Matrix $\mathbf C$ such that $\mathbf A = \mathbf B \mathbf C=\mathbf C\mathbf B$ Let 
$$\mathbf A =\begin{pmatrix}1&1&1\\1&2&2\\1&2&3\end{pmatrix} \text{and }\mathbf B=\begin{pmatrix}1&0&0\\1&1&0\\1&1&1\end{pmatrix}$$
Then which of following is true


*

*There exists a matrix $\mathbf C$ such that $\mathbf A=\mathbf B\mathbf C=\mathbf C\mathbf B$

*There exists no matrix $\mathbf C$ such that $\mathbf A=\mathbf B\mathbf C$

*There exists a matrix $\mathbf C$ such that $\mathbf A=\mathbf B\mathbf C$ but $\mathbf A \ne  \mathbf C\mathbf B$

*There exists no matrix $\mathbf C$ such that $\mathbf A=\mathbf C\mathbf B$.
My attempt: since $\mathbf B$ is invertible I see option 1 to be correct. But it can't be that easy. Also, $\mathbf A$ is symmetric and $\mathbf B$ is triangular so think there some trick here. So I need suggestions. Thanks.

Yes, it's that easy: both matrices $\;A\,,\,\,B\;$ are invertible. and thus
$$A=BC\implies C=B^{-1}A$$
Now, if also for the same $\;C\;$ we'd have $\;A=CB\;$ , then
$$B^{-1}A=C=AB^{-1}\implies AB=BA$$
which is false as you can easily check, and thus you can already solve all four points.
