# Systems of ODE's - Complex Eigenvalues

## Homework Statement

Find the general solution of the given system.

The given matrix is X' = (1st row (1,-1,2) 2nd row (-1,1,0) 3rd row (-1,0,1))X

2. The attempt at a solution

The eigenvalue determinant = (1st row (1-λ,-1,2) second row (-1,1-λ,0) 3rd row (-1,0,1-λ)

Solving the determinant gives -(λ-1)(λ^2-2λ+2)

So λ1,2 = 1+i and 1-i and λ3 = 1

Now my question is how do I solve the 3x3 matrix with copmlex numbers in it? I used gaussian elimination to solve for the real value λ3.

(1st row (-i,-1,2) 2nd row (-1,-i,0) 3rd row (-1,0,-i)) multiplied by the column (k1,k2,k3) = 0

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gabbagabbahey
Homework Helper
Gold Member
Solving the determinant gives -(λ-1)(λ^2-2λ+2)

So λ1,2 = -1+i and -1-i and λ3 = 1

Your two complex eigenvalues have a sign error. The roots of $\lambda^2-2\lambda+2=0$ are $\lambda=1\pm i$.

Now my question is how do I solve the 3x3 matrix with copmlex numbers in it? I used gaussian elimination to solve for the real value λ3.

(1st row (2-i,-1,2) 2nd row (-1,2-i,0) 3rd row (-1,0,2-i)) multiplied by the column (k1,k2,k3) = 0

You can still use Gaussian elimination to solve for the eigenvectors.

Your two complex eigenvalues have a sign error. The roots of $\lambda^2-2\lambda+2=0$ are $\lambda=1\pm i$.

You can still use Gaussian elimination to solve for the eigenvectors.

Can someone show me how?

gabbagabbahey
Homework Helper
Gold Member
You do it exactly the same way you did it for your real eigenvalue....try it out and post your work if you get stuck.

You do it exactly the same way you did it for your real eigenvalue....try it out and post your work if you get stuck.

Is multiplying a row by a complex number a valid operation?

gabbagabbahey
Homework Helper
Gold Member
Yes. It's exactly the same as multiplying both sides of an equation by a complex number.

The corrected eigenvalue matrix is
[-i,-1,2]
[-1,-i,0]
[-1,0,-i]

using the operations 1) i*R1, 2) R1+R2, 3) R1+R3, 4) (-1/2)R2+R3, 5) -2*R2+R1 gives

[1,-i,0]
[0,-i, i]
[0,0,0]

which says k2=k3; picking k1=1 and k2=k3= i gives one solution,k=
[1]

which is split into the real and imaginary parts B1 and B2, respectively

[1]
[0]
[0]

[0]
[1]
[1]

The general solution is c1[B1*cos(t)-B2sin(t)]e^t + c2[B2*cos(t)-B1sin(t)]e^t +c3
([0])e^t
([1])
([2])

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