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If the system of linear equationsx+ky+3z=03x+ky-2z=02x+4y-3z=0has a non-zero solution (x,y,z) then `(xz)/(y^2)` is equal toA. 30B. -10C. 10D. -30 |
Answer» Correct Answer - 3 Given system of equations `x+ky+3z=0` `3x+ky -2x =0` `2x+4y -3z =0` Eliminating from first two equations we get `2x-5z=0 " or " 2x=5z` ` :. X=5lambda , z= 2lambda` from equation (iii) we have `10 lambda +4y-6lambda= 0 " or " y=-lambda` `:. (xz)/(y^(2)) =((5lambda)(2lambda))/((-lambda)^(2)) =10` |
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