Answer
True
Work Step by Step
Obtain $x'=Ax$ where $X(t)=x_1,x_2,...x_n$
then $e^{At}=X(t)X^{-1}(0)=X_0(t)$
The transition matrix $X_0(t)$ for the vector differential equation $x′ = Ax$ is precisely the same as the matrix exponential function $e^{At}$.