Let $\mathcal{O}$ be a complete Noetherian local rings with residue field $k$ and suppose $A \to B$ is a surjective morphism of Noetherian local $\mathcal{O}$-algebras (with residue field $k$) with principal kernel annihilated by $m_A$ (the maximal ideal of $A$. Suppose $\rho \colon G \to \textrm{GL}_2(A)$ is a continuous representation of a profinite group $G$ such that the only matrices commuting with the image of $G$ in $\textrm{GL}_2(B)$ or $\textrm{GL}_2(k)$ are scalars. I'd like to show that similarly, $M \in \textrm{GL}_2(A)$ commutes with the image of $\rho$ if and only if $M$ is scalar. This is essentially Lemma 3.8 in "Deformations of Galois Representations" by Fernando Q. Goueva, but the proof in the book seems to contain a mistake.
Suppose $M \in \textrm{GL}_2(A)$ commutes with the image of $\rho$. Then its image $\overline{M}$ in $\textrm{GL}_2(B)$ must be a scalar by hypothesis, so $M = r + tM'$, where $r \in \textrm{GL}_2(A)$ is a scalar matrix, $t \in \textrm{GL}_2(A)$ is a scalar matrix whose diagonal entry generates $\ker(A \to B)$, and $M' \in M_2(A)$. We have $$(r + tM')\rho(g) = \rho(g)(r + tM')$$ for all $g \in G$. Since $r$ is scalar, this implies $$tM'\rho(g) = \rho(g)tM'$$ for all $g \in G$. Here is where the book seems to go wrong. It's claimed that this means $M'$ commutes with $\rho(g)$. But $t$ is not invertible, so this doesn't seem to follow. The book then goes on to say that the image of $M'$ in $\textrm{GL}_2(k)$ is a scalar by the hypothesis, but this again seems to be erroneous as we don't know that $M'$ is invertible. Is this salvageable?
What we can say is that $$t(M'\rho(g) - \rho(g)M') = 0$$ for all $g \in G$. If $t = 0$, then $M$ is scalar and we're done. If not, then we can say that $M'\rho(g) - \rho(g)M'$ has entries in $m_A$ for all $g \in G$.
Small note: $t \notin GL_2(A)$, $t$ is a scalar matrix in $M_2(A)$ (but in fact, its determinant is zero).
Anyway, we indeed have $t(M’\rho(g)-\rho(g)M’)=0$. This implies that the images of $M’$ and $\rho(g)$ in $k$ commute, because $m_A$ is the annihilator of $t$.
(In other words: let $N=M’\rho(g)-\rho(g)M’$: then $tN=0$. So no coefficient of $N$ can be invertible, so the image of $N$ in $M_2(k)$ is zero.)
It follows that the image of $M’$ in $k$ is a scalar matrix, so $M’=a+M’’$, where $a \in AI_2$ and $M’’ \in m_AM_2(A)$. Thus $tM’=ta$ and $r+tM’=r+ta$ is scalar.