Matrices that satisfy $A^2=0$ and anticommutation. Suppose that there is a set of matrices.
1) $A^2 =0, C^2=0, E^2=0 .....$ and $AB+BA=0$, $CD+DC=0$... where $A,B,C..$ are matrices.
2) Matrices in the set either anticommute or commute.
3) $AD+BC \neq 0,... $
Is there any set of matrices that satisfies all of the aforementioned?
Edit: all matrices are not equal to each other.
 A: $
N:=
\begin{pmatrix}
0&1\\
0&0
\end{pmatrix},
\ \ 
\sigma:=
\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix}\ \ \ $
$\Longrightarrow \ \ \ N^2=0, \ \ \ \sigma^2=1,\ \ \ \sigma N=N=-N\sigma$
So define
$A:=a\ N, \ \ \ C:=c\ N, \ \ \ D:=d\ N,$
$B:=b\  N+\beta\ \sigma$
$\Longrightarrow$ products among $\{A,C,D,...\}$ are zero, while $B^2=\beta^2\ 1$ and e.g. $BA=ab\ \sigma$.
The relation $\sigma N=N=-N\sigma$ makes $N$ an "eigenvector" of $\sigma$ from left and right so as to fulfill the anti-commutation relation $AB+BA=0$, without $AB$ already being zero. This way the expression involving $B$ is the only one which survives in your last condition:
$AD+BC=ad\ N^2+bc\ \sigma\propto \sigma\ne 0.$
A: Finding $n$ matrices $A_1, ...  A_n$ satisfying the commutation relations
$$A_i A_j + A_j A_i = 0 \forall 1 \le i, j \le n$$
is equivalent to finding a finite-dimensional module over the exterior algebra of an $n$-dimensional vector space. The exterior algebra has dimension $2^n$, hence is itself such a module (acting on itself by left multiplication). The exterior algebra acts faithfully on itself, hence the corresponding matrices satisfy only relations which follow from the relations above. 
