Determine $\phi(x,0)$ for $A(x)=\begin{pmatrix} -1 & \cos(x) \\ 0 & -1\end{pmatrix}$, where $\phi(x,0)t_{0}$ is a solution of $\frac{d}{dx}t(x)=A(x)t(x)$.
I am not entirely sure as to how to proceed when $A$ is not constant. What I know is that if $A$ had been constant then I would first begin by determining the eigensystem of $A$ and multiply this by its inverse, from the right, to get $\phi$. Now, since $A$ is not constant, I presume one still can do this, but that the inverse of the eigensystem will depend on $x_{0}=0$. If so, then I get the following result:
Edit - Wrong matrix
I noticed I calculated the eigenvalues and their respective eigenvectors on the wrong matrix $\begin{pmatrix} 1 & \cos(x) \\ 0 &-1 \end{pmatrix}$.
\begin{equation} \begin{pmatrix} e^{x}-\frac{e^{-x}}{2}\cos(x) &-\frac{e^{-x}}{2}\cos(x) \\ -1 &-1\end{pmatrix}=\begin{pmatrix} e^{x}&-\frac{\cos(x)}{2}e^{-x} \\ 0 &e^{-x} \end{pmatrix}\begin{pmatrix} 1 &0 \\ 1 & 1\end{pmatrix} \end{equation}