general solution of a matrix in vector form This is my first general solution and I want to put it in vector form. 
\begin{cases}
x1 &= 1\\
x2 &= -1\\
x3 & \text{is free}\\
\end{cases}
Here is the matrix I pulling the information from just in case you need it. 
\begin{bmatrix}
1 & 0 & -1\\ 0 & 1 & 1\\ 0 & 0 & 0
\end{bmatrix}
This is what I am thinking. 
\begin{bmatrix}
1\\ 0\\1
\end{bmatrix}
\begin{bmatrix}
0\\ -1\\1
\end{bmatrix}
This is my second general solution and I want to put it in vector form. 
\begin{cases}
x1 &= -x2 -x3\\
x2 & \text{is free}\\
x3 & \text{is free}\\
\end{cases}
Here is the matrix I pulling the information from just in case you need it. 
\begin{bmatrix}
1 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0
\end{bmatrix}
This is what I am thinking. 
\begin{bmatrix}
1 - 1\\ 0\\0
\end{bmatrix}
so 
\begin{bmatrix}
0\\ 0\\0
\end{bmatrix}
The first one I thought for sure was right but the second one I don't think is right. 

Fixed my typo in the first matrix.
This is my first general solution and I want to put it in vector form. 
\begin{cases}
x1 &= .5x4\\
x2 &= x4\\
x3 &= .5x4\\
x4 & \text{is free}\\
\end{cases}
Here is the matrix I pulling the information from just in case you need it. 
\begin{bmatrix}
1 & 0 & 0 & -.5 \\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & -.5\\ 0 & 0 & 0 & 0
\end{bmatrix}
This is what I am thinking. 
$$\begin{bmatrix}
x_1\\ 
x_2\\ 
x_3\\
x_4\end{bmatrix}=\begin{bmatrix}
.5x_4\\ 
x_4\\ 
.5x_4\\ 
x_4\end{bmatrix}=x_4\begin{bmatrix}
.5\\ 
1\\
.5\\ 
1\end{bmatrix}, x_4\in\mathbb{R}$$
Are you not allowed to have a half number in a basis for an eigenspace? The book shows this. 
\begin{bmatrix}
1\\ 2\\1\\2
\end{bmatrix}
 A: Perhaps you don't understand how to write the solution set.
Consider \begin{bmatrix}
1 & 0 & -1\\ 0 & 1 & 1\\ 0 & 0 & 0
\end{bmatrix}
You can only write the solution set once you are in reduced row echelon form. So here what we see is:
$$x_1-x_3=0 \to x_1 = x_3$$
$$x_2+x_3=0\to x_2=-x_3$$
$$x_3=x_3$$
So the solution set is: 
$$\begin{bmatrix}
x_1\\ 
x_2\\ 
x_3\end{bmatrix}=\begin{bmatrix}
x_3\\ 
-x_3\\ 
x_3\end{bmatrix}=x_3\begin{bmatrix}
1\\ 
-1\\ 
1\end{bmatrix}, x_3\in\mathbb{R}$$

As for your second matrix:
\begin{bmatrix}
1 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0
\end{bmatrix}
This is already in reduced row echelon form, and so we can find the solution set. Notice how $x_1$ is the only one that has a pivot. So here we have the solution set:
$$x_1 = -x_2 -x_3$$
$$x_2=x_2$$
$$x_3 = x_3$$
Here it's tricky, but what you do is the same steps.
$$\begin{bmatrix}
x_1\\ 
x_2\\ 
x_3\end{bmatrix}=\begin{bmatrix}
-x_2-x_3\\ 
x_2\\ 
x_3\end{bmatrix}=x_2\begin{bmatrix}
-1\\ 
1\\ 
0\end{bmatrix}+x_3\begin{bmatrix}
-1\\ 
0\\ 
1\end{bmatrix},x_2,x_3\in\mathbb{R}$$
A: Actually, I think your question is referring to finding the null space of the two matrices. Your general solution to first matrix is wrong, whereas for the second matrix, it is correct. The general solution for first matrix is $\begin{cases} x_1=x_3\\x_2=-x_3\\x_3\ \text{is free}\end{cases}$ in which case, the null space is spanned only by one free vector, i.e. $x_3\begin{bmatrix}1\\-1\\1\end{bmatrix}$. Similarly, the null space for the second matrix is spanned by two free vectors, i.e.$x_2\begin{bmatrix}-1\\1\\0\end{bmatrix}+x_3\begin{bmatrix}-1\\0\\1\end{bmatrix}$.
I think it is better to solve solution set to homogenous equations usin the pivot and free variables technique, which, although same as the previous answer, gives you immediate information regarding free variables. See the second solution here
