Romanian Mathematical Olympiad

Let $K\in {\mathcal{M}}_{n,1}(\mathbb{C})$ and $L\in {\mathcal{M}}_{1,n}(\mathbb{C})$ be such that $KL=B$. Then $O_n=(AK)(LB)$ with $AK\in {\mathcal{M}}_{n,1}(\mathbb{C})$, $LB\in {\mathcal{M}}_{1,n}(\mathbb{C})$. Let $AK={}^t(a_1,a_2,\dots ,a_n)$ (where ${}^{t}X$ denotes the transpose of $X$), and let $LB=(b_1,b_2,\dots ,b_n)$. We infer $$O_n=\left(\begin{array}{cccc}{a}_1{b}_1& a_1{b}_2& \cdots & a_1{b}_n\\ ⋮& ⋮& \cdots & ⋮\\ a_n{b}_1& a_n{b}_2& \cdots & a_n{b}_n\end{array}\right)$$ If $LB\ne O_{1,n}$ then there exists $1\le i\le n$ with $b_i\ne 0$. It follows $a_1=a_2=\cdots =a_n=0$, hence $AK=O_{n,1}$, and so at least one of the matrices $AK$ and $LB$ is null. In the former case $AB=(AK)L=O_{n,1}L=O_n$, while in the latter $BC=K(LC)=K{O}_{1,n}=O_n$.