Solving a first order linear system matrix I am attempting to solve a system of equations using a method from a textbook.
The problem comes in when I find multiple copies of the same eigenvalue, and I am struggling to find the 3 eigenvectors.
Here is the question:
Solve
$x'(t) = Bx(t)$
where $B$ is
\begin{bmatrix} -1 & 2 & -1 \\ 0 & -1 & 2  \\ 0 & 0 & -1\end{bmatrix}
The answer is stated to be:
$x = (a + (2b - c)t + 2ct^2)e^{-t}$
$y = (b + 2ct)e^{-t}$
$z = ce^{-t}$
$\textbf{Attempt at solution:}$
Since $B$ is an upper triangular matrix it's clear that the determinant is $(-1 -
 \lambda)^3 $ which gives us $\lambda_{1,2,3} = -1$ as eigenvalues.
After plugging in $\lambda = -1$ for $(B - \lambda I)$
The resulting matrix is
\begin{bmatrix} 0 & 2 & -1 \\ 0 & 0 & 2  \\ 0 & 0 & 0\end{bmatrix}
Therefore, after multiplying this matrix by-
 \begin{bmatrix} x_1  \\ x_2  \\ x_3\end{bmatrix} 
and setting this equal to 0, I get that:
$2x_2 - x_3 = 0$  and  $2x_3 = 0$
I then get the eigenvector $v_1 = (1,0,0)$ (I believe this is correct)
I am not sure how to proceed from this point, in terms of finding 2 more eigenvectors to reach that solution.  If someone could provide a walkthrough from this point and show how the answer is reached, that would be incredibly helpful! This is for self-study. 
 A: Method 1: Notice that the third equation is decoupled from the other two.
We have $$z' = -z \implies z(t) = c e^{-t}$$
Substituting $z(t)$ into the second equation, we have
$$y' = -y + 2 z = -y + 2 c e^{-t} \implies y(t) =  (b + 2ct)e^{-t}$$
Substituting $y(t)$ and $z(t)$ into the first equation, we have
$$x' = -x + 2 y - z = -x + 2(b + 2ct)e^{-t} - c e^{t}$$
Solving we get
$$x(t) = (a + (2b - c)t + 2ct^2)e^{-t}$$
Method 2: Eigenvalues / Eigenvectors
This is a deficient matrix (as you discovered). Are you familiar with generalized eigenvectors and the Jordan form?
The eigenvalue (triple) is $\lambda = -1$ and the RREF of $[A + I]v _1 = 0$ gives
$$\begin{bmatrix}
 0 & 1 & 0 \\
 0 & 0 & 1 \\
 0 & 0 & 0 \\
\end{bmatrix}v_1 = 0 \implies v_1 = (1, 0, 0)$$
Unfortunately, we cannot get any more linearly independent eigenvectors, so we need to find generalized ones. Following these Jordan Matrix Notes, we have the RREF of $[A + I]v_2 = v_1$ of (shown as augmented matrix)
$$\begin{bmatrix}
 0 & 1 & 0 & \frac{1}{2} \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 \\
\end{bmatrix}$$
We can choose 
$$v_2 = \left(0, \dfrac{1}{2}, 0 \right)$$
Repeating for the RREF of $[A + I]v_3 = v_2$, we get
$$\begin{bmatrix}
 0 & 1 & 0 & \frac{1}{8} \\
 0 & 0 & 1 & \frac{1}{4} \\
 0 & 0 & 0 & 0 \\
\end{bmatrix}$$
We can choose 
$$v_3 = \left(0, \dfrac{1}{4}, \dfrac{1}{8} \right)$$
Referring to the notes, we can now write
$$X(t) = e^{-t}\left[c_1 v_1 + c_2 (v_2 + t v_1) + c_3 \left(v_3 + t v_2 + \dfrac{t^2}{2!} v_1\right)\right]$$
