I think so, because I need to use the relationship between t and dt in the future. I have an idea, if MatrixLog cannot handle vectors, can I find the relationship between the MatrixLog and the Log in MMA, and then use Log instead of MatrixLog in the formula to handle vectors. But at present, I have not found a document that specifically introduces the calculation process of the MatrixLog command.

Yes, the division of vectors doesn't seem to exist. So how should I simplify next? emmm

Yes, then I shouldn't have made any derivation errors. But the next step is to use the MatrixLog instruction on both sides of the equation, and there will be a problem.

But there's nothing wrong with my manual simplification process, right? Emmm

I want to transform the formula into the relationship between t and dt (i.e. the left side is the variable t, and the right side is the formula containing all variables dt), so I have manually derived the following: The code is as follows:

{r = 22, l = 2 10^-1, c = 1 10^-4, vi = 24};
\[CapitalDelta]vi = 0;
A = {{0, -1/l}, {1/c, -1/(r c)}};
G = Inverse[A];
B = {1/l, 0};
S1 = Inverse[DiagonalMatrix[{1, 1}] - dt*A + 1/2*dt*dt*A . A];
S2 = Inverse[DiagonalMatrix[{1, 1}] - dt*A];
STAR1 = S1 . G . (B*vi + G . B*\[CapitalDelta]vi/dt) - 
   G . (B*(vi + \[CapitalDelta]vi) + G . B*\[CapitalDelta]vi/dt);
STAR2 = S2 . G . (B*vi + G . B*\[CapitalDelta]vi/dt) - 
   G . (B*(vi + \[CapitalDelta]vi) + G . B*\[CapitalDelta]vi/dt);
X0 = {0, 0};
t = MatrixLog[(MatrixLog[S2] . X0/dt - 
        Inverse[DiagonalMatrix[{1, 1}] - S2] . STAR2 . MatrixLog[S2]/
         dt) . (MatrixLog[S1] . X0/dt - 
        Inverse[DiagonalMatrix[{1, 1}] - S1] . STAR1 . MatrixLog[S1]/
         dt)] . Inverse[MatrixLog[S1 . Inverse[S2]]]*dt;
t /. dt -> 0.00006

Obviously, the formula contains X0, which is a vector that prevents MatrixLog from working properly

I see. Where can I find the calculation method for MatrixLog? And I want to try using Log to replicate the solution of MatrixLog because my formula contains non square matrice.

Yes, I found the problem, thank you! But it seems that MatrixLog can only calculate square matrix. What if the matrix is non square matrix?

Just like the following example：

{a = 1, b = 2};
S = {a, b};
MatrixLog[S]