Find three nonzero vectors $x, y, b \in R^3$ than $Ax = b, By = b$, where
$A=\begin{pmatrix} 0 & 1 & 2\\ 1 & 1 & 1\\ 2 & 1 & 0 \end{pmatrix}$
$B=\begin{pmatrix} 1 & -1 & 0\\ 0 & 0 & 1\\ 1 & 1 & 0 \end{pmatrix}$
My answer: Using that $Ax=b$ & $By=b$, $$x=By$$ $$A(By)=b$$ $$(AB)y=b$$ Now solving the new system of equations: $$AB=\begin{pmatrix} 0 & 1 & 2\\ 1 & 1 & 1\\ 2 & 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & -1 & 0\\ 0 & 0 & 1\\ 1 & 1 & 0 \end{pmatrix}=\begin{pmatrix} 2 & 2 & 1\\ 2 & 0 & 1\\ 2 & -2 & 1 \end{pmatrix} $$
which implies that $\det AB=0$. What it implies does not exist and such that it solves the system. And therefore that does not exist three nonzero vectors $x, y, b \in R^3$ than $Ax = b, By = b$.
I am not sure that my answer is correct, it will help me a lot to know if I do it well.