= Solution
Since $AB=0$, the <image of a linear map> represented by $B$ lies in the <kernel> of $A$. The <matrix> $B$ is nonzero, so this <image of a linear map> contains a nonzero <vector>; consequently $\dim\ker A\geq1$. On the other hand, $A\neq0$ gives $\operatorname{rank}A\geq1$. The <rank-nullity theorem> in dimension two therefore gives
$$
\dim\ker A=1,\qquad\operatorname{rank}A=1.
$$
Thus \b[the image of $A$ is a line through the origin], and $A$ maps onto that line.
Reversing the order of the <matrix multiplication> need not give zero. For example,
$$
A=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad B=\begin{pmatrix}0&0\\1&0\end{pmatrix}
$$
gives $AB=0$ but $BA=B\neq0$. This also shows why information about one composition of <linear maps> does not establish the corresponding property of the reversed composition.
Write $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. Its <rank> is one, so its <determinant> is zero. Choose its <adjugate matrix>:
$$
\boxed{C=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.}
$$
Direct <matrix multiplication> gives $AC=CA=(ad-bc)I=0$. Since the entries of $C$ are the entries of the nonzero <matrix> $A$, rearranged with possible sign changes, $C\neq0$. This establishes both required annihilation properties.
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