The system of three DE I want to solve the following system of DE:
$$
\begin{cases} \dot{x} = 2x+6y -15z,  \\ \dot{y} =x+y-5z,\\ \dot{z} = x+2y-6z, \end{cases}
$$
First, I rewtite the coefficents in the matrix form:
$$A = \begin{bmatrix}
2 & 6 & -15\\
1 & 1 & -5\\
1 & 2 & -6
\end{bmatrix}$$Then, I find $$det(A-\lambda I) = \begin{vmatrix}
2 -\lambda& 6 & -15\\
1 & 1 -\lambda& -5\\
1 & 2 & -6-\lambda
\end{vmatrix}
=-(\lambda+1)^3
$$
$\lambda=-1$ is of multiplicity $3$ and I don't know how to continue.
 A: When the coefficient matrix $A$ has only one (repeated) eigenvalue $\lambda$, you’re in luck: the exponential $e^{tA}$ is easily computed without having to find any eigenvectors, generalized or otherwise. If the eigenvalue’s algebraic and geometric multiplicities are equal, then it must be a multiple of the identity matrix, and the exponential is trivially $e^{\lambda t}I$. Otherwise, for a $3\times3$ matrix, $A-\lambda I$ is nilpotent of index at most $3$. Moreover, $\lambda I$ and $A-\lambda I$ commute, therefore $$e^{tA} = e^{\lambda t}e^{t(A-\lambda I)} = e^{\lambda t}\left(I+t(A-\lambda I)+\frac{t^2}2(A-\lambda I)^2\right).$$ You can save yourself a bit of work by examining $A-\lambda I$: it will be obvious if this is a rank-1 matrix, in which case $(A-\lambda I)^2=0$.  
In this case, you’ve found that $\lambda = -1$. We then have $$A-\lambda I = \begin{bmatrix}3&6&-15\\1&2&-5\\1&2&-5\end{bmatrix},$$ which is clearly a rank-one matrix. Therefore, $$e^{tA} = e^{-t}\begin{bmatrix} 1+3t & 6t &-15t \\ t & 1+2t & -5t \\ t & 2t & 1-5t \end{bmatrix}.$$ The general solution to the system of differential equations is then obtained by multiplying this matrix by a vector of arbitrary constants.  
It’s likely, though, that you’re meant to compute the Jordan decomposition of $A$ and use that to produce the solution to the system. This is a tedious and unnecessary process for this particular matrix.
A: For $\lambda =-1$, the rank of $A- \lambda I $ is 1, so the dimension of the corresponding eigenspace is 2, which is not equal to the multiplity of the eigenvalue in its characteristic equation, which is 3. Thus the matrix is not diagonalizable. You need to learn about the Jordan form and generalized eigenvectors, which are solutions to $$(A- \lambda I)^nv=0$$ for $n>1.$ I'm not going to do the details for you, but using generalized eigenvectors you can construct a matrix $P$ such that $P^{-1}AP$ has -1 on the leading diagonal and either one  or two 1's on the super-diagonal. Thus exactly one of the following will be true: (i) there exists a non-singular $P$ such that $$P^{-1}AP=\begin{bmatrix}-1&1&0\\0&-1&1\\0&0&-1 \end{bmatrix}$$ or (ii) there exists a non-singular $P$ such that $$P^{-1}AP=\begin{bmatrix}-1&0&0\\0&-1&1\\0&0&-1\end{bmatrix}$$ In either case, let $$\begin{bmatrix}x\\y\\ z\end{bmatrix}=P\begin{bmatrix}\eta\\ \zeta \\ \xi\end{bmatrix}$$. Then $$\begin{bmatrix}\dot{\eta}\\ \dot{\zeta}\\ \dot{\xi}\end{bmatrix}=P^{-1}AP\begin{bmatrix}\eta\\\zeta\\\xi\end{bmatrix}$$ Then you solve for first $\xi$, then $\zeta$ and finally $\eta$. Finally you can recover $x,y \text { and }z.$
