Use what we can linear cominations to eliminate one variable.
Start with your equations, but you multiply in a factor #c# for one of them.
#c(\Lambda-2)x+cy=0#
#x+(\Lambda-2)y=0#
And add them up:
#(c(\Lambda-2)+1)x+(c+\Lambda-2)y=0#
Now we see that if we put in
#c=2-\Lambda#
Then the combined equation has only #x# not #y#. We eliminate #y# with that choice:
#((2-\Lambda)(\Lambda-2)+1)x=0#
Then we have
#(-\Lambda^2+4\Lambda-3)x=0#
We can factor that:
#-(\Lambda-1)(\Lambda-3)x=0#
So we have one of the following:
#x=0#, #\Lambda=1#, #\Lambda=3#
Suppose #x=0#. Then your original equations become:
#y=0#
#(\Lambda-2)y=0#
They both have the same solution #y=0#, but #x=y=0# is just the trivial solution. That did not work. In most cases setting a variable to zero does not work; you need to get the right value for your parameter, in this case #\Lambda#, instead. That's what the other two possibilities we got from the linear combination are designed to do.
So try #\Lambda=1# next.
#(-1)x+y=0#
#x+(-1)y=0#
We see now that instead of just #x=y=0#, we have solutions with #x=y# for any value of #x# or #y#. Thus #\Lambda=1# does give nontrivial solutions.
Finally try #\Lambda=3#. You can work out that with this value of #\Lambda# both of your original equations become #x+y=0#. Again we get nontrivial solutions, this time with #y=-x#.
