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JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    If the system of linear equations
    \[x+ky+3z=0\]
    \[3x+ky-2z=0\]
    \[2x+4y-3z=0\]
    Has a non-zero solution (x, y, z), then \[\frac{xz}{{{y}^{2}}}\] is equal to:    [JEE Main Online 08-04-2018]

    A)  \[-30\]                        

    B)  \[30\]

    C)  \[-10\]                        

    D)  \[10\]

    Correct Answer: D

    Solution :

    For non-zero solution, \[\therefore x+11y=-3z\]             ?(1) \[3x+11y=2z\]                   ?(2) \[-2x=-5z\]                          ?(1)-(2) \[\Rightarrow x=\frac{5}{2}z\] \[\therefore 11y=-3z-\frac{5}{2}z=-\frac{11}{2}z\Rightarrow y=-\frac{z}{2}\] \[\therefore \frac{xz}{{{y}^{2}}}=\frac{\frac{5}{2}z.z}{\frac{{{z}^{2}}}{4}}=10\]


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