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Consider $\Bbb Z_3$ the field with $3$ elements .Let $\Bbb Z_3\times \Bbb Z_3$ be the vector space over $\Bbb Z_3$.

Find the number of distinct linearly dependent sets of the form $\{u,v\}$ where $u,v\in \Bbb Z_3\times \Bbb Z_3\setminus \{(0,0)\}$.

My try:

If $\{u,v\}$ is linearly dependent then $u=av$ for some $a\in \Bbb Z_3$.

Then the possible options are $\,\{0,1\},\{0,2\},\{1,2\}.$ So it will be $3$ .But the answer given is $4$ .Where am I wrong?

Please help.

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You have correctly found the 2-element linearly dependent subsets of $\mathbb{Z}_3$. But that's not what the problem asks for: it asks for linearly dependent 2-element subsets of $\mathbb{Z}_3\times\mathbb{Z}_3\setminus\{(0,0)\}$. So $u$ and $v$ should each be an ordered pair of elements of $\mathbb{Z}_3$ (different from $(0,0)$).

(Also, the problem statement is slightly incorrect: it should also require that $u\neq v$.)

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  • $\begingroup$ Thank you so much Now I got $4$ $\endgroup$ – Learnmore Feb 4 '17 at 5:48
  • $\begingroup$ i got 8 as an answer why is it 4? $\endgroup$ – Upstart Feb 4 '17 at 6:18
  • $\begingroup$ @Upstart: You're probably counting ordered pairs $(u,v)$ instead of sets. $\endgroup$ – Eric Wofsey Feb 4 '17 at 6:19
  • $\begingroup$ if the {($0,0$)} is included then i think it will be $8$ $\endgroup$ – Upstart Feb 4 '17 at 6:29
  • $\begingroup$ @EricWofsey,,,(1,0), (0,2),(0,1) and (2,0) here im getting 4,,,Is it correct or not ? pliz tell me $\endgroup$ – lomber Oct 31 '17 at 14:40

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