The question is:
"Decide which elements of $Z/12Z$ have inverses, and for each such element find its inverse, then solve $[11]_{12} X=[6]_{12}$"
I belive that the only inverse of $Z/12Z$ are $[1]$ and $[11]$ ($[1]_{12} * [11]_{12} = [-1]_{12}$
How do I solve for $X$?
$[11]_{12} * [1]_{12} * X=[6]_{12}* [1]_{12}$
Does that mean that $X=6$ or $X=[6]_{12}$?
Also, $5$ and $7\ (\equiv -5 \pmod{12})$ have multiplicative inverses, as they are also relatively prime to $12$.
Your solution seems ok. After you observe that $11\equiv -1\pmod{12}$, you can freely interchange $11$ to $-1$ in modulo $12$ context, just like with equations.
So that, we look for $$6\equiv 11X\equiv -X \pmod{12}$$ and the solution is $X\equiv -6 \ \equiv 6\pmod{12}$, or, with your notation, $X=[-6]_{12}=[6]_{12}$.
The equation also makes sense with $X\in\Bbb Z$, as that, it has infinitely many solutions ($6+12k,\ k\in\Bbb Z$), but if you regard the unknown as $X\in\Bbb Z/12\Bbb Z$, then it has a unique solution. That's why we mostly prefer the latter interpretation.
