I have been given a problem to find the Jordan Canonical form of the following $ 3 \times 3 $ matrix

$$ A=
\begin{bmatrix} 
3 & 0 & 1 \\
-4 & 1 & -2 \\
-4 & 0 & -1 
\end{bmatrix}
$$

I have been told to do this using the transformation matrix way, wherein $A$ can be written as

$$
A = TJT^{-1}
$$
where $T$ is the modal matrix analogue which contains the eigenvectors and/or the generalized eigenvectors as columns and $J$ is Jordan Canonical form of A. Doing some matrix operations on the above equation, $J$ can be obtained as

$$
J = T^{-1}AT
$$

Coming back to the problem, I have found the characteristic equation and got all three eigenvalues as $1$. I found eigenvectors by

$$
Ax\_1 = \lambda x\_1
$$

Solving this I got two linearly independent eigenvectors, since I was getting only 1 pivot after row transformation operations on $A$ the above equation must give me two solutions. I got these

$$
x\_1 = \begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}, \begin{bmatrix}
1 \\ 0 \\ -2
\end{bmatrix}
$$

For the next step, according to my textbook, I should get a generalized eigenvector by solving for $x\_2$

$$
Ax\_2 = \lambda x\_2 + x\_1
$$

The textbook example only shows one possible value for $x\_1$, so clearly I've got a problem regarding which one of the two eigenvectors I got, should be chosen for the next step.

But even without considering that, I have yet another problem

$$
Ax\_2 = \lambda x\_2 + x\_1 \\
(A - \lambda I)x\_2 = x\_1
$$

If I find out the matrix $A - \lambda I = A - I$, I get

$$
\begin{bmatrix}
2 & 0 & 1\\
-4 & 0 & -2\\
-4 & 0 & -2
\end{bmatrix}
$$

which forms an inconsistent set of equations to solve, because of the repeated row but different values in the rows of (either) $x\_1$. So I can't proceed further to get $x\_2$.

Have I done something wrong, or is there some pre-condition to be checked on $A$ that I missed?