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

  • question_answer
    If the system of linear equations \[x+ay+z=3\] \[x+2y+2z=6\] \[x+5y+3z=b\] has no solution, then                                                                     [JEE Online 15-04-2018 (II)]

    A)                     \[a=1,b\ne 9\]                  

    B) \[a\ne -1,b=9\]

    C)          \[a=-1,b=9\]

    D)          \[a=-1,b\ne 9\]

    Correct Answer: D

    Solution :

    If the system of equations has no solution then \[\Delta =0\] and at least one of \[{{\Delta }_{1}},{{\Delta }_{2}}\]and \[{{\Delta }_{3}}\] is not zero. \[\Delta =\left| \begin{matrix}    1 & a & 1  \\    1 & 2 & 2  \\    1 & 5 & 3  \\ \end{matrix} \right|=0\] \[\Rightarrow -a-1=0\Rightarrow a=-1\] \[{{\Delta }_{2}}=\left| \begin{matrix}    1 & 1 & 3  \\    1 & 2 & 6  \\    1 & 3 & b  \\ \end{matrix} \right|\ne 0\] \[\Rightarrow b\ne 9\]


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