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Asume in $\mathbb{R}^4$ $$AX=0 \,(\text{mod } 11)$$ with $$A=\begin{bmatrix}1 & 2 & 3 & 4\\ 1 & 2 & 3 & 4\\1 & 2 & 3 & 4\\1 & 2 & 3 & 4\end{bmatrix}$$

because $\text{rank}\,A+\dim\langle A \rangle=n$ $$\dim\langle A\rangle=4-1=3$$ so we need $3$ independent vectors to generate linear space $\langle A\rangle$ and there are three independent solutions. In my book they are given as $$X1=\begin{bmatrix}9 & 1 & 0 & 0\end{bmatrix}$$ $$X2=\begin{bmatrix}8 & 0 & 1 & 0\end{bmatrix}$$ $$X3=\begin{bmatrix}7 & 0 & 0 & 1\end{bmatrix}$$

OK, but what about $X4=\begin{bmatrix}0 & 2 & 1 & 1\end{bmatrix}$ ? It is not in linear space $\langle A \rangle$, is it? it is not linear combination of $X1$, $X2$ and $X3.$ So do we need a $4$th vector? I know not, because $\dim \langle A \rangle=3$ so how could it be?

edit: ok, it is! we are in $\text{mod }11$ and $X4=\begin{bmatrix}0 \\ 2 \\ 1 \\ 1\end{bmatrix}=2\begin{bmatrix}9 \\ 1 \\ 0 \\ 0\end{bmatrix}+1\begin{bmatrix}8 \\ 0 \\ 1 \\ 0\end{bmatrix}+1\begin{bmatrix}7 \\ 0 \\ 0 \\ 1\end{bmatrix}$ $(\text{mod }11)$

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Yup, you got it. –  Gerry Myerson Feb 27 '13 at 12:17
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great, thank you all in all, if I haven't decided to post it I think I still would don't know –  bits_international Feb 27 '13 at 12:23
    
@cf16 you can also add the 'edit' part as an asnwer to your own question. Also, the notation $VA$ (at least to me) is not familiar. –  Berci Feb 27 '13 at 12:24
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A bit misleading that we have two different $X1$'s. The second one could be $X4$ or $Y$.. –  Berci Feb 27 '13 at 12:35
    
yes, sure, changed –  bits_international Feb 27 '13 at 15:43

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up vote 3 down vote accepted

ok, it is! we are in $mod 11$ and $X4=\begin{bmatrix}0 \\ 2 \\ 1 \\ 1\end{bmatrix}=2\begin{bmatrix}9 \\ 1 \\ 0 \\ 0\end{bmatrix}+1\begin{bmatrix}8 \\ 0 \\ 1 \\ 0\end{bmatrix}+1\begin{bmatrix}7 \\ 0 \\ 0 \\ 1\end{bmatrix}$ $mod(11)$

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Maybe replacing $-2$ for $9$ and $-3$ for $8$ makes it even clearer. (These replacements can be done modulo $11$.) –  Berci Feb 27 '13 at 12:37

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