Order 3 square matrix property

Question

I found this problem in an old Romanian GM and I don't even know how to approach it:

Let $A$ be a square matrix of order $3$ with real numbers such that $detA=0$. Prove that there exists a square matrix $B$ of order $3$ such that $B\neq0$ and $$AB=BA=0$$

I asked myself (don't know if it is relevant to this problem) if for any square matrix there exists another one (of the same order) such that they commute. Is this true? (for order $3$ it looks like it is true, at least when determinant is $0$)

Also, how should I approach this problem?

Answer 1

There will be a nonzero column vector $u$ with $Au=0$, and also a nonzero row vector $v$ with $vA=0$. Can you make a matrix from $u$ and $v$?