Matrices: $AB=0 \implies A=0 \text{ or } \ B=0$ When A and B are square matrices of the same order, and O is the zero square matrix of the same order, prove or disprove:-
$$AB=0 \implies A=0 \text{ or } \ B=0$$
I proved it as follows:-
Assume $A \neq O$ and $ B \neq O$:
 then, $$ |A||B| \neq 0 $$
$$ |AB| \neq 0 $$
$$ AB \neq O $$
$$ \therefore A \neq O\ and\ B \neq O \implies AB \neq O $$
$$ \neg[ AB \neq O] \implies \neg [ A \neq O\ and\ B \neq O ] $$
$$AB=O \implies A=O \text{ or } \ B=O$$
But when considering, 
A := \begin{pmatrix} 1&1 \\1&1
\end{pmatrix} and B:= \begin{pmatrix} -1& 1\\ 1 &-1
\end{pmatrix}then, AB=O and A$\neq $O and B $\neq$ O
I can't figure out which one and where I went wrong.

The problem with your first approach is that there exist non-zero matrices with zero determinant. For instance, $A$ and $B$ in your example.
