With the matrix $A$ given by $\left( \begin{array}{ccc} 0 & -1 & 0 \\ 0 & 1 & a \\ 1 & 0 & 1 \end{array} \right)$
the solution to the initial value problem $x'=Ax$, $x(0) = \left( \begin{array}{ccc} 0 \\ -1\\ b \end{array} \right)$ is $x = \left( \begin{array}{ccc} -\cos t-\sin t+e^{ct} \\ \cos t-\sin t -2e^{ct} \\ \cos t+e^{ct} \end{array} \right)$
Here $a,b,$ and $c$ are real constants. What are they?
I found the determinant of the matrix $A = -\lambda^2+2\lambda^2-\lambda-a$, but I'm not sure how I find out what the eigenvalues are supposed to be from the given solution. Is there anything I can understand about the process to make the simplification easier?