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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$Let $G_2(\mathbf{Z}_p):=\begin{pmatrix} 1+p\mathbf{Z}_{p} & \mathbf{Z}_{p}\\ p\mathbf{Z}_{p} & 1+p \mathbf{Z}_{p} \end{pmatrix}$. Then it is a Sylow pro-$p$ subgroup of $\GL_2(\mathbf{Z}_p$) and $S_2(\mathbf{Z}_p):=G_2(\mathbf{Z}_p) \cap \SL_2(\mathbf{Z}_p)$ is a Sylow pro-$p$ subgroup of $\SL_2(\mathbf{Z}_p$). It's well-known that $S_2(\mathbf{Z}_p)$ is a $2$-generator pro-$p$ group. For exmaple, the set $\{\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} ,\begin{pmatrix} 1& 0 \\ p& 1 \end{pmatrix} \}$ is a topological generating set. Moreover, in Prop. 3.1.1 of The image of Galois representations attached to elliptic curves with an isogeny it's proved that if $A,B\in S_2(\mathbf{Z}_p)$ such that the image of $A$ in $\GL_2(\mathbf{F}_p)$ is non-trivial and the image of $B$ in $\GL_2(\mathbf{F}_p)$ is trivial and the image of $B$ in $\GL_2(\mathbf{Z}/p^2\mathbf{Z})$ is not upper triangular, then the set $\{A,B\}$ topologically generates $S_2(\mathbf{Z}_p)$.

Question: Can $S_2(\mathbf{Z}_p)$ be generated by two matrices $A,B$ such that the images of $A,B$ in $\GL_2(\mathbf{F}_p)$ are both non-trivial?

As pointed out in the answer, the answer to question is No in a trivial way. What about the case assuming further that the eigenvalues of $A,B$ are all $1$?

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If $x$ and $y$ generate a group then so do $x$ and $xy$, so take $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 0 \\ p & 1 \end{pmatrix}$.

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