Give a basis for the solutionspace (i.e. the Nullspace) of this system of equations and express the basis in terms of its basis elements Suppose that we have the system:
$$-x_1 + x_2 + x_3 + x_4 = 0$$
$$x_2 + x_4 + x_5 = 0$$
$$x_1-x_3+x_5 = 0$$
Give a basis for the solutionspace (i.e. the Nullspace) of this system of equations and express the basis in terms of its basis elements.
My attempt at a solution: (please tell me if I'm going wrong somewhere)
So first of all I used row reduction to simplify this system into 1 equation. I'll just tell you the steps where $R_1, R_2, R_3$ are Row $1, 2, 3$.
$$-R_1$$
$$R_3-R_1$$
$$R_2-R_3, R_3 - R_2$$
The system simplifies to this equation:
$$x_1 -x_2-x_3-x_4+ 0x_5= 0$$
$$=>x_1 -x_2-x_3-x_4 = 0$$
Let $C_1, C_2, C_3, C_4, C_5$ denote Column $1,2,3,4,5.$
$$\langle C_1,C_2,C_3,C_4,C_5 \rangle \subseteq \mathbb R^3 \text { is the column space}$$
$$\langle R_1,R_2,R_3 \rangle \subseteq \mathbb R^5 \text { is the row space}$$
Let $S$ denote the nullspace
$S= $ {$(x_1,x_2,x_3,x_4,x_5) \in \mathbb R^5: x_1 = x_2+x_3+x_4$ }
$$(x_2+x_3+x_4, x_2, x_3, x_4, x_5) = x_1(0,0,0,0,0) + x_2(1,1,0,0,0) + x_3(1,0,1,0,0) + x_4(1,0,0,1,0) +x_5(0,0,0,0,0)$$
 A: In your row reduction process, you can not do 
$$R_2-R_3, R_3-R_2$$
one after another. After $R_2-R_3$, $R_2$ is eliminated. So you end up with
$$x_1-x_2-x_3-x_4=0\\x_2+x_4+x_5=0$$
So your null space can be represented by $x_3,x_4,x_5$.
A: Row reduction on the matrix goes as follows:
\begin{align}
\begin{bmatrix}
-1 & 1 & 1 & 1 & 0 \\
0 & 1 & 0 & 1 & 1 \\
1 & 0 & -1 & 0 & 1
\end{bmatrix}
&\to
\begin{bmatrix}
1 & -1 & -1 & -1 & 0 \\
0 & 1 & 0 & 1 & 1 \\
1 & 0 & -1 & 0 & 1
\end{bmatrix}
&&R_1\gets -R_1\\[6px]
&\to
\begin{bmatrix}
1 & -1 & -1 & -1 & 0 \\
0 & 1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 & 1
\end{bmatrix}
&&R_3\gets R_3-R_1\\[6px]
&\to
\begin{bmatrix}
1 & -1 & -1 & -1 & 0 \\
0 & 1 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
&&R_3\gets R_3-R_2\\[6px]
&\to
\begin{bmatrix}
1 & 0 & -1 & 0 & 1 \\
0 & 1 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
&&R_1\gets R_1+R_2\\[6px]
\end{align}
The free variables are $x_3$, $x_4$ and $x_5$, with
$$
\begin{cases}
x_1=x_3-x_5\\
x_2=-x_4-x_5
\end{cases}
$$
so a basis of the null space is given by
$$
\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}
\qquad
\begin{bmatrix} 0 \\ -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}
\qquad
\begin{bmatrix} -1 \\ -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}
$$
Since the first two columns are the dominant ones (those which correspond to the nonfree variables), a basis for the column space is
$$
\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}
\qquad
\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}
$$
Note that the first two columns are easily seen to be linearly independent, so the rank of the matrix is at least $2$; therefore you can't have four free variables as you'd get with your (wrong) row reduction, because $2+4=6$ would violate the rank-nullity theorem.
