I am currently reading a book on control theory and I have to calculate $e^{At}$ where $ A=\begin{pmatrix} 1 & 1 \\ 0 & -2 \end{pmatrix}$
I computed : $e^{At}=\begin{pmatrix} e^t & \frac{1}{3}(e^t-e^{-2t})\\ 0 & e^{-2t} \end{pmatrix} $ whereas the writer has: $e^{At}=\begin{pmatrix} 1 & \frac{1}{2}(1-e^{-2t})\\ 0 & e^{-2t} \end{pmatrix} $
I worked by both Laplace method and Caley - Hamilton theorem and in both cases I got the first result I wrote. Am I missing something?
Most likely the typo is the top left element and the $A$ matrix should be
$$ A = \begin{pmatrix} 0 & 1 \\ 0 & -2 \end{pmatrix} $$
because then you get
$$ e^{A t} = \begin{pmatrix}1 & \frac{1}{2}(1 - e^{-2 t}) \\ 0 & e^{-2 t} \end{pmatrix} $$
which is the same as in your book.
If $A=\begin{pmatrix} 1 & 1\\ 0 & -2 \end{pmatrix}$ then your solution is correct.