+8 Find all real constants such that the system
\(\begin{align*} x + 3y &= kx, \\ 3x + y &= ky. \end{align*}\)
has a solution other than \($(x,y) = (0,0).$ \)
+310 I'm assuming that you mean "all real constants k"
First add the two equations together.
\(4(x+y)=k(x+y)\). k = 4
Then subtract the two.
\(-2(x-y)=k(x-y)\). k = -2.
So our two solutions are k = 4, -2
