Finding the Exponential of a Matrix that is not Diagonalizable

this is my first answer on this site so if anyone can help to improve the quality of this answer, thanks in advance.

That said, let us get to business.

1. Compute the Jordan form of this matrix, you can do it by hand or check this link. (or both).
2. Now, we have the following case: $$ A = S J S^{-1}.$$ You will find $S$ and $S^{-1}$ on the previous link. For the sake of simplicity, $J$ is what actually matters, $$ J = \begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{pmatrix} $$ because:
3. $e^A = e^{SJS^{-1}} = e^J$ And the matrix $J$ can be written as: $J = \lambda I + N$, where $I$ is the identity matrix and $N$ a nilpotent matrix.
4. So, $e^J = e^{\lambda I + N} = \mathbf{e^{\lambda} \cdot e^N}$ By simple inspection, we get that: $$ J = \lambda I + N = 1 \cdot \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix} + \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{pmatrix} $$ where you can check that $\lambda =1$ and N is $$ \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{pmatrix} $$
5. So, $e^A = e \cdot e^N$ we just apply the definition $ e^N \equiv \sum^{\infty}_{k=0} \frac{1}{k!} N^k$. And, of course, it converges fast: $N^2 \neq 0$ but $N^3=0$.
6. Finally: $$ e^A = e \cdot \left[ 1 \cdot I + 1 \cdot N^1 + \frac{1}{2} N^2 \right] $$ where $$ N^2 = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{pmatrix} $$ then: $$ \mathbf{ e^A = e \cdot \begin{pmatrix} 1 & 1 & 1/2\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{pmatrix} } $$ Last but not least, $$ e^{At} = e^{A \cdot t} = e^{\lambda \cdot t} \cdot e^{N \cdot t} = e^t \cdot \begin{pmatrix} 1 & t & 1/2 t^2\\ 0 & 1 & t\\ 0 & 0 & 1\\ \end{pmatrix} $$ You replace $N$ by $At$ in the exp definition and that's it.

Hint:

Write your matrix $A$ as $I+N$ where $I$ is the identity matrix and $N$ is a nilpotent matrix. Then use the definition of $e^{At}$ as a power series, noting that $N^k=0$ for some $k$.

(This question was edited a lot, I'm referring to this revision.)

Yes, this is correct. Note however that:

1. You've used $e^{(M+N)t}=e^{Mt}e^{Nt}$. Note that this is only valid if $M$ and $N$ commute (that is, $MN=NM$). In this case it's ok because $M$ is scalar and commutes with everything, but you should mention it explicitly.
2. In general, it may be easier to find the Jordan form of $A$ and use that. You can calculate the exponential in blocks, and there is an elegant expansion for each block.