$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\Ann}{\text{Ann}}$ I'm having an issue which boils down to basic linear algebra but for the life of me I can't justify it.
Problem: Let $M$ be an $R$-module and $N$ a cyclic $R$-submodule with generator $x$. Give an example to show that if $R$ is not commutative, then $\Ann_R(N) \neq \{a \in R \mid ax = 0\}$.
A hint was to consider the ring $R = M_n(\R)$ as a left $R$-module over itself and $E_1$ the matrix with $1$ in the $a_{11}$ entry and zero's elsewhere. I know that any matrix with a zero first column will annihilate $E_1$ and so that part isn't difficult, and I imagine that $\Ann(\langle E_1 \rangle) = \{0\}$ but I just can't justify the second part. I know that $\langle E_1 \rangle $ consists of all $n \times n$ matrices that have arbitrary first column and zeros elsewhere, so then if $A = (a_{ij}) \in M_n(\R)$ and $(B_1\mid 0 \mid \dots \mid 0) \in \langle E_1 \rangle$, then requiring $A(B_1\mid 0\mid \dots \mid 0) = 0$ amounts to $$ \sum_{j=1}^n a_{ij}b_j = 0 $$ for all $1 \leq i \leq n$, where $b_j$ are the elements of the $B_1$ column. But even in the $2 \times 2$ case this just gives $a_{11}a_{22} = a_{21}a_{12}$ and isn't useful.
Obviously any element of $\langle E_1 \rangle$ is singular, I know that $A$ must also be singular, but this doesn't let me force that $A$ needs to be zero. I'm absolutely sure there's an easy computation or fact from linear algebra I'm missing, so sorry in advance, but thank you for the clarification.
Edit: The definition of annihilator being used in the text is $\Ann_R(N) = \{a \in R \mid an = 0 ,\forall n \in N\}$, however in the case where $R$ is commutative and $N$ is a cyclic submodule, one can show that $\Ann_R(N) = \{a \in R \mid ax = 0\}$ since if $R$ is commutative it suffices to annihilate the generator of $N$.
If you take $B_1$ to be the $k$-th standard basis vector (i.e. the $k$th column of the identity matrix), then $AB_1$ will be the $k$th column of $A$.
Thus, if $AB_1 = 0$ for all choices of $B_1$, it follows that each column of $A$ is zero, hence $A = 0$.