I think the first problem was from the textbook
$
\mathbf x' = \left(\begin{array}{ccc}
1 & 1 & 1 \\
2 & 1 & -1 \\
0 & -1 & 1 \\
\end{array}\right) \mathbf x \\
$
The characteristic polynomial is
$
-4 + 3k^2 - k^3 = 0 \\
k = 2, 2, -1
$
The eigenvectors are
$
\lambda = -1,
\left(\begin{array}{c}
-3 \\
4 \\
2
\end{array}\right),
\lambda = 2,
\left(\begin{array}{c}
-0 \\
-1 \\
1
\end{array}\right)
$
Jordan decomposition yields the similarity transform of
$
\mathbf T = \left(\begin{array}{ccc}
-3 & 0 & -1 \\
4 & -1 & -1 \\
2 & 1 & 0
\end{array}\right)
$
Thus the solution is
$
\mathbf x = c_1 \left(\begin{array}{c} -3 \\ 4 \\ 2 \end{array}\right) e^{-t} + c_2 \left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right) e^{2t} + c_3 \left( \left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right) t e^{2t} + \left(\begin{array}{c} -1 \\ -1 \\ 0 \end{array}\right) e^{2t}\right)
$
