Question 658850

Add all three equations together. You'll get:

{{{x+y+kz=9}}}...........1

{{{x+ky+z=6}}}............2

{{{kx+y+z=5}}}.................3
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{{{x+x+kx+y+ky+y+kz+z+z=9+6+5}}}

{{{2x+2y+2z+kx+ky+kz=20}}}

{{{2(x+y+z)+k(x+y+z)=20}}}

{{{(2+k)(x+y+z)=20}}}....when {{{(2+k)=0}}}, then {{{(2+k)(x+y+z)=0}}}, and that will be if {{{k=-2}}}


{{{(2-2)(x+y+z)=20}}}

{{{0(x+y+z)=20}}}

{{{0*x+0*y+0*z=20 }}}

{{{0=20}}} which is {{{impossible}}}, meaning there are {{{no}}} {{{solutions}}}, and therefore {{{no}}} {{{unique}}} solutions if {{{highlight(k=-2)}}}