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- #1

\[

\mathbf{A} =

\begin{bmatrix}

0 & 1 & 0\\

0 & 0 & 1\\

-4 & -5 & -4

\end{bmatrix}

\]

Then I want to find \(e^{\mathbf{A}t}\).

\[

\mathbf{I} + \mathbf{A}t +\frac{\mathbf{A}^2t^2}{2!} + \frac{\mathbf{A}^3t^3}{3!} + \cdots

\]

I have up to the 6th term but I can't identify the series.

\[

\begin{bmatrix}

1 - \frac{2t^3}{3} + \frac{2t^4}{3} - \frac{11t^5}{30} + \frac{7t^6}{45} &

t - \frac{5t^3}{6} + \frac{2t^4}{3} - \frac{13t^5}{40} + \frac{2t^6}{15} &

\frac{t^2}{2} - \frac{2t^3}{3} + \frac{11t^4}{24} - \frac{7t^5}{30} + \frac{73t^6}{720}\\

-2t^2 + \frac{8t^3}{3} - \frac{11t^4}{6} + \frac{14t^5}{15} - \frac{73t^6}{180} &

1 - \frac{5t^2}{2} + \frac{8t^3}{3} - \frac{13t^4}{8} + \frac{4t^5}{5} - \frac{253t^6}{720} &

-\frac{49 t^6}{180} + \frac{73 t^5}{120} - \frac{7 t^4}{6} + \frac{11 t^3}{6} - 2t^2 + t\\

\frac{49 t^6}{45} - \frac{73 t^5}{30} + \frac{14 t^4}{3} - \frac{22 t^3}{3} + 8t^2 - 4t &

\frac{43 t^6}{45} - \frac{253 t^5}{120} + 4t^4 - \frac{13 t^3}{2} + 8t^2 - 5t &

\frac{59 t^6}{80} - \frac{49 t^5}{30} + \frac{73 t^4}{24} - \frac{14t^3}{3} + \frac{11 t^2}{2} - 4t + 1

\end{bmatrix}

\]