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Let $R$ be a ring, and let $A$ be an $n\times n$ matrix with coefficients from $R$. Suppose for $r\in R^n$ we have $Ar=r$. Prove that $\det (A-I)\cdot r=0$.

It is actually part of a bigger problem where $r_i$ generate ring $B$ which is module finite over $R$, and we need to prove that $\mathfrak{a} B\neq B$, where $\mathfrak{a}$ is a proper ideal of $T$. (matrix entries are picked from $\mathfrak{a}$). The rest of proof is straightforward, just this fact is unclear to me.

ralleee
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    In your first part, do you mean that $;Ar=r;$ for all $;r\in R;$ or just one? It looks weird since if the ring is a field, say $;R=\Bbb R;$ , the claim is false big time: just take $;A=I_n;$ . Prhaps some other data is given? – DonAntonio Apr 08 '16 at 21:06
  • i have edited, i wrote it incorrectly, sorry – ralleee Apr 09 '16 at 08:19

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This follows from $(A-I_n)r=0$ by multiplying (on the left) with the classical adjoint of $A-I_n$.

user26857
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