How do I integrate dA(t)A^-1(t)=Bdt?

[exmachina]

I have the following two equations

#1

d(A(t))/dt=A(t)B

where A is some matrix that depends on parameter t, and B is another matrix, d is the differential

this can be simplified to by multiplying both sides by the left inverse of A(t),

A^-1(dA(t))=B*t

which allows me to solve A(t) = Ce^(Bt)

#2

d(A(t))/dt=BA(t) note that A and B do not necessarily commute

I'm asked to once again find A(t)

and I get

dA(t)A^-1(t)=Bdt

but how do I integrate this thing?

 

[trambolin]

How about this?

$0 = \frac{d}{dt}I = \frac{d}{dt}\left(A(t)A^{-1}(t)\right) = \dot A(t)A^{-1}(t) + A(t)\dot{A^{-1}}(t) \ \ \forall t\in\mathbb{F}$

Then,

$-A^{-1}(t)\dot A(t)A^{-1}(t) = \dot{A^{-1}(t)}$

and

$\dot A(t) = -A\dot{A^{-1}}(t)A$