Repeated eigenvalues in 3x3 matrix homogeneous equation

[Yuetong Ling]

How to find the solution for this 3 by 3 matrix (I think it has repeated eigenvalues):
x' =  \begin{bmatrix}1 & -3 & 3 \\3 & -5 & 3 \\6 & -6 & 4\end{bmatrix} x

[Chonghan Ma]

Set det\begin{bmatrix}1-\lambda & -3 & 3 \\3 & -5-\lambda & 3 \\6 & -6 & 4-\lambda\end{bmatrix}=0
then,
We have  λ³-12λ-16 = 0
Therefore, the eigenvalues are λ = 4 , -2
λ = 4:
Null\begin{bmatrix}-3 & -3 & 3 \\3 & -9 & 3 \\6 & -6 & 0\end{bmatrix} = span\begin{bmatrix}1\\1\\2\end{bmatrix}
λ = -2:
Null\begin{bmatrix}3 & -3 & 3 \\3 & -3 & 3 \\6 & -6 & 6\end{bmatrix}=span{\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}-1\\1\\0\end{bmatrix}}

Then the solution should be
y(t) = c₁e^4t\begin{bmatrix}1\\1\\2\end{bmatrix} + c₂e^-2t\begin{bmatrix}1\\0\\1\end{bmatrix} + c₃e^-2t\begin{bmatrix}-1\\1\\0\end{bmatrix}