If we have the following, where A is a $n \times n$ matrix: $$x'(t)=Ax(t)$$
And I then write it as $x'(t)-Ax(t)=0$ and calculate the integration factor of $-A$ and the solution being $x(t)=e^{-At}$ (its the integrating factor). Testing this solution now yields $x'(t) = -Ae^{-At}$ which is not same as $x'(t)=Ax(t)$ from the problem.
The answer is $x(t)=e^{At}x(0)$ but I don't understand why there is no minus before the exponential matrix A. Could you help me out?
$$x^\prime - A x =0$$ $$e^{-A t}x^\prime - A e^{-At} x =0$$ $$\frac{d}{dt}\left( e^{-At} x \right)=0$$ $$ e^{-At} x = \text{const.}$$ $$ x (t) = \text{const.} \cdot e^{At} = x(0) e^{At} $$