Simple eigenvalue problem (x'=[2x2]x)

Hi, I've got a homework problem that should be very simple, the previous problem was identical (different numbers) and was a breeze, but this one is giving me problems. Here's what we're asked for:

Apply the eigenvalue method to find the particular solution to the system of differential equations

$\displaystyle x'=\begin{bmatrix}

4 & 4\\

3 & 5

\end{bmatrix}x$

which satifies the initial conditions

$\displaystyle x(0)=\begin{bmatrix}

-4\\

2

\end{bmatrix}$

So I begin by finding the eigenvalues from the characteristic equation:

$\displaystyle (4-\lambda)(5-\lambda)-12=(\lambda^2-9\lambda+8)=(\lambda-1)(\lambda-8)$

$\displaystyle \lambda=1,8$

To get the matrices:

$\displaystyle \begin{bmatrix}

3 & 4\\

3 & 4

\end{bmatrix}\begin{bmatrix}

a\\

b

\end{bmatrix}=0$

and

$\displaystyle \begin{bmatrix}

-4 & 4\\

3 & -3

\end{bmatrix}\begin{bmatrix}

a\\

b

\end{bmatrix}=0$

To get

$\displaystyle \begin{bmatrix}

a\\

b

\end{bmatrix}=\begin{bmatrix}

-4\\

3

\end{bmatrix}$ and $\displaystyle \begin{bmatrix}

a\\

b

\end{bmatrix}=\begin{bmatrix}

1\\

1

\end{bmatrix}$

Which gives the particular solution $\displaystyle x_p=a\begin{bmatrix}

-4\\

3

\end{bmatrix}e^t+b\begin{bmatrix}

1\\

1

\end{bmatrix}e^{8t}$

But at this point, no values for a and/or b will satisfy $\displaystyle x(0)=\begin{bmatrix}

-4\\

2

\end{bmatrix}$

Can anyone help me out, please?

Thanks!