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Conjecture: Given a $M(R)$ the set of matrices under a commutative ring with identity, $R$, a matrix $A$ in $M(R)$ is a unit if and only if for any vector $b$ there exists a vector $x$ such that $Ax = b$.

I have conjectured the previous statement, and could prove the forward implication;

If $A$ is a unit, $A^{-1}$ exists such that $AA^{-1} = I.$ For any $b$, $Ib = b$. Then $(AA^{-1})b = b$. Then $A(A^{-1}b) = b$.

Now suppose for any $b$, there exists $x$ such that $Ax = b$. Take $e_1, ... e_n$, the unit vectors of $I$. then $AB = I$ exists.

However, I am not able to show $AB = I$ implies $BA = I$, because the existence of $B$ such that $BA = I$ does not guarantee existence of $C$ such that $CA = I$. If C exists, the result is immediate.

Is the conjecture true, or does there exist a counterexample?

rr01
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1 Answers1

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A matrix in $M(R)$ is invertible if and only if its determinante is a unit of $R$. (Argument for example here). If $AB=I$, we get from $$1=\det(I)=\det(AB)=\det(A)\det(B)$$ that $\det(A)$ is a unit of $R$, so $A$ is invertible and so there is a matix $C$ with $CA=I$.