I'm required to solve this system through the variation of parameters method: $$\begin{bmatrix}x'\\y'\\z'\end{bmatrix}=\begin{bmatrix}1&1&0\\1&1&0\\0&0&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}+\begin{bmatrix}e^t\\e^{2t}\\e^{3t}\end{bmatrix}$$ I found its eigenvalues, which are $0,2,3$. I got some problems with $\lambda=3$. The first two equations did not match and the last one was undetermined.
Anyway, if anyone can save me, I'd appreciate it.
Given
$$\begin{bmatrix}x'\\y'\\z'\end{bmatrix}=Ax + g(t) = \begin{bmatrix}1&1&0\\1&1&0\\0&0&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}+\begin{bmatrix}e^t\\e^{2t}\\e^{3t}\end{bmatrix}$$
We find the eigenvalues and eigenvectors as
$$\lambda_1 = 0, v_1 = (-1,1,0), \lambda_2 = 2, v_2 = (1,1,0), \lambda_3 = 3, v_3 = (0,0,1)$$
Using Variation of Parameters Example 2, the complimentary solution is
$$x_c =c_1 e^{0t} \begin{bmatrix}-1\\1\\0\end{bmatrix}+ c_2 e^{2 t}\begin{bmatrix}1\\1\\0\end{bmatrix} + c_3 e^{3t}\begin{bmatrix}0\\0\\1\end{bmatrix}$$
We have
$$X = \begin{bmatrix}-1&e^{2t}&0\\1&e^{2t}&0\\0&0&e^{3t}\end{bmatrix}, X^{-1} = \begin{bmatrix} -\dfrac{1}{2} & \dfrac{1}{2} & 0 \\ \dfrac{e^{-2 t}}{2} & \dfrac{e^{-2 t}}{2} & 0 \\ 0 & 0 & e^{-3 t} \\ \end{bmatrix}, X^{-1} g = \begin{bmatrix}\dfrac{e^{2 t}}{2}-\dfrac{e^t}{2}\\\dfrac{e^{-t}}{2}+\dfrac{1}{2}\\1\end{bmatrix}$$
Next
$$\int X^{-1}g~ dt = \begin{bmatrix}\dfrac{1}{4} e^t \left(e^t-2\right)\\\dfrac{1}{2} \left(t-e^{-t}\right)\\t\end{bmatrix}$$
We can now write the particular solution
$$x_p = X \int X^{-1}g~ dt $$
I will let you fill in the blanks.
The final solution is
$$x(t) = x_c(t) + x_p(t)$$
Note: It is worth noting that it is easier to solve for $z(t)$ straight off since it is decoupled and this reduces the problem to a $2 \times 2$ (try it).
Update
From the previous note, here is the process.
For $z(t)$, we have a linear equation and can solve it using an Integrating factor
$$z' = 3 z + e^{3t} \implies z(t) = e^{3t}(c + t)$$
We can now write our system as
$$\begin{bmatrix}x'\\y'\end{bmatrix}=Ax + g(t) = \begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}e^t\\e^{2t}\end{bmatrix}$$
We now solve for $x$ and $y$ from this reduced system.