If the characteristic polynomial $\det (A-kI)$ has a multiple root (say root $k$ of multiplicity $m$) then it can have any number $l=1,\ldots, m$ of linearly independent eigenvectors, associated with this eigenvalue (we do not distinguish here real or complex eigenvalues). Then one needs to find Jordan (echelon?) form.
In particular, let $m=2$. Then either there are two l.i. eigenvectors $\mathbf{e}_1$, $\mathbf{e}_2$; then the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \mathbf{e}_1+C_2\mathbf{e}_2\Bigr)
$$
or there exists a vector $\mathbf{f}$ s.t. $(A-k)^2\mathbf{f}=0$, but $\mathbf{e}=(A-k)\mathbf{f}\ne 0$; then the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \bigl[t\mathbf{e}+\mathbf{f}\bigr]+C_2\mathbf{e}\Bigr)
$$
If $m=3$ then there are three cases: either there are three l.i. $\mathbf{e}_1$, $\mathbf{e}_2$, $\mathbf{e}_3$, in which case solution is similar to the former case of $m=2$, or there are a vector $\mathbf{f}$ s.t. $(A-k)^2\mathbf{f}=0$, but $\mathbf{e}_1=(A-k)\mathbf{f}\ne 0$, and $\mathbf{e}_2$, also eigenvector but not proportional $\mathbf{e}_1$; then the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \bigl[t\mathbf{e}_1+\mathbf{f}\bigr]+C_2\mathbf{e}_1+C_3\mathbf{e}_2\Bigr);
$$
or there exists a vector $\mathbf{g}$ s.t. $(A-k)^3\mathbf{g}=0$, but $\mathbf{e} =(A-k)^2\mathbf{g}\ne 0$; let $\mathbf{f} =(A-k)\mathbf{g}$ and
the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \bigl[\frac{t^2}{2}\mathbf{e}_1+t\mathbf{f}+\mathbf{h}\bigr]+C_2\bigl[t\mathbf{e} +\mathbf{f}\bigr]+C_3\mathbf{e}\Bigr).
$$
and so on...
