Beginner troubleshooting an eigenvector calculation I am having some difficulty identifying the error in my eigenvector calculation. I am trying to calculate the final eigenvector for $\lambda_3 = 1$ and am expecting the result $ X_3 = \left(\begin{smallmatrix}-2\\17\\7\end{smallmatrix}\right)$
To begin with, I set up the following equation (for the purpose of this question I will refer to the leftmost matrix here as A). 
$$
\begin{bmatrix}
   1 - \lambda & 0 & 0 \\ 
   3 & 3 - \lambda  & -4\\ 
   -2 & 1 & -\lambda -2 \\
\end{bmatrix}
\begin{bmatrix}
  x_1 \\ 
   x_2\\ 
   x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0\\ 
0\\ 
0\\
\end{bmatrix}
$$
I) Substitute $\lambda_3 = 1$
$$
\begin{bmatrix}
   0 & 0 & 0 \\ 
   3 & 2  & -4\\ 
   -2 & 1 & -3 \\
\end{bmatrix}
\begin{bmatrix}
  x_1 \\ 
   x_2\\ 
   x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0\\ 
0\\ 
0\\
\end{bmatrix}
$$
II) Reduce the matrix with elementary row operations.
$R_2 \leftarrow R_2 - 2R_3$
$$
A = 
\begin{bmatrix}
   0 & 0 & 0 \\ 
   7 & 0  & 2\\ 
   -2 & 1 & -3 \\
\end{bmatrix}
$$
$R_3 \leftarrow  3R_2 +  2R_3$
$$
A = 
\begin{bmatrix}
   0 & 0 & 0 \\ 
   7 & 0  & 2\\ 
   17 & 2 & 0 \\
\end{bmatrix}
$$
$R_2 \leftarrow  \frac{1}{7} R_2$
$R_3 \leftarrow \frac{1}{17} R_3$$
$$
A = 
\begin{bmatrix}
   0 & 0 & 0 \\ 
   1 & 0  & 2/7\\ 
   1 & 2/17 & 0 \\
\end{bmatrix}
$$
III) multiply matrices to get a series of equations equal to 0 and rearrange them in terms of a common element.
$x_1 + \frac{2}{7}x_3 = 0 \rightarrow x_1 = -\frac{2}{7}x_3$
$x_1 + \frac{2}{17}x_2 = 0 \rightarrow x_1 = -\frac{2}{17}x_2$
IV) Substitute a value into the vector to get an eigenvector.
Let $\ x_1 = 1 \rightarrow X_3 =  \left(\begin{smallmatrix}1\\-2/17\\-2/7\end{smallmatrix}\right)
$
Which at this point we can see is not a multiple of the expected $X_3$. Can anyone highlight my error for me?
Many thanks in advance.
 A: The error is right in the end after you obtain the system of equations. You have
$$
x_1 + \frac{2}{7}x_3 = 0 \\ x_1 + \frac{2}{17}x_2 = 0
$$
You want to write $x_2$ and $x_3$ in terms of the common element $x_1$ so
$$
x_2 = -\frac{17}{2}x_1 \\ x_3 = -\frac{7}{2}x_1
$$
This means that the vector you are looking for is 
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix} = x_1\begin{pmatrix} 1 \\ -17/2 \\ -7/2\end{pmatrix}$$
A: Uh, that's just bad algebra- your fractions are upside down!  You have $x_1= -\frac{2}{7}x_3$ and $x1= -\frac{2}{17}x_2$.  Setting $x_1= 1$ gives $-\frac{2}{7}x_3= 1$ so $x_3= -\frac{7}{2}$ and $-\frac{2}{17}x_2= 1$ so $x_2= -\frac{17}{2}$, not what you have. that gives $X_3= \begin{pmatrix} 1 \\ -\frac{17}{2} \\ -\frac{7}{2} \end{pmatrix}$.   That is a multiple of $\begin{pmatrix}2 \\- 17 \\ -7\end{pmatrix}$.
A: You have solved the problem correctly up to $$x_1 + \frac{2}{7}x_3 = 0 \rightarrow x_1 = -\frac{2}{7}x_3$$ and $$x_1 + \frac{2}{17}x_2 = 0 \rightarrow x_1 = -\frac{2}{17}x_2$$ Now when you let $x_1=1$ you get $x_3=-\frac {7}{2}$ and $x_2=-\frac {17}{2}$ which give you the correct eigenvector.  
