Question: Does there exist $A \in M_{2}(\mathbb{Z})$ such that every element of $M_{2}(\mathbb{Z})$ can be represented as a linear combination of powers of $A$ with integer coefficients? In other words,
$$\exists A \in M_{2}(\mathbb{Z}) \, \,s.t. M_{2}(\mathbb{Z})=\left.\left\{\sum_{i}a_{i}A^{i} \;\right| \; a_{i} \in \mathbb{Z}, a_j=0\text{ for almost all }j\right\}$$
Motivation: I want to construct a surjective ring homomorphism $\varphi: \mathbb{Z}\left[x\right] \rightarrow M_{2}(\mathbb{Z})$ by letting $\varphi: \displaystyle \sum_{i}a_{i}x^{i} \mapsto \sum_{i}a_{i}A^{i}$. The above property is sufficient for surjectivity.
What I've tried: It's sufficient to find four linear combinations that sum to $\left\{ \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right] \right\}$. I haven't been able to find a magic $A$ that does this. For example, powers of $\left[ \begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array} \right]$ generate Fibonacci numbers, except that off diagonal elements are the same, so this will never work.
This is not homework. I have a feeling it may be impossible, but I'm not sure why. If for some reason it is impossible over $\mathbb{Z}$, I would be interested to know if it's possible over $\mathbb{Z}_{p}$ for a prime $p$. Thanks for any help.