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JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    Let A be a \[3\times 3\] matrix such that \[{{A}^{2}}-5A+7I=0\]. Statement. - I : \[{{A}^{-1}}=\frac{1}{7}(5I-A)\]. Statement -II : The polynomial \[{{A}^{3}}-2{{A}^{2}}-3A+I\] can be reduced to \[5(A-41)\]. Then   JEE Main Online Paper (Held On 10 April 2016)

    A) Statement- I is false, but Statement-II is true.

    B) Both the statements are false.

    C) Both the statements are true.

    D) Statement-I is true, but Statement-II is false.

    Correct Answer: C

    Solution :

    \[{{A}^{2}}-5A+7I=0\,\left| \left. A \right|\pm 0 \right.\]              \[\Rightarrow \,\,\,A-5I=-7{{A}^{-1}}\]              \[\Rightarrow \,\,\,{{A}^{-1}}=\frac{1}{7}(5I-A)\]              Hence statement 1 is true Now \[{{A}^{3}}-2{{A}^{2}}-3A+I=A{{(}^{2}}A)-3A+I\]                    \[=A(5A-7I)-2{{A}^{2}}-3A+I\]                    \[=3{{A}^{3}}-10A+I\]                   \[=5A-20I\,\,\,\,\,=3(5A-7I)-10A+I\] Statement 2 also correct


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