to perform this action.
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