Finding the Multiplicity of Eigenvalues for a 2x2 Matrix with a Variable Element

[snoggerT]

For which value of k does the matrix

A=
|4 k|
|-7 -5|

have one real eigenvalue of multiplicity 2?

The Attempt at a Solution

- I tried by setting this problem up with det(A-(lambda)I) and trying to solve like that, but I can't seem to get it that way either. I am getting (lambda)^2+(lambda)-20+7k=0, but I don't know what to do from there. Please help.

 

[Pere Callahan]

Well, what you have there is quadratic equation lambda .. do you know how to solve such an equation (hint; there is a formula ...)

This formula will give you (in general) two different solutions (which might be complex). However, when you've figured out the formula for the eigenvalues (depending on k) you can think about what k has to be, such that these two eigenvalues are actually the same.

So first you need to find the lambda's which fulfill your equation. And don't worry about k. It's just some constant number.

 

[ircdan]

look at the discriminant

 

[snoggerT]

so the discriminant is negative, so that means there are no real roots and k would have to be something that would have the discriminant come out to 1, right? is it even possible to get the discriminant to equal 1 in this situation with b=1? am I not seeing what you all are pointing out?

 

[Dick]

The discriminant is b^2-4ac. You want it to be zero to get a double root.

 

[snoggerT]

in order to get zero you would need the discriminant to come out to 1-1, but I don't know how you would get the 4ac to be equal to 1 with this problem. If you ignore the 7k, the discriminant is 81, but in order for the discriminant to be zero, wouldn't -20+7k have to equal 1/4?

 

[Dick]

Yes. 7k-20 should equal 1/4. Why is that a problem?

 

[snoggerT]

I just made a simple mistake in my math. Thanks for the help.