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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?

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    @Robert: You could write that as an answer.2012-12-14

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The $\cos(t)$ and $\sin(t)$ (which come from the complex exponentials $e^{it}$ and $e^{-it}$) say that $\pm i$ are eigenvalues. The characteristic polynomial (which as joriki noted, should have $-\lambda^3$, not $-\lambda^2$) must be divisible by $(\lambda+i)(\lambda-i) = \lambda^2+1$, and the quotient will tell you what the third eigenvalue is.