Diagonalization of a Matrix. Having Trouble.. This is for Continuous-time Markov chains but I'm having trouble with the linear algebra.
Formal Definitions:
If Q is diagonalizable, then so is $e^{tQ}$, and the transition function can be expressed in terms of the eigenvalues and eigenvectors of $Q$. Write $Q=SDS^{-1}$, where $D$ is a diagonal matrix whose diagonal entries are the eigenvalues of $Q$, and $S$ is an invertible matrix whose columns are the corresponding eigenvectors. This gives,
$$e^{tQ} = Se^{tD}S^{-1}$$
The Problem:
A Markov Chain has generator matrix,
$$Q=
    \begin{pmatrix}
    -1 & 1 & 0 \\
    0 & -2 & 2 \\
    3 & 0 & -3 \\
    \end{pmatrix}
$$
Find the transition function by diagonalizing the generator and finding the matrix exponential.

My pr0fessor seems to be getting eigenvalues $\lambda = -4, \lambda = -2, \lambda = 0$.
I, on the other hand, seem to be getting $\lambda = -3, \lambda = 0$.
Edit: I recalculated but still am getting $\lambda = 0$.
This gets him a completely different set of eigenvectors.. Am I miscalculating something?
EDIT:
In the back of the book I have,
$$P(t)=
    \begin{pmatrix}
    -1 & 0 & 1 \\
    -3 & 1 & 1 \\
    3 & 0 & 1 \\
    \end{pmatrix}
\begin{pmatrix}
    e^{-4t} & 0 & 0 \\
    0 & e^{-2t} & 0 \\
    0 & 0 & 1 \\
    \end{pmatrix}
\begin{pmatrix}
    -1/4 & 0 & 1/4 \\
    -3/2 & 1 & 1/2 \\
    3/4 & 0 & 1/4 \\
    \end{pmatrix}
$$
Where $P(t)$ is the transition function asked for.

There's a typo.  Reverse engineering the given answer we have:
$$
\begin{align}
Q&=PDP^{-1}\\
&=\begin{pmatrix}
    -1 & 0 & 1 \\
    -3 & 1 & 1 \\
    3 & 0 & 1 \\
    \end{pmatrix}
\begin{pmatrix}
    -4 & 0 & 0 \\
    0 & -2 & 0 \\
    0 & 0 & 0 \\
    \end{pmatrix}
\begin{pmatrix}
    -1/4 & 0 & 1/4 \\
    -3/2 & 1 & 1/2 \\
    3/4 & 0 & 1/4 \\
    \end{pmatrix}\\
&=\begin{pmatrix}
    -1 & 0 & 1 \\
    0 & -2 & 2 \\
    3 & 0 & -3 \\
    \end{pmatrix}
\end{align}
$$
Now knowing that 
$$
Q=\begin{pmatrix}
    -1 & 0 & 1 \\
    0 & -2 & 2 \\
    3 & 0 & -3 \\
    \end{pmatrix}
$$
You can find your characteristic polynomial, e.g. via cofactor expansion down the second column to get $$p(\lambda)=-\lambda(\lambda+2)(\lambda+4)$$ and proceed from there.
