+231 Find all real constants k, such that the system
x + 3y = kx,
3x + y = ky.
has a solution other than (x,y) = (0,0)
+23252 x + 3y = kx ---> divide by x ---> 1 + 3(y/x) = k
3x + y = ky ---> divide by y ---> 3(x/y) + 1 = k
Setting these two equations equal to each other: 1 + 3(y/x) = 3(x/y) + 1
3(y/x) = 3(x/y)
y/x = x/y
y2 = x2
So: either x = y or x = -y
If x = y, x can be any number and y will be the same number. (k = 4)
If x = -y, x can be any number and y will be its additive inverse. (k = -2)
