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JCECE Engineering JCECE Engineering Solved Paper-2014

  • 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) \[nA-(n-1)l\]                     

    B) \[nA-l\]

    C) \[{{2}^{n-1}}A-(n-1)l\]  

    D) \[{{2}^{n-1}}A-l\]

    Correct Answer: A

    Solution :

    Given\[,\]           \[{{A}^{2}}=2A-l\]                            ? (i) On multiplying by \[A\] both sides, we get                 \[{{A}^{2}}\cdot A=(2A-l)A\] \[\Rightarrow \]               \[{{A}^{3}}=2{{A}^{2}}-lA\]                 \[=2(2A-l)-lA\]                   [from Eq. (i)]                 \[=4A-2l-A\]       \[(\because \,\,lA=A)\]                 \[=3A-2l\] Similarly,\[{{A}^{4}}=4A-3l\]                 \[{{A}^{5}}=5A-4l\] Hence,  \[{{A}^{n}}=nA-(5-1)l\]


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