Assume $$AX=0$$ is a linear system where $A$ is $n\times n$ matrix over ring $\Bbb r$ and $X$ is a length $n$ vector of variables if $rank_{\Bbb r}(A)=n-1$ how do we solve for the unique solution up to scale?
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Read MacCoy, Rings an Ideals https://books.google.fr/books/about/Rings_and_Ideals.html?id=D5_EngEACAAJ&redir_esc=y OR necessary and sufficient condition for trivial kernel of a matrix over a commutative ring
Let $R$ be a commutative ring with unit and $A\in M_{n,m}(R)$ with $n\leq m$.
Mac Coy defines $rank(A)$ as follows. If all the entries of $A$ have a common non-zero annihilator, then $rank(A)=0$. Otherwise its rank is the greatest positive integer $r\leq n$ such that the set of all $r \times r$ subdeterminants of $A$ does not have a common non-zero annihilator.
Theorem (Mac Coy). If $Ax=b$ has a solution for a given vector $b$, then the solution is unique iff $rank(A)=m$.
Corollary. When $m=n$ the NS condition is "$\det(A)$ is not a zero-divisor in $R$".
Assume that $rank(A)<n$ and let $adj(A)=[C_1,\cdots,C_n]$ (its columns) be the $m\times n$ adjoint matrix of $A$. Then $Ax=0$ admits non-zero solutions and there is $r\not= 0$ s.t. $r\det(A)=0$. We can choose for $x$ any non-zero vector amongst $\{rC_1,\cdots,rC_n\}$ (there is at least one).
EDIT. For instance, let $R=\mathbb{Z}/216\mathbb{Z}$, $m=n=3$, $A=\begin{pmatrix}57&140&184\\27&144&142\\123&214&212\end{pmatrix}$. Then $\det(A)=108$, $r=2$ and $adj(A)=\begin{pmatrix}140&192&80\\78&36&114\\162&54&108\end{pmatrix}$. For instance $2C_1=[64,156,108]^T$ is a solution of $Ax=0$.
Note that $rank(A)=1$ because $108$ "kills" all the $2\times 2$ minors of $A$.
Maple gives the following (not free) generator $[8,60,0]^T,[0,108,0]^T,[0,0,108]^T$. Note that $\ker(A)$ is a not a free module and, consequently, we cannot define its rank; since $\mathbb{Z}/216\mathbb{Z}$ is a PIR, $\ker(A)$ is generated by $3$ or fewer elements. I do not know if there exists a generator with $2$ elements.
In particular, in a commutative ring, the rank-theorem for a matrix does not exist.
When $R$ is a PID, $\ker(A)$ is a free module and its rank can be defined.
