Basis for an eigenspace Let $A=\begin{pmatrix}
6&3&-8\\
0&-2&0\\
1&0&-3\\
\end{pmatrix}$. I found that the two distinct eigenvalues are $\lambda=5,-2$. Now, I am asked to find a basis for each eigenspace corresponding to each eigenvalue. I found the basis for the eigenspace corresponding to  $\lambda=5$ to be $(8,0,1)$. For the eigenspace corresponding to $\lambda=-2$, I get a basis different than the key; below is my work.
We say $\vec{x}$ is an eigenvector of a given matrix A if and only if A$\vec{x}=\lambda\vec{x}$ for some eigenvalue $\lambda$. Thus, we solve for $\vec{x}$. $$(A-\lambda I)\vec{x}=\vec{0}$$ $$\begin{pmatrix}
6&3&-8\\
0&-2&0\\
1&0&-3\end{pmatrix}\vec{x}-\begin{pmatrix}
-2&0&0\\
0&-2&0\\
0&0&-2\\
\end{pmatrix}\vec{x}=\vec{0}$$ $$\begin{pmatrix}
8&3&-8\\
0&0&0\\
1&0&-1\\
\end{pmatrix}\vec{x}=\vec{0}$$ This equates to to solving the linear system: $$\begin{pmatrix}
8&3&-8&0\\
0&0&0&0\\
1&0&-1&0\\
\end{pmatrix}$$
Interchanging rows $1$ and $3$: $$\begin{pmatrix}
1&0&-1&0\\
0&0&0&0\\
8&3&-8&0\\
\end{pmatrix}$$
Eliminating below $1$ in first column: $$\begin{pmatrix}
1&0&-1&0\\
0&0&0&0\\
0&3&0&0\\
\end{pmatrix}$$
Interchanging rows $2$ and $3$ and rescaling: $$\begin{pmatrix}
1&0&-1&0\\
0&1&0&0\\
0&0&0&0\\
\end{pmatrix}$$
Thus I would conclude my basis for the eigenspace corresponding to $\lambda=-2$ is $(1,0,0)$. However, the book insists the basis is $(1,0,1)$. Are there any errors in my work? Thanks in advance!
Edit: I now see my error. The basis is indeed $(1,0,1)$. 

From here, 
$$\begin{pmatrix}
8&3&-8\\
0&0&0\\
1&0&-1\\
\end{pmatrix}\vec{x}=\vec{0}$$
we see that $\vec{x}=(1,0,1)$ is the correct solution.
Indeed from here
$$\begin{pmatrix}
8&3&-8&0\\
0&0&0&0\\
1&0&-1&0\\
\end{pmatrix}\stackrel{R_1-8R_3}\to \begin{pmatrix}
0&3&0&0\\
0&0&0&0\\
1&0&-1&0\\
\end{pmatrix}
\implies x_2=0 \quad x_1=x_3$$
