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 $k$.

If you know about the Jordan Canonical Form you can use that.

Another method, probably more elementary, was mentioned in a comment. The comment was deleted; I don't know why. Note that $A=I+N$, where $N^3=0$. It follows that $$A^k=I+kN+\frac{k(k-1)}{2}N^2.$$You can use that to calculate $e^{At}=\sum t^kA^k/k!$.

Edit: Oh, that comment was converted to an answer. I'll leave this here anyway, being more detailed (at least regarding one approach).