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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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Hint: What does the third row of x(t) tell you about b from x(0)? –  Amzoti Dec 14 '12 at 3:30
    
It says that $b=2$. –  Bailor Tow Dec 14 '12 at 3:35
    
Okay, now look at all of the information you have. Can you see the next step you can take from all of this? –  Amzoti Dec 14 '12 at 3:36
    
Hint: you have $A$, you know $x$, you can calculate $x'$ and you can you setup $x'$ = $Ax$ and then see if you can multiply out the right side and try to figure out a and c? Make sense? –  Amzoti Dec 14 '12 at 3:42
    
Shouldn't I be able to tell what the eigenvalues are of the system based on the solution given? Then I can just equate my determinant of A to one that solves the system in c? –  Bailor Tow Dec 14 '12 at 3:53

1 Answer 1

up vote 0 down vote accepted

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.

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