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The equations x+ky+3z=0,3x+ky-2z=0,2x+3y...

The equations x+ky+3z=0,3x+ky-2z=0,2x+3y-4z=0 possess a nontrivial solution then the value of `(2k)/33` is

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STATEMENT-1 : The system of equations x + ky + 3z =0, 3x + ky - 2z =0, 2x+3y-z=0, possesses a non-trival solution. then value of k is 31/2 STATEMENT -2 Three linear equations in x, y, z can never have no solution if it is homogeneous, hence exactly two types of possible solution.

The value of k for which the set of equations x + ky + 3z=0, 3x + ky – 2z=0, 2x + 3y - 4z=0 has a non-trivial solution over the set of rationals is

Knowledge Check

  • If the system of equations x-ky+3z=0, 2x+ky-2z=0 and 3x-4y+2z=0 has non - trivial solutions, then the value of (10y)/(x) is equal to

    A
    3
    B
    `-(15)/(2)`
    C
    `(5)/(7)`
    D
    `-(5)/(7)`

  • If the system of equations x-ky-z=0 , kx-y-z=0 , x+y-z=0 has a non-zero solution , then possible values of k are :

    A
    `-1,2`
    B
    `1,2`
    C
    `0,1`
    D
    `-1,1`

  • If the system of equations x - ky – z=0, kx- y - x=0,x+y - z=0 has a non-zero solution, then the possible values of k are

    A
    `-1,2`
    B
    `1,2`
    C
    `0,1`
    D
    `-1,1`

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    If the system of equation x+ky+3z=0,3x+ky-2z=0,2x+3y-4z=0 has non trivial solution, then (xy)/(z^2) =

    If system of homogenous equations x+ky-2z=0,2x+y-3z=0 and 4x+2y-kz=0 has non-trivial solution thenthe integral value of k is

    If 2ax-2y+3z=0,x+ay+2z=0, and 2+az=0 have a nontrivial solution,find the value of a.

    If the system of equations x-ky-z=0,kx-y-z=0,x+y-z=0 has a nonzero solution,then the possible value of k are -1,2 b.1,2 c.0,1 d.-1,1

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