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Find a basis for the submodule $\mathbb Z^3$ consisting of all integer solutions of the system of equations $x+2y+3z=0, x+4y+9z=0$.

$$\begin{align*} x +2y +3z &=0\\ x+4y+9z &=0\\ \end{align*}$$

Then the augmented matrix is:

$$ \left[\begin{array}{ccc|c} 1 & 2 & 3 &0\\ 1 & 4 & 9 &0\\ \end{array}\right] $$

Reduced Row Echelon Form $\to$

$$ \left[\begin{array}{ccc|c} 1 & 0 & -3 &0\\ 0 & 1 & 3 &0\\ \end{array}\right] $$

So a basis for the solution set is $$\{\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}\}$$

I don't think this is the answer they want though because they talk about integer solutions.

I would appreciate some help/guidance pushing me in the right direction.

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  • $\begingroup$ You computed it well, but the conclusion is not correct. Solution set it generated by $[3,-3,1]$, not by $[1,2,3]$. $\endgroup$ – SMM Apr 29 '18 at 18:53
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Rewrite your solution in equations. You have obtained $$ x=3z,\quad y=-3z. $$ So you can choose any integer $z$. It determines $(x,y,z)=(3z,-3z,z)$. So the basis is $(3,-3,1)$.

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