If the system of equations x+ky+3z=0, 3x+ky-2z=0 2x+3y-4z=0 has non-trivial solutions , then xy/z^2=?

Answers (1)

Solution:    We have ,    x+ky+3z=0,

                      3x+ky-2z=0   and   2x+3y-4z=0

                        eginvmatrix 1 &k &3 \2 &k &-2 \2 &3 &-4 endvmatrix=0Rightarrow k=frac332

             By cross multiplications  rule using the last  2 equations ,

       Rightarrow           fracx6-4k=fracy8=frac39-2k

     Rightarrow                          fracx15=fracy-2=fracz6

  Hence ,           fracxyz^2=frac-3036=-frac56.   

                        ,   

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