Answer
$B=\begin{bmatrix}-2&6\\-1&3\end{bmatrix}$
Work Step by Step
We wish to find a matrix $B$ such that:
$\begin{align}AB&=\begin{bmatrix}3&-6\\-1&2\end{bmatrix}\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}\\&=\begin{bmatrix}3b_{11}-6b_{21}&3b_{12}-6b_{22}\\-b_{11}+2b_{21}&-b_{12}+2b_{22}\end{bmatrix}\\&=\begin{bmatrix}0&0\\0&0\end{bmatrix}\end{align}$
Each column of the matrix product is itself a system of equations (the first in $b_{11}$ and $b_{21}$, the second in $b_{12}$ and $b_{22}$), each with the same coefficients. Since $\text{rref}(\begin{bmatrix}3&-6&0\\-1&2&0\end{bmatrix})=\begin{bmatrix}1&-2&0\\0&0&0\end{bmatrix}$, we are left with the solutions $b_{11}=2b_{21}$ and $b_{12}=2b_{22}$ for any real numbers $b_{21}$ and $b_{22}$. Choosing, for example, $b_{21}=-1$ and $b_{22}=3$ yields the solution above. More generally, any matrix of the form $\begin{bmatrix}2x&2y\\x&y\end{bmatrix}$ will suffice, where $x$ and $y$ are real numbers.
