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BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    Matrix A is such that \[{{A}^{2}}=2A-I,\]where\[I\]is the identity matrix, then for \[n\ge 2,{{A}^{n}}\]is equal to:

    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, \[{{A}^{n}}-nA-(n-1)I\]


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