$a_1=-1$, $a_2=2$, and $a_n=\frac {a_{n-1}}{a_{n-2}}$ for $n\geq 3$. What is $a_{2006}$?

$ 
\textbf{(A)}\ -2
\qquad\textbf{(B)}\ -1
\qquad\textbf{(C)}\ -\frac 12
\qquad\textbf{(D)}\ \frac 12
\qquad\textbf{(E)}\ 2
$
1. First, we list out the initial terms given in the problem:
   \[
   a_1 = -1, \quad a_2 = 2
   \]
2. Using the recurrence relation \(a_n = \frac{a_{n-1}}{a_{n-2}}\) for \(n \geq 3\), we calculate the next few terms:
   \[
   a_3 = \frac{a_2}{a_1} = \frac{2}{-1} = -2
   \]
   \[
   a_4 = \frac{a_3}{a_2} = \frac{-2}{2} = -1
   \]
   \[
   a_5 = \frac{a_4}{a_3} = \frac{-1}{-2} = \frac{1}{2}
   \]
   \[
   a_6 = \frac{a_5}{a_4} = \frac{\frac{1}{2}}{-1} = -\frac{1}{2}
   \]
   \[
   a_7 = \frac{a_6}{a_5} = \frac{-\frac{1}{2}}{\frac{1}{2}} = -1
   \]
   \[
   a_8 = \frac{a_7}{a_6} = \frac{-1}{-\frac{1}{2}} = 2
   \]
   \[
   a_9 = \frac{a_8}{a_7} = \frac{2}{-1} = -2
   \]
   \[
   a_{10} = \frac{a_9}{a_8} = \frac{-2}{2} = -1
   \]
3. Observing the terms, we see that the sequence starts repeating every 6 terms:
   \[
   a_1 = -1, \quad a_2 = 2, \quad a_3 = -2, \quad a_4 = -1, \quad a_5 = \frac{1}{2}, \quad a_6 = -\frac{1}{2}
   \]
   \[
   a_7 = -1, \quad a_8 = 2, \quad a_9 = -2, \quad a_{10} = -1, \quad a_{11} = \frac{1}{2}, \quad a_{12} = -\frac{1}{2}
   \]
4. To find \(a_{2006}\), we note that the sequence repeats every 6 terms. Therefore, we need to find the remainder when 2006 is divided by 6:
   \[
   2006 \div 6 = 334 \text{ remainder } 2
   \]
   This means \(a_{2006}\) is the same as \(a_2\).
5. From the initial terms, we know:
   \[
   a_2 = 2
   \]

The final answer is \(\boxed{2}\).