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let $$ A=\begin{pmatrix} 1& 2\\ 3& 6\end{pmatrix}$$ $A$ is a $2\times 2$-matrix. Find a $2\times 3$-matrix which is such that $AB=0$ and $B \ne 0$.

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$$ B = \left( \begin{array}{ccc} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{array} \right) \\ AB =\left( \begin{array}{ccc} b_{11} + 2b_{21} & b_{12} + 2b_{22} & b_{13} + 2b_{23} \\ 3b_{11} + 6b_{21} & 3b_{12} + 6b_{22} & 3b_{13} + 6b_{23} \end{array}\right) = \left( \begin{array}{ccc} 0 & 0& 0 \\ 0 & 0 & 0 \end{array} \right) \\ b_{11} = -2b_{21}; \quad b_{12} = -2b_{22}; \quad b_{13} = -2b_{23} $$ So any matrix of form $$ B = \left( \begin{array}{ccc} -2a & -2b & -2c \\ a & b & c \end{array} \right) $$ will satisfy the condition

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More generally, the columns of $B$ must be a linear combination of the elements of the (right) nullspace of $A$.

As others have mentioned, the second column is twice the first, so the vector $v=[2,-1]^T$ spans the right nullspace of $A$. Thus the columns of $B$ must be a linear combination (in this case a scalar multiple) of $v$.

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What about $$ B=\begin{pmatrix} 2& 2 & 2\\ -1& -1 & -1\end{pmatrix}. $$

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Looking at $A$ we see that the second column is two times the first. So $A$ anihilates $\begin{pmatrix} -2 \\ 1\end{pmatrix}$. Adding some zero columns does it then $$ B = \begin{pmatrix} -2 & 0 & 0\\ 1 & 0 & 0\end{pmatrix} $$

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