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I have been trying to prove linear independence of rows implies linear independence of columns in kind of an abstract setting of modules following some exercises and notes from an old course. The following fact is something I think we are supposed to use but I did not see any way to prove it so I would like to understand the proof so I can potentially adapt it to a couple of similar problems.

Let $A = (a_{i,j})$ be a square matrix with $n$ rows and columns and with entries in a commutative ring with identity $R$ and let $M$ be an $R$-module. Let $x_1, \ldots ,x_n \in M$ and suppose that$\sum_{j=1}^{n}a_{i,j} x_j = 0$ for $1 \leq i \leq n$ . Then how do you show $\det(A)x_i = 0$ for every $i = 1, \ldots, n$.

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Given you are referring to the determinant of $A$, you mean $R$ is a commutative ring (with $1$). Correct? –  Bill Cook Nov 6 '11 at 23:59
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This question may be of help: math.stackexchange.com/questions/71740/… –  Bill Cook Nov 7 '11 at 0:06
    
@BillCook - Thank you so much that post is very helpful! –  user7980 Nov 7 '11 at 0:15
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1 Answer

The equation you have got is in matrix form $$ A\cdot\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}= \begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}. $$ Left-multiply by the adjugate matrix $\def\Adj{\operatorname{Adj}}\Adj A$ of$~A$: $$ \Adj A\cdot A\cdot\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}=\Adj A\cdot \begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}= \begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}. $$ The left hand side simplifies $$ (\det A)I_n\cdot\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}= \begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}, $$ which means $(\det A)x_n=0$ for $i=1,2,\ldots,n$.

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