Find all matrices that commute with $A$ 
Given $$A = \begin{bmatrix}
3 & 1 &0 \\ 
0 &3  & 1\\ 
0 &0  & 3
\end{bmatrix}$$ find matrices $B$ such that $AB=BA$.

Trivially $B=A^{-1}$ and $B=kI$ are the solutions
Also we have Characteristic Polynomial as
$$A^3-9A^2+27A-27I=0$$ $\implies$
$$(A-3I)^3=0$$
Is it possible to find other $B's$ using above Nilpotency of $A-3I$?

Since$$A=3\operatorname{Id}_3+\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$and every matrix commutes with $3\operatorname{Id}_3$, you're after the matrices that commute with$$\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}.\tag1$$A simple computation shows that\begin{multline}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}-\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}=\\=\begin{pmatrix}d & e-a & f-b \\ g & h-d & i-e \\ 0 & -g & -h\end{pmatrix}\end{multline}and therefore the matrix$$\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$$commutes with $(1)$ if and only if$$\left\{\begin{array}{l}d=g=h=0\\a=e=i\\f=b.\end{array}\right.$$Therefore, the answer to your question is:$$\left\{\begin{pmatrix}a&b&c\\0&a&b\\0&0&a\end{pmatrix}\,\middle|\,a,b,c\in\mathbb{R}\right\}.$$
