Again, look at x' = Ax. If x1 and x2 are independent solutions, we call the matrix Phi(t) = [x1 x2] a "fundamental matrix" for x' = Ax. The general solution is x = Phi(t)[a;b]. In particular, x(0) = Phi(0)[a;b], so [a;b] = Phi(0)^{-1}x(0). This implies that
x(t) = Phi(t)Phi(0)^{-1}x(0).
This is exactly analogous to the one-variable case: x' = ax ==> x = e^{at}x(0). So we'll write
e^{At} = Phi(t)Phi(0)^{-1}.
That is, e^{At} is the matrix whose first column is the solution to the IVP x' = Ax, x(0) = [1;0] and whose second column is the solution to the IVP x' = Ax, x(0) = [0;1]. This way of saying it makes it clear that e^{At} doesn't really depend upon the choice of independent solutions that went into Phi(t). In fact, e^{At} is itself a fundamental matrix. It is the only fundamental matrix which is "normalized" in the sense that e^{A0} = I, the identity matrix.
For example, suppose A = [1 1;0 2]. The characteristic polynomial lambda^2 - 3 lambda + 2 has roots lambda1 = 1, lambda2 = 2 (so the phase portrait is an unstable proper node) and corresonding eigenvectors c1 = [1;0], c2 = [1;1]. The corresponding fundamental matrix is Phi(t) = [e^t, e^{2t} ; 0, e^{2t}]. Phi(0) = [1 1;0 1], and we compute that
e^{At} = Phi(t)Phi(0)^{-1} = [e^t, -e^t+e^{2t} ; 0, e^{2t}].
Notice that if x = e^{At}x(0) then
(d/dt)e^{At}x(0) = x' = Ax = Ae^{At}x(0)
which implies that (d/dt)e^{At} = Ae^{At}. Also, e^{A0} = I. This is precisely analogous to the characterization of e^{at}, where a is a (perhaps complex) number, and justifies the notation.
Second example: A = [-2 1;-2 0]. The characteristic polynomial lambda^2 + 2 lambda + 2 has roots -1+i and -1-i (so the phase portrait is a stable clockwise spiral) and corresponding eigenvectors [1;1+i] and its complex conjugate vector. The corresponding fundamental matrix is Phi(t) = [e^{(-1+i)t, (1+i)e^{(-1+i)t}; conjugate of this]. We could go on, compute Phi(0)^{-1} and so on; but let's get into the real world first, by coming up with a different and real fundamental matrix: we can take for the columns the real and imaginary parts of Phi(t):
e^{-t}[cos(t), cos(t)-sin(t); sin(t), cos(t)+sin(t)].
To compute e^{At} we evaluate this at t=0 and find the inverse; it's [1 0;1 1], with invers [1 0;-1 1]. Thus
e^{At} = e^{-t}[cos(t)-sin(t), sin(t) ; -2sin(t), cos(t)+sin(t)].