Computing eigenvalues for specific initial conditions

Take the following equation $-x'' = \gamma x$ which has the known solution $$x = A\cos(\sqrt{\gamma}t ) + B \sin(\sqrt{\gamma}t).$$Now assume that $x(0) = x(\alpha)$ and $x'(0) = -x'(\alpha)$. I am looking for the eigenvalues $\gamma$.

Approach: I tried constructing the system that results from the two equations: $$\begin{cases}A = A\cos(\sqrt{\gamma}\alpha) + B\sin(\sqrt{\gamma}\alpha),\\B\sqrt{\gamma} = A\sqrt{\gamma}\sin(\sqrt{\gamma}\alpha) - B\sqrt{\gamma}\cos(\sqrt{\gamma}\alpha),\end{cases}$$which can then be transformed into a matrix when solving for $A$ and $B$, $$M = \begin{bmatrix}1- \cos(\sqrt{\gamma}\alpha) & - \sin(\sqrt{\gamma}\alpha)\\ - \sqrt{\gamma} \sin(\sqrt{\gamma}\alpha) & \sqrt{\gamma}+ \sqrt{\gamma}\cos(\sqrt{\gamma}\alpha) \end{bmatrix}.$$ But this matrix has determinant zero. I am stuck now.

Question: How can I get the eigenvalue $\gamma$ from these conditions?