A) \[{{2}^{n-1}}A-(n-1)I\]
B) \[{{2}^{n-1}}A-I\]
C) \[nA-(n-1)I\]
D) \[nA-I\]
Correct Answer: C
Solution :
[c] Given \[{{A}^{2}}=2A-I\] Now \[{{A}^{3}}=A({{A}^{2}})=A(2A-I)\] \[=2{{A}^{2}}-A=2(2A-I)-A=3A-2I\] \[{{A}^{4}}=A({{A}^{3}})=A(3A-2I)\] \[=3{{A}^{2}}-2A=3(2A-I)-2A=4A-3I\] Following this, we can say \[{{A}^{n}}=nA-(n-I)I\]You need to login to perform this action.
You will be redirected in
3 sec