$ A=\begin{pmatrix} 0 & 2 & 0\\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix}$, calculate $e^A$ I have encounter a question in my book , it was For $ A=\begin{pmatrix}
0 & 2 & 0\\
0 & 0 & 3 \\
0 & 0 & 0
\end{pmatrix}$, calculate $e^A$
My solution way : I tried to find its eigenvalues , so i found that the only eigenvalue is $0$ and the eigenspace is $(1,0,0)$ .Hence , it cannot be diagonalized.
Then , i tried to use taylor exponential and it gives me $ A=\begin{pmatrix}
1 & 2 & 0\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .However the answer is $ A=\begin{pmatrix}
1 & 2 & 3\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .
What am i missing ,can you help me?

Note that $$A^2=\begin{pmatrix}
0 & 0 & 6\\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},$$ which is where the $3$ in the upper-right corner comes from.
Since $A^3$ is just the $3\times 3$ matrix of zeroes, $e^A=I+A+\frac12 A^2.$
