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.

I consider $$C=\left(\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i \end{array}\right),$$ then 
$$BC=\left(\begin{array}{ccc}a&b&c\\a+d&b+e&c+f\\a+d+g&b+e+h&c+f+i \end{array}\right),$$
and
$$CB=\left(\begin{array}{ccc}a+b+c&b+c&c\\d+e+f&e+f&f\\a+d+g&h+i&i \end{array}\right).$$
If, we want $A=BC$, we get 
$$C=\left(\begin{array}{ccc}1&1&1\\0&1&1\\0&0&1 \end{array}\right),$$
in the case $A=CB$ we have
$$C=\left(\begin{array}{ccc}0&0&1\\-1&0&2\\-1&-1&3 \end{array}\right).$$
Therefore:
1) False
2) False
3) True
4) False
Regards!! 
