I have a problem about skew polynomial ring.

If I have $R=M_2(\mathbb{R})$ and $\sigma$ is an endomorphism in $R$ with $\sigma \left(\begin{bmatrix} a & b \\ c& d \end{bmatrix}\right) = \begin{bmatrix} a& -b \\ -c&d \end{bmatrix}$. Also let $\delta = 0$, then we have $R[x;\sigma]$ is skew polynomial ring. Consider that, in this case we have $xa = \sigma(a)x$.

Problem : I have an intuition about the non-zero zero divisor of $R[x;\sigma]$, i.e

$P = \{\sum_{i=0}^{n} A_ix^i : det(A_i)=0, A_i = \begin{bmatrix} a_i & b_i \\ c_i&d_i \end{bmatrix}, a_i,b_i,c_i,d_i \in \mathbb{R}\backslash \{0\} \}$.

Let $Z_0(R[x;\sigma])$ is non-zero divisor of $R[x;\sigma]$. I have already proved $P \subseteq Z_0(R[x;\sigma])$. But I am confused with the $Z_0(R[x;\sigma]) \subseteq P$.

Any advice? Thanks in advance.