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Let $G$ be an abelian group with generating set ${a_1, a_2, a_3}$ and relation: $$12a_1 + 15a_2 + 18a_3 = 0 = 20a_1 + 19a_2 - 2a_3$$

Find $ord(2a_2 + 16a_3)$.

I tried to make manipulations with with the matrix, but it gave me odd relations like $7a_2+58a_3=283a_1$. Thanks.

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Here is a start: fives times the first relation minus three times the second gives $18a_2+96a_3=0$, so we have $6(2a_2+16a_3)=-6a_2$. So we know that $ord(6(2a_2+16a_3))=ord(-6a_2)=ord(6a_2)$.

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    It would be nice if you explain first two steps, because I don't understand why we can multiply first relation on 5 (would it change the order?) – 2017-02-13
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    Multiplying by $5$ is here to raise to the power of $5$, but written additively. And of course you can raise a relation to some power. Example. If $a_1^2=1$, then $a_1^{10}=1$. Written additively: if $2a_1=0$, then $10a_1=0$. So $5$ times the first relations gives $60a_1+75a_2+90a_3=0$. The order is not considered here. This comes afterwards, namely $ord(2a_2+16a_3)=ord(-a_2)=ord(a_2)$. – 2017-02-13
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    OK, it's clear, but how can we be sure that $3a_2+16a_3 = 0$? Can order of this element divide 6? – 2017-02-13
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    Well, I understood that $18a_2+96a_3=0$ as a "linear combination" of relations, I just don't understand why we can say that $18a_2+96a_3=0 \Rightarrow 3a_2+16a_3=0$. Why element $3a_2+16a_3$ can't have order 2 or 3 and then $18a_2+96a_3=0$? – 2017-02-13
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    Yes, you are right. We only know the order of $6(2a_2+16a_3)$ from this. – 2017-02-13