Find the number of distinct linearly dependent sets of the form $\{u,v\}$

Question

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.

Answer 1

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 $u\neq v$.)

Answer 2

Fourth pair will be {(1,2),(2,1)} as (2,1)can be written by -1(1,2)=(-1,-2) and field is F3 or can say Z3 and in Z3 -1 is equivalent to 2 and -2 is equivalent to 1