Finding a matrix M of rank 2 such that AM = 0 
Let
$$ A = \begin{bmatrix} 1 & 0 & -1 & 2 &1\\
 -1 & 1 & 3 & -1 & 0\\
 -2 & 1 & 4 & -1 & 3\\ 3 & -1 & -5 & 1 & -6\\ \end{bmatrix} $$
Find a $ 5 \times 5$ matrix $M$ with rank 2 such that $AM =0_{4\times5}$

My logic was to row reduce $A$ into a matrix $B$ such that:
$$
A = E_n \dots E_2 \cdot E_1 \cdot B
$$
therefore $AM = 0$ becomes
$$
AM = (E_n \dots E_2 \cdot E_1 \cdot B) \cdot M = E_n \dots E_2 \cdot E_1 \cdot (B \cdot M) = 0$$
This allows me to find a matrix M that, multiplied left by B, will give me the zero matrix. I got that
$$ B = \begin{bmatrix} 1 & 0 & -1 & 0 &0\\
 0 & 1 & 2 & 0 & 0\\
 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{bmatrix} $$
However, the only column I can think of that would give me a zero is in the form:
 $$ x = \begin{bmatrix} 
c\\
-2c\\
c\\
0\\
0\\
\end{bmatrix} $$
Setting that as one of the columns of M and filling the other entries with zeroes would only give me a matrix of rank 1.
Am I missing something or is my logic totally flawed?

We have that
$$A = \begin{bmatrix} 1 & 0 & -1 & 2 &1\\
 -1 & 1 & 3 & -1 & 0\\
 -2 & 1 & 4 & -1 & 3\\ 3 & -1 & -5 & 1 & -6\\ \end{bmatrix}\to  \begin{bmatrix} 1 & 0 & -1 & 2 &1\\
 0 & 1 & 2 & 1 & 1\\
0 & 1 & 2 & 3 & 5\\ 0 & -1 & -2 & -5 & -9\\ \end{bmatrix}\to$$
$$\begin{bmatrix} 
1 & 0 & -1 & 2 &1\\
 0 & 1 & 2 & 1 & 1\\
0 & 0 & 0 & 2 & 4\\ 
0 & 0 & 0 & -2 & -4\\ \end{bmatrix}\to
\begin{bmatrix} 
1 & 0 & -1 & 2 &1\\
 0 & 1 & 2 & 1 & 1\\
0 & 0 & 0 & 2 & 4\\ 
0 & 0 & 0 & 0 & 0\\ \end{bmatrix}$$
therefore since the rank of the matrix is equal to $r=3$ the null space has dimension $\dim(N(A))= n-r=5-3=2\,$ and we can find $x_1$ and $x_2$ linearly independent such that
$$Ax_1=Ax_2=0$$
