Question

How to find the values of `Lambda` for which the homogeneous linear system has nontrivial solutions?

`(Lambda-2)x+y=0`
`x+(Lambda-2)y=0`
please help me

Answer 1

To get nontrivial solutions you need `\Lambda=1` or `\Lambda=3`. See below for how you find that.

Explanation:

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`.