Given a system
$$x' = 3x +y$$ $$ y'=-x+y $$
Thus, the associated coefficients are
\begin{bmatrix} 3&1\\ -1&1\\ \end{bmatrix}
If the complete solution for a $2\times2$ matrix is $C_1e^{\lambda t} \vec{u}+C_2e^{\lambda t}(t\vec{v}+\vec{w})$ where $\vec{w}$ solves as $(A-\lambda I)\vec{w}=\vec{v}$
How can we define the complete solution for a $3\times3$ matrix with eigenvalues of multiplicity 2 or three and its associated eigenvectors?
On
It seems "over-kill" to do this with "matrices", "eigenvalues", and "eigenvectors"! Given the system of equations x'= 3x+ y and y'= -x+ y, then x''= 3x'+ y'= 3x'- x+ y. From x'= 3x+ y. y= x'- 3x so x''= 3x'- x+ x'- 3x= 4x'- 4x.
The equation, x''- 4x'+ 4x= 0 is a linear homogeneous second order equation with constant coefficients. It has associated characteristic equation $r^2- 4r+ 4= (r- 2)^2= 0$ which has r= 2 as double root. The general solution is $x(t)= Ae^{2t}+ Bte^{2t}$. $y= x'- 3x= 2Ae^{2t}+ Be^{2t}+ 2Bte^{2t}- 3Ae^{2t}- 3Bte^{2t}= (B- 2A)e^{2t}- Bte^{2t}$.
The general solution to the system is
$x(t)= Ae^{2t}+ Bte^{2t}$
$y(t)= (B- 2A)e^{2t}- Bte^{2t}$
Let me give a simpler take on this on the question for systems in $\mathbb{R}^3$, and adding on to Will Jagy's comment. We can express this using the Jordan normal form if the matrix $\boldsymbol{A}$ is a defective matrix.
Given the first order ODE $$\dot{\boldsymbol{x}} = \boldsymbol{Ax} $$ with $\boldsymbol{A} = \boldsymbol{PJ}\boldsymbol{P}^{-1}$ where $\boldsymbol{J}$ represents the Jordan block and $\boldsymbol{P}$ consists of the (generalized) eigenvectors.
The general solution is given by $$\boldsymbol{x}=e^{\boldsymbol{A}t}\boldsymbol{x}_0=\boldsymbol{P}e^{\boldsymbol{J}t}\boldsymbol{P}^{-1}\boldsymbol{x}_0 = \boldsymbol{P}e^{\boldsymbol{J}}\boldsymbol{c}_0$$ for $\boldsymbol{x}_0$ is a constant vector, and $\boldsymbol{c}_0 = \boldsymbol{P}^{-1}\boldsymbol{x}_0$.
To answer your question, let us consider an example in $\mathbb{R}^3$. Assuming the eigenvalues are all the same, ie. $\lambda_1 = \lambda_2 = \lambda_3 = \lambda$, for $\lambda \in \mathbb{R}$, the general solution will be: $$ \boldsymbol{x} = \boldsymbol{P}e^{\boldsymbol{J}t}\boldsymbol{c}_0 $$ $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \boldsymbol{P} \begin{bmatrix} e^{\lambda t} & te^{\lambda t} & \frac{t^2}{2}e^{\lambda t} \\ 0 & e^{\lambda t} & te^{\lambda t}\\ 0 & 0 & e^{\lambda t}\\ \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ \end{bmatrix} $$ for $\boldsymbol{c}_0 = [c_1 \space c_2 \space c_3]^T$. You can take a look at this link https://math24.net/general-solution-system-differential-equations-jordan-form.html for a better idea.