Solve the following system: $$$\left . \begin{array} {rcl} x' & = & -x-y+z \\ y' & = & -y -z \\ z' & = & y-3z \end{array}\right\}$$$
Development:
We observe that we are dealing with a linear systems with constant coefficients, where $$$A=\begin{pmatrix} -1 & -1 & 1 \\ 0 & -1 & -1 \\ 0 & 1 & -3 \end{pmatrix}$$$ The eigenvalues of this matrix are:
$$\lambda_1=-1 \\ \lambda_2=\lambda_3=-2$$
and a basis of eigenvectors is: $$$v_1=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \ v_2=\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \ v_3=\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$$$ Notice that this matrix does not diagonalize. Therefore we already calculated: $$$J=\begin{pmatrix} -1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 1 & -2 \end{pmatrix}; \ \ S=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$$ We know that the fundamental matrix of the system is: $$$\phi(t)=S\cdot e^{tJ}=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix}e^{-t} & 0 & 0 \\ 0 & e^{-2t} & 0 \\ 0 & t\cdot e^{-2t} & e^{-2t} \end{pmatrix} =$$$ $$$= \begin{pmatrix} e^{-t} & e^{-2t} & 0 \\ 0 & e^{-2t}+t\cdot e^{-2t} & e^{-2t} \\ 0 & t\cdot e^{-2t} & e^{-2t} \end{pmatrix}=$$$ $$$= \begin{pmatrix} e^{-t} & e^{-2t} & 0 \\ 0 & (t+1)e^{-2t} & e^{-2t} \\ 0 & t\cdot e^{-2t} & e^{-2t} \end{pmatrix}$$$
Multiplying by a vector of constants, we obtain our solution.
Solution:
$$$\begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix}= \begin{pmatrix} e^{-t} & e^{-2t} & 0 \\ 0 & (t+1)e^{-2t} & e^{-2t} \\ 0 & t\cdot e^{-2t} & e^{-2t} \end{pmatrix}\cdot\begin{pmatrix} c_1 \\ c_2 \\ c_3 \end{pmatrix} $$$