Creating a zero matrix through matrix multiplication From an assignment:

Let $A = \left[ \begin{matrix} 3 & -6 \\ -2 & 4 \end{matrix}\right] $ Construct a $2 * 2$ matrix $B$ such that $AB$ is the zero matrix. Use two different nonzero columns for $B$.

The value of $AB$ would be:
$$
AB = \left[
\begin{matrix}
    3b_{11}  -6b_{12} & 3b_{21}  -6b_{22} \\
    -2b_{11} + 4b_{12} & -2b_{21} + 4b_{22}
\end{matrix}\right]
$$
I was thinking of using substitution, but the following equations just result in the variables equalling $0$:
$$\begin{align*}
3b_{11} -6b_{12} &= 0\\
 -2b_{11} + 4b_{12} &= 0
\end{align*}$$
Any hints on how I can solve this?
 A: Consider the solutions to $A\mathbf{x}=\mathbf{0}$. These can be found easily by Gaussian elimination:
$$\left(\begin{array}{rr}
3 & -6\\
-2 & 4
\end{array}\right)\rightarrow \left(\begin{array}{rr}
1 & -2\\
1 & -2
\end{array}\right) \rightarrow \left(\begin{array}{rr}
1 & -2\\
0 & 0
\end{array}\right).$$
So $\mathbf{x}=\left(\begin{array}{c}r\\s\end{array}\right)$ satisfies $A\mathbf{x}=\mathbf{0}$ if and only if $r-2s = 0$, if and only if $r=2s$. 
Now notice that if $B=[\mathbf{b}_1|\mathbf{b}_2]$, where $\mathbf{b}_1$ is the first column and $\mathbf{b}_2$ is the second column, then 
$$AB = [A\mathbf{b}_1|A\mathbf{b}_2].$$
A: I think for small matrices as the one above, it s good to just look for linear dependence of the columns in such a problem. (i.e. here you can quickly notice that in matrix $A$ column 2 is just column 1 multiplied by -2)
Looking at this kind of linear dependence between the columns gives you a quick way to understand why $B$ looks the way it does (i.e. take 1 of column 1 and -2 of column two and you arrive at the zero matrix, and this is exactly what multiplication by $B$ does )! 
A: Your equations $$3b_{11}-6b_{12}=0,\qquad -2b_{11}+4b_{12}=0$$ are both telling you $b_{11}=2b_{12}$, which means you can let $b_{12}$ be your favorite nonzero number, and then you also know $b_{11}$. You also get two equations for $b_{21}$ and $b_{22}$; you can let $b_{22}$ be your second favorite nonzero number, and work out $b_{21}$. Then you're done. 
