$\newcommand{\Neg}{\phantom{-}}$When solving a system of equations: $$ \frac{dX}{dt}= \begin{bmatrix} 1 & \Neg0 & \Neg1\\ 1 & -1 & \Neg0 \\ 0 & \Neg0 & \Neg1 \end{bmatrix}X, $$ I find the characteristic equation to be $(1-\lambda)^2(-1-\lambda)$.
Does this imply that the general solution is $X=(2)C_1K_1e^{\lambda_1t}+C_2K_2e^{\lambda_2t}$? Or are there only two eigenvectors? Making the solution $X=C_1K_1e^{\lambda_1t}+C_2K_2e^{\lambda_2t}$?
Suppose $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda$, $\lambda$, and $\mu$.
If $A$ has three linearly independent eigenvectors $K_{1}$, $K_{2}$, and $K_{3}$, the general solution of $X' = AX$ is $$ X(t) = e^{\lambda t}(c_{1}K_{1} + c_{2}K_{2}) + c_{3} e^{\mu t}K_{3}. $$
Otherwise, there exist two linearly independent eigenvectors $K_{1}$ and $K_{3}$ of $A$ and a vector $K_{2}$ such that $AK_{2} = K_{1} + \lambda K_{2}$, and the general solution is $$ X(t) = e^{\lambda t}\bigl[(c_{1} + c_{2}t)K_{1} + c_{2}K_{2}\bigr] + c_{3} e^{\mu t}K_{3}. $$