JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    Let \[S\] be the set of all real values of \[k\] for which the system of linear equations \[x+y+z=2\] \[2x+y-z=3\] \[3x+2y+kz=4\] has a unique solution. Then S is                                                                      [JEE Online 15-04-2018]

    A) An empty set   

    B) Equal to \[R-\{0\}\]     

    C) Equal to \[\{0\}\]

    D)        Equal to \[R\]

    Correct Answer: B

    Solution :

    Given are the system of linear equations. \[x+y+z=2\] \[2x+y-z=3\] \[3x+2y+kz=4\] We know that, this system has a unique solution. Therefore, coefficient determinant is non-zero. \[\left| \begin{matrix}    1 & 1 & 1  \\    2 & 1 & -1  \\    3 & 2 & k  \\ \end{matrix} \right|\ne 0\] \[\Rightarrow k+2-(2k+3)+1\ne 0\] \[\Rightarrow k\ne 0\] Therefore, \[k\in R-\{0\}\equiv S\]


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