The questions states solve for y such that $$y' = \begin{bmatrix} -4 & 2 & 1 \\ 1 & -3 & 1 \\ 3 & -3 & -2 \\ \end{bmatrix}y , y(0)= c = \begin{bmatrix} 1\\5\\3 \end{bmatrix}$$
The general solution is $y=ce^{At}$ where A is the above 3x3 matrix. We find the eigenvalues and corresponding eigenvectors:
$e_{-5} = \begin{bmatrix} -1\\0\\1 \end{bmatrix}$ $e_{-2} = \begin{bmatrix} 1\\1\\0 \end{bmatrix}$
and the generalised eigenvector for -2
$e_{-2} = \begin{bmatrix} 0\\0\\1 \end{bmatrix}$
We then want to find c as a linear combination of our eigenvalues, and the result is:
$ c = 5 \begin{bmatrix} 1\\1\\0 \end{bmatrix} - \begin{bmatrix} 0\\0\\1 \end{bmatrix} + 4\begin{bmatrix} -1\\0\\1 \end{bmatrix}$
Therefore soln = $ce^{At}$ = $e^{At} (5t \begin{bmatrix} 1\\1\\0 \end{bmatrix} - \begin{bmatrix} 0\\0\\1 \end{bmatrix} + 4\begin{bmatrix} -1\\0\\1 \end{bmatrix})$
Adding a t above to the one vector because we have a generalised eigenvector.
And since $e^{At}v = e^{\lambda t}v$, we can write our solution as $ y = 4e^{-5t}\begin{bmatrix} -1\\0\\1 \end{bmatrix} -e^{-2t}\begin{bmatrix} 0\\0\\1 \end{bmatrix} + 5te^{-2t}\begin{bmatrix} 1\\1\\0 \end{bmatrix}$. However my solution solution is wrong. The correct one is
$ y = 4e^{-5t}\begin{bmatrix} -1\\0\\1 \end{bmatrix} + e^{-2t}\begin{bmatrix} 5\\5\\-1 \end{bmatrix} - te^{-2t}\begin{bmatrix} 1\\1\\0 \end{bmatrix}$
I suspected that there may be multiple correct answers, however i'm not sure. What have i done wrong?