If the product of two matrices, $A$ and $B$ is zero matrix, prove that matrices $A$ and $B$ don't have to be zero matrices I can give an example where product of two non-zero matrices is zero matrix,
$$
A=
        \begin{bmatrix}  
        3 & 6 \\
        2 & 4 \\
        \end{bmatrix}
$$
$$
B=
        \begin{bmatrix}
        2 & -8 \\
        -1 & 4 \\
        \end{bmatrix}
$$
the product is zero matrix.
I don't know how to prove this. Could someone help?
 A: The claim you want to prove is equivalent to: 
there exist nonzero $A$ and $B$ such that $AB$ is zero. 
To prove an existential claim it suffices to exhibit examples and to verify that the example do what they are supposed to do. 
You have found $A$ and $B$ that are clearly nonzero,  and their product is zero. 
That's all there is to it. 
A: For an easier example:
Consider any diagonal $D$: $2 \times 2$ matrix with $a_{11}= a \neq 0$ and $0$ otherwise. Can you see how to find a non-zero $M$ with $DM=0$?
A: Nilpotent matrices are a good (and often used) counterexample:
$$
A = \left[
\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}
\right]\qquad \Rightarrow AA = O
$$
A: whilst exhibiting a single value $a$ for which $P(a)$ is true is sufficient to establish the existential claim $\exists x\cdot P(x)$, i sympathize with OP's feeling that a single case doesn't give us much grasp of the why's and wherefores of the presence of zero divisors in matrix rings. there is, after all, considerably more that might be pointed out in this context. here just a couple of observations will suffice to offer a little contextualization of OP's question.
in the first place $1 \times 1$ matrices, whilst lacking the characteristics of the paradigmatic matrix, are nevertheless valid matrices. for the matrix ring $M_1(R)$ there are zero divisors if and only if the matrix elements are drawn from a ring $R$ which fails to be an integral domain. in particular if this ring is in fact a field, then there are no zero divisors. let us now focus on the (finite-dimensional) ring $M_n(F)$ of $n \times n$ matrices over a field $F$
1. all zero-divisors are singular (have no multiplicative inverse)
its proof is exceedingly simple, but this result is basic knowledge nonetheless. here is the left-hand-side case. the other case is easily dealt with in the same way (exercise!)
$$
(\exists A \cdot AB=I) \Rightarrow (BC=0 \Rightarrow ABC=0 \Rightarrow IC=0 \Rightarrow C=0)
$$
as is often the case to prove the converse offers greater difficulty. 
2. all singular matrices are zero-divisors 
the difference between showing this to be true and showing the converse (as above) is a good illustration of the power of 'strategic' mathematical ideas. instead of regarding a matrix merely as an array of numbers (or other field elements), we examine the connections between $M_n(F)$ and the endomorphism ring of the free module $F^n$. establishing this connection is the role of the concept of a basis for $F^n$.
any endomorphism $\alpha:V \to V$ has no inverse exactly when it fails to be both injective and surjective. in finite dimensions these two conditions are equivalent. such an endomorphism must annihilate some non-zero $v \in F^n$, which means that any matrix $A$ which represents $\alpha$ wrt some basis has zero as an eigenvalue. hence its characteristic equation must take the form $\lambda  f(\lambda)=0$. now, by the Cayley-Hamilton theorem it follows that $Af(A)=0$, i.e. $A$ is a zero-divisor.
A: Like
$$ \begin{vmatrix} x^2 & x \\ x & 1 \end{vmatrix} \begin{vmatrix} -x & -1 \\ x^2 & x \end{vmatrix} = \begin{vmatrix} 0 & 0 \\ 0 & 0 \end{vmatrix} $$
for any finite $x$.
A: Let, $B = \begin{bmatrix} a & b\\ c & d\end{bmatrix}$ 
$ A*B= \begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix}$ results in a system of linear equations (in this case 4 equations and 4 variables). 
We know (0,0,0,0) is always a solution but it is also possible that the above set of equations are over determined (true in your example) in which case you have multiple solutions. Hence A and B don't have to be zero matrices. Also note that if you have linearly dependent rows or columns, the resultant system of equations is over determined.
