This question has a solution presented in this paper even if with the jargon and notation of theoretical physics. So, I will use a somewhat different notation and I will redefine
$${\bf A}(t)\rightarrow -i{\bf A}(t).$$
Then, I will compute the eigenvalues and eigenvectors of ${\bf A}(t)$ through
$${A}(t)|n;t\rangle=\lambda_n(t)|n;t\rangle.$$
Now, you get a series with a leading order term
$${\bf B}(r)=\sum_n e^{i\gamma_n}e^{-ir\int_{-1}^1 dt\lambda_n(t)}|n;1\rangle\langle n;-1| \qquad r\rightarrow\infty$$
being $\gamma_n=\int_{-1}^1dt\langle n;t|i\partial_t|n;t\rangle$ known as geometric phase. Then, an expansion in the inverse of $r$ can be obtained with the matrix
$$\tilde {\bf A}(t)=-\sum_{n,m,n\ne m}e^{i(\gamma_n(t)-\gamma_m(t))}e^{-i\int_{t_0}^tdt[\lambda_m(t)-\lambda_n(t)]}\langle m;t|i\partial_t|n;t\rangle|m;t_0\rangle\langle m;t_0|$$
being in this case
$$\tilde {\bf B}(r)=\prod_{-1}^1e^{\tilde {\bf A}(t)dt}$$
so that
$$B(r)=\sum_n e^{i\gamma_n}e^{-ir\int_{-1}^1 dt\lambda_n(t)}|n;1\rangle\langle n;-1|\tilde {\bf B}(r).$$
This represents a solution of the Schroedinger equation
$$-ir{\bf A}(t)B(r;t,t_0)=\partial_tB(r;t,t_0)$$
in the interval $t\in [-1,1]$ and $r\rightarrow\infty$.