Finding eigenvalues and eigenvectors and then determining their geometric and algebraic multiplcities I have the following matrix:
$A = \begin{bmatrix} 
    1 && 7 && -2 \\ 
    0 && 3 && -1 \\
    0 && 0 && 2 \end{bmatrix}$
and I am trying to find the eigenvalues and eigenvectors followed by their respective geometric multiplicity and algebraic multiplicity. 
What I have so far:
$\det(A - \lambda I) = \det\begin{bmatrix} 
    1-\lambda && 7 && -2 \\ 
    0 && 3-\lambda && -1 \\
    0 && 0 && 2-\lambda \end{bmatrix}$
I see that it is an upper triangular matrix so determinant is just the diagonal. Which gives me 
$(1-\lambda)(3-\lambda)(2-\lambda)$ which gives me $\lambda = 1,3,2$. I also notice that all three have the algebraic multiplicity of 1 (their exponents were 1).
Following that I move on to the geometric multiplicity:
$ A - 3I = \begin{bmatrix} 
    -2 && 7 && -2 \\ 
    0 && 0 && -1 \\
    0 && 0 && -1 \end{bmatrix}$ which has RRE of $\begin{bmatrix} 
    1 && -\frac{7}{2} && 0 \\ 
    0 && 0 && 1 \\
    0 && 0 && 0 \end{bmatrix}$ which yields 
$\begin{bmatrix} 
    x \\ 
    y \\
    z \end{bmatrix} = \begin{bmatrix} 
    \frac{7}{2}\\ 
    1 \\
    0\end{bmatrix} s$ which has a geometric multiplicity of 1.
$ A - 2I = \begin{bmatrix} 
    -1 && 7 && -2 \\ 
    0 && 1 && -1 \\
    0 && 0 && 0 \end{bmatrix}$ which has RRE of $\begin{bmatrix} 
    1 && 0 && -5 \\ 
    0 && 1 && -1 \\
    0 && 0 && 0 \end{bmatrix}$ which yields 
$\begin{bmatrix} 
    x \\ 
    y \\
    z \end{bmatrix} = \begin{bmatrix} 
    5\\ 
    1 \\
    1\end{bmatrix} s$ which has a geometric multiplicity of 1.
Finally, 
$ A - I = \begin{bmatrix} 
    0 && 7 && -2 \\ 
    0 && 2 && -1 \\
    0 && 0 && 1 \end{bmatrix}$ which has RRE of $\begin{bmatrix} 
    0 && 1 && 0 \\ 
    0 && 0 && 1 \\
    0 && 0 && 0 \end{bmatrix}$
This is where I am stuck, I'm not sure what is the resulting  $\begin{bmatrix} 
    x \\ 
    y \\
    z \end{bmatrix}$ that column of 0's is confusing me.
Any help would be appreciated.
Edit: Would I just say that column of zeros is a free variable? Thus giving me $\begin{bmatrix} 
    1 \\ 
    0 \\
    0 \end{bmatrix}$ ? or is that wrong?

That RRE $\begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix}$ means $y=0$ and $z=0$, so $$\begin{bmatrix} x\\y\\z \end{bmatrix} =\begin{bmatrix} x\\0\\0 \end{bmatrix} =x\begin{bmatrix} 1\\0\\0 \end{bmatrix}.$$
