Explicitly computing the monodromy matrix of a system of ODEs

I am trying to get my head around the idea of Monodromy matrices, but I am having issues. Suppose we are considering the following period 1 system, ˙x(t)=A(t)x(t)\dot{x}(t)=A(t)x(t) where
A(t)={A1=(110−1) if t∈[0,.5)A2=(−1011) if t∈[.5,1).A(t)=\left\{ \begin{matrix}A_1=\left(\begin{matrix}1 &1\\0 &-1\end{matrix}\right) &\textrm{ if } t\in[0,.5)\\
A_2=\left(\begin{matrix}-1 &0\\1 &1\end{matrix}\right) &\textrm{ if } t\in[.5,1)\end{matrix}\right. .

The goal is to compute the monodromy matrix of this system. The questions I have are:

Is my current attempt correct?
If we are going to use the monodromy matrix to see whether we are stable or not, then does the initial time actually matter, or can we assume it to be 0 without loss of generality?

Current attempt:
If I start at t0∈[0,.5),t_0\in[0,.5), then I need to first consider what the solution would be at t=1/2.t=1/2. This is is given by eA1(1/2−t0)x0.e^{A_1(1/2-t_0)}x_0. Likewise, the solution at time t=1t=1 should be eA2(1−1/2)eA1(1/2−t0)x0.e^{A_2(1-1/2)}e^{A_1(1/2-t_0)}x_0. To complete our trip, we have eA1(t0)eA2(1−1/2)eA1(1/2−t0)x0e^{A_1(t_0)}e^{A_2(1-1/2)}e^{A_1(1/2-t_0)}x_0.

So my monodromy matrix would be given by
M(t0)=[eA1(t0)eA2(1−1/2)eA1(1/2−t0)e1,eA1(t0)eA2(1−1/2)eA1(1/2−t0)e2]M(t_0)=[e^{A_1(t_0)}e^{A_2(1-1/2)}e^{A_1(1/2-t_0)}e_1,\qquad e^{A_1(t_0)}e^{A_2(1-1/2)}e^{A_1(1/2-t_0)}e_2]

The case when t0∈[.5,1)t_0\in[.5,1) can be approached similarly to obtain
M(t0)=[eA2(t0−1/2)eA1(1/2)eA2(1−t0)e1,eA2(t0−1/2)eA1(1/2)eA2(1−t0)e2].M(t_0)=[e^{A_2(t_0-1/2)}e^{A_1(1/2)}e^{A_2(1-t_0)}e_1,\qquad
e^{A_2(t_0-1/2)}e^{A_1(1/2)}e^{A_2(1-t_0)}e_2].

Original attempt:
I think the above method is better, but I’ve included this just to show I have put in some effort. before asking for help. First, consider the systems

{˙x1(t)=A1x(t) for t∈[0,.5)x1(t0,1)=x1,0, and {˙x2(t)=A2x(t) for t∈[.5,1)x2(t0,2)=x2,0\left\{\begin{matrix}\dot{x_1}(t)=A_1x(t) \textrm{ for } t\in[0,.5)\\ x_1(t_{0,1})=x_{1,0}\end{matrix}\right., \textrm{ and }
\left\{\begin{matrix}\dot{x_2}(t)=A_2x(t) \textrm{ for } t\in[.5,1)\\ x_2(t_{0,2})=x_{2,0}\end{matrix}\right.

have the solutions \phi_1(t,t_{0,1},x_{0,1})=e^{A_1(t-t_{0,1})}x_{0,1},\phi_1(t,t_{0,1},x_{0,1})=e^{A_1(t-t_{0,1})}x_{0,1}, and \phi_2(t,t_{0,2},x_{0,2})=e^{A_2(t-t_{0,2})}x_{0,2},\phi_2(t,t_{0,2},x_{0,2})=e^{A_2(t-t_{0,2})}x_{0,2}, respectively.

So for our original system, we have that
\phi(t,t_0,x_0) = \left\{\begin{matrix}
\left\{\begin{matrix}\phi_1(t,t_0,x_0) &\textrm{if } t\in[0,.5)\\
\phi_2(t,1/2,\phi_1(1/2,t_0,x_0))&\textrm{if }t\in[5,1)\end{matrix}\right.
&\textrm{ if } t_0\in[0,.5)\\
\left\{\begin{matrix}\phi_1(t,1/2,\phi_2(1/2,t_0,x_0))&\textrm{if } t\in[0,.5)\\
\phi_2(t,t_0,x_0))&\textrm{if }t\in[5,1)\end{matrix}\right.
&\textrm{ if } t_0\in[.5,1)\\
\end{matrix}\right.,\phi(t,t_0,x_0) = \left\{\begin{matrix}
\left\{\begin{matrix}\phi_1(t,t_0,x_0) &\textrm{if } t\in[0,.5)\\
\phi_2(t,1/2,\phi_1(1/2,t_0,x_0))&\textrm{if }t\in[5,1)\end{matrix}\right.
&\textrm{ if } t_0\in[0,.5)\\
\left\{\begin{matrix}\phi_1(t,1/2,\phi_2(1/2,t_0,x_0))&\textrm{if } t\in[0,.5)\\
\phi_2(t,t_0,x_0))&\textrm{if }t\in[5,1)\end{matrix}\right.
&\textrm{ if } t_0\in[.5,1)\\
\end{matrix}\right.,
constructed so that our solutions are continuous at t=1/2.t=1/2. Now the monodromy matrix should be
M(t_0)=\Pi(t_0+1,t_0)=\left[\begin{matrix} \phi(t_0+1,T_0,e_1) &\phi(t_0+1,T_0,e_2)\end{matrix}\right]M(t_0)=\Pi(t_0+1,t_0)=\left[\begin{matrix} \phi(t_0+1,T_0,e_1) &\phi(t_0+1,T_0,e_2)\end{matrix}\right]

right? However, this seems unfortunate since I will need to find the eigenvalues of this matrix and there are so many cases. So I decided to try another route (above).

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