I am studying some properties of the initial value problem \begin{equation*} \begin{cases} \frac{d}{dt}X(t)=A(t)X(t),\\ X(t_0)=I. \end{cases} \end{equation*} where $A(t)$ is a real-valued $n\times n$ matrix.
If we denote as $X(t,t_0)$ the solution of this problems, we have the following properties:
$X(t,r)=X(t,s)X(s,r)$
$X(s,t)=X(t,s)^{-1}$
I want to show this two properties and I know how to proof the first one using the second. Basicly, if you define $Y(t)=X(t,s)X(s,r)$, you can do the following calculation:
\begin{equation*} \begin{split} \frac{d}{dt}Y(t)&=\frac{d}{dt}X(t,s)X(s,r) \\ &=A(t)X(t,s)X(s,r) \\ &=A(t)Y(t). \end{split} \end{equation*}
It is also easy to see that $Y(r)=I$ using the second property and we have the first property. My problem is, how can I prove the second property?
We can show these properties by allowing the both the time and value in the intial conditions to vary. Define $X(t,r,B)$ as solution to the initial value problem $$\begin{cases} \frac{d}{dt}X(t,r,B)=A(t)X(t,r,B) \\ X(r,r,B)=B \end{cases}$$ Since $\frac{d}{dt}(X(t,r,I)B)=\frac{d}{dt}(X(t,r,I))B$, we see by right multiplying the entire IVP by $B$ that $X(t,r,B)=X(t,r,I)B$. Now, by assuming solutions exist for all time and are unique, any solutions that agree at one time agree everywhere. This implies $$ X(t,r,I)=X(t,s,X(s,r,I)) $$ Applying the multiplicative property $X(t,r,B)=X(t,r,I)B$ gives the first property $$ X(t,r,I)=X(t,s,I)X(s,r,I) $$ The second property follows immediately from the first, since $X(t,s,I)X(s,t,I)=X(t,t,I)=I$.