On a linear $3\times 3$ system of differential equations with repeated eigenvalues.

Following: Method 1 linked by Git Gud and suggested by Amzoti.

I Now know that a generalized eigenvector can be found by:

$$(A-\lambda I)v_2=v_{1}$$

I obtain $ v_2 = \begin{bmatrix} 0.2 \\ 0.4\\ 0\end{bmatrix}$ so an other solution is:

$$x_2(t) =e^t \left( t \begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix} + \begin{bmatrix} 0.2 \\ 0.4\\ 0\end{bmatrix} \right) $$

But I need three solutions so I must find a $v_3$, as before:

$$(A-\lambda I)v_3=v_{2}$$

I now obtain $v_3 = \begin{bmatrix} 0.08 \\ 0.4\\ 0\end{bmatrix}$ and the last solution needed is: $$ x_3(t) = e^{ t}\left(\dfrac{t^2}{2!}\begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}+t\begin{bmatrix} 0.2 \\ 0.4\\ 0\end{bmatrix}+ \begin{bmatrix} 0.08 \\ 0.4\\ 0\end{bmatrix} \right)$$

So my general solution would look like:

$$X(t) = C_1e^t\begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}+ C_2 e^t \left( t \begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix} + \begin{bmatrix} 0.2 \\ 0.4\\ 0\end{bmatrix} \right) + C_3e^{ t}\left(\dfrac{t^2}{2!}\begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}+t\begin{bmatrix} 0.2 \\ 0.4\\ 0\end{bmatrix}+ \begin{bmatrix} 0.08 \\ 0.4\\ 0\end{bmatrix} \right) $$

Correct?