The question asks to find the solution of the system x'(t)=Ax(t)

given my A is

$\displaystyle \begin{bmatrix}

-7 & 0 & 6 \\

0 & 5 & 0 \\

6 & 0 & 2

\end{bmatrix}

$

After finding my det($\displaystyle \lambda I -A)$=>$\displaystyle \lambda ^3 -75\lambda +250 =0$

I found three real roots:

$\displaystyle \lambda 1=-10$

$\displaystyle \lambda 2=5$

$\displaystyle \lambda 3=5$

Now starting with the first root equal to -10 and plugging into the det matrix

$\displaystyle \begin{bmatrix}

-3 & 0 & -6 \\

0 & -15 & 0 \\

-6 & 0 & -12

\end{bmatrix}

$

*

$\displaystyle \begin{bmatrix}

x\\

y\\

z

\end{bmatrix}

$

=

$\displaystyle \begin{bmatrix}

0\\

0\\

0

\end{bmatrix}

$

I found my U1 to be <2, 0, -1>

Now I'm confused what to do with the last two lambdas, I get the feeling that there should be more than just two Cn's

$\displaystyle x(t)=C1 e^-10t\begin{bmatrix}

2\\

0\\

1

\end{bmatrix}+C2...+C3...

$

My question is can someone show me how to solve for u2 and u3 (if there is a u3), the way my textbook explains solving for eigenvectors is not helping very much considering the questions are getting harder and the method I've been using doesn't seem to be working.