Finding a principal fundamental matrix The question is as follows: 

Let $$A=\begin{pmatrix} \sin(x) & 0 \\ 0 & -1\end{pmatrix} $$ This is periodic with period $T = 2\pi$. Find a principal fundamental matrix $\phi(x)$.

I am currently able to do problems similar to this where it is just constants in the matrix, but continue to struggle with this question. $($I have a feeling I need to use Floquet's theorem but I am not sure$)$. Any help would be much appreciated. Thanks.
 A: I take it you're solving the differential equation system
$$\dfrac{dY}{dx} = \pmatrix{\sin(x) & 0\cr 0 & -1\cr} Y $$
Hint: the system decouples into $$ \eqalign{\dfrac{dy_1}{dx} &= \sin(x) y_1\cr 
\dfrac{dy_2}{dx} &= - y_2\cr}$$
A: Let's see now . . . 
If I recall correctly, a principal fundamental matrix corresponding to the coefficient matrix
$A(x) = \begin{bmatrix} \sin x & 0 \\ 0 & -1 \end{bmatrix} \tag 1$
is the $2 \times 2$ matrix function $\phi(x, x_0)$ of $x$ satisfying
$\phi'(x, x_0) = \dfrac{d \phi(x, x_0)}{dx} = A(x) \phi(x, x_0), \; \phi(x_0, x_0) = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}; \tag 2$
if we write
$\phi(x, x_0) = \begin{bmatrix} \phi_{11}(x, x_0) & \phi_{12}(x, x_0)  \\ \phi_{21}(x, x_0) & \phi_{22}(x, x_0) \end{bmatrix}, \tag 3$
then we see that the $\phi_{ij}(x, x_0)$ satisfy
$\phi'_{1j}(x, x_0) = (\sin x) \phi_{1j}(x, x_0), \tag 4$
$\phi'_{2j}(x, x_0) = - \phi_{2j}(x, x_0), \tag 5$
where $j = 1, 2$; the solution to (5) is easily recognized to be
$\phi_{2j}(x, x_0) = \phi_{2j}(x_0, x_0) e^{-(x - x_0)} = \phi_{2j}(x_0, x_0)e^{x_0 - x}, \tag 6$
whereas that to (4) is only slightly less simple:
$\phi_{1j}(x, x_0) = \phi_{1j}(x_0, x_0) e^{-(\cos x - \cos x_0)} = \phi_{1j}(x_0, x_0) e^{\cos x_0 - \cos x}; \tag 7$
assembling (6) and (7) into $\phi(x, x_0)$ whilst taking the initial conditions given in (2) into account yields
$\phi(x, x_0) = \begin{bmatrix} e^{\cos x_0 - \cos x} & 0 \\ 0 & e^{x_0 - x} \end{bmatrix} \tag 8$
as the desired fundamental matrix.
