What are the eigenvalues of the 3x3 matrix [2 2 1; 1 3 1; 2 2 2]?

[mkay123321]

Ive been trying for 3 hours now and can't seem to find the eigenvalues, the long polynomials are getting me confused, the matrix is [2 2 1:1 3 1:1 2 2]




So far i did [2-L 2 1:1 3-L 1:1 2 2-L] then I do the normal way to find the determinant but after that I get a horrible polynomial. Please help anyone!

Thanks

 

[Dick]

What horrible polynomial did you get? Multiply it out. It will be a cubic, but you can factor it. There's no way anyone can help until you show more of your work.

 

[Mark44]

Ive been trying for 3 hours now and can't seem to find the eigenvalues, the long polynomials are getting me confused, the matrix is [2 2 1:1 3 1:1 2 2]




So far i did [2-L 2 1:1 3-L 1:1 2 2-L] then I do the normal way to find the determinant but after that I get a horrible polynomial. Please help anyone!

Thanks

After taking the determinant, I get
$(2 - \lambda)(6 - 5\lambda + \lambda^2 - 2) - (4 - 2\lambda - 2) + 2 - (3 - \lambda)$

$= (2 - \lambda)(\lambda^2 - 5\lambda + 4) + 3\lambda - 3$

Instead of multiplying all that stuff out, factor the quadratic and the last two terms and notice that there is a common factor.

I get $\lambda = 1~and~\lambda = 5$.

 

[mkay123321]

I get (2-L)((3-L)(2-L)-2) - 2((2-L)-1) + 2-(3-L)

2-L(4 - 3L - 2L - L^2) + 3L - 3 I tried doing all sorts of stuff to this, just can't get it.

 

[mkay123321]

After taking the determinant, I get
$(2 - \lambda)(6 - 5\lambda + \lambda^2 - 2) - (4 - 2\lambda - 2) + 2 - (3 - \lambda)$

$= (2 - \lambda)(\lambda^2 - 5\lambda + 4) + 3\lambda - 3$

Instead of multiplying all that stuff out, factor the quadratic and the last two terms and notice that there is a common factor.

I get $\lambda = 1~and~\lambda = 5$.

Ahh I see now, thanks a lot.