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JEE Main & Advanced Mathematics Determinants & Matrices Question Bank Types of matrices, Algebra of matrices

  • question_answer
    If \[A=\left[ \begin{matrix}    1 & 0  \\    1 & 1  \\ \end{matrix} \right]\] and \[I=\left[ \begin{matrix}    1 & 0  \\    0 & 1  \\ \end{matrix} \right]\], then which one of the following holds for all \[n\ge 1\], (by the principal of mathematical induction) [AIEEE 2005]

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

    B) \[{{A}^{n}}={{2}^{n-1}}A+(n-1)I\]

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

    D) \[{{A}^{n}}={{2}^{n-1}}A-(n-1)I\]

    Correct Answer: C

    Solution :

    \[{{A}^{2}}=\left[ \begin{matrix}    1 & 0  \\    1 & 1  \\ \end{matrix} \right]\,\left[ \begin{matrix}    1 & 0  \\    1 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}    1 & 0  \\    2 & 1  \\ \end{matrix} \right]\] \[{{A}^{3}}=\left[ \begin{matrix}    1 & 0  \\    2 & 1  \\ \end{matrix} \right]\,\left[ \begin{matrix}    1 & 0  \\    1 & 1  \\ \end{matrix} \right]=\left[ \begin{matrix}    1 & 0  \\    3 & 1  \\ \end{matrix} \right]\] \ \[{{A}^{n}}=\left[ \begin{matrix}    1 & 0  \\    n & 1  \\ \end{matrix} \right]\] \[nA=\left[ \begin{matrix}    n & 0  \\    n & n  \\ \end{matrix} \right],(n-1)I=\left[ \begin{matrix}    n-1 & 0  \\    0 & n-1  \\ \end{matrix} \right]\] \[nA-(n-1)I=\left[ \begin{matrix}    1 & 0  \\    n & 1  \\ \end{matrix} \right]={{A}^{n}}\].


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