Consider the following 2-dimensional system $$ \dot{x}(t) = A(t)x(t) \quad x(0)\in\mathbb{R}^2, $$ where $A(t)$ is a 2-dimensional time-varying matrix. Suppose that the origin of the above system is an unstable equilibrium.
Now consider the following "perturbed" system $$ \dot{z}(t) = (A(t)+\Delta(t))z(t) \quad z(0)\in\mathbb{R}^2, $$ where $\Delta(t)$ is a $2\times 2$ matrix whose entries are zero-mean periodic functions of $t$.
My question: Is the origin of the "perturbed" system unstable for every choice of $\Delta(t)$ as above?
My feeling is that the answer is no, but I couldn't find so far an explicit counterexample.
Markus and Yamabe showed that the origin in the system $$ \dot{x}(t) = \begin{pmatrix} -1 + \frac{3}{2} \cos^2(t) & 1 - \frac{3}{2} \sin(t) \cos(t) \\ -1 - \frac{3}{2} \sin(t) \cos(t) & -1 + \frac{3}{2} \sin^2(t) \end{pmatrix} x(t) $$ is unstable (see, e.g., p. 121 in Hale Ordinary Differential Equations). If we take $$ A(t) = \begin{pmatrix} -1 + \frac{3}{2} \cos^2(t) & 1 - \frac{3}{2} \sin(t) \cos(t) \\ -1 - \frac{3}{2} \sin(t) \cos(t) & -1 + \frac{3}{2} \sin^2(t) \end{pmatrix}, \quad \Delta(t) = \begin{pmatrix} - \frac{3}{4} \cos(2t) & \frac{3}{4} \sin(2t) \\ \frac{3}{4} \sin(2t) & \frac{3}{4} \cos(2t) \end{pmatrix} $$ ($\Delta(t)$ has zero-mean periodic entries) then the autonomous system $$ \dot{z}(t) = \begin{pmatrix} -\frac{3}{4} & 1 \\ -1 & -\frac{3}{4} \end{pmatrix} z(t), $$ having eigenvalues with negative real parts, is asymptottically stable.