The question asks: do the vectors $[1,1,0], [0,1,1], [1,0,1]$ form a basis for $Z^2$?
I know how to check whether vectors form a basis normally: see if an arbitrary vector in $R^n$ can be written as a linear combination of the vectors (ex. $[1,1,0], [0,1,1], [1,0,1]$ forms a basis for $R^3$ because the matrix with those three as rows has a solution, meaning any vector can be written as a linear combination of them).
But how do I apply that idea to $Z$? I simplified the matrix under $Z$ and found that one of the three vectors was a combination of the other two, but I'm not sure how that really helps or how to apply that information.
The definition of a basis is the same as before. The coefficients are now in $\Bbb{Z}$ instead of a field $K$. So $\{e_1,\ldots ,e_n\}$ is a basis of $\Bbb{Z}^n$ if $a_1e_1+\cdots +a_ne_n=0$ implies that all $a_i=0$, and the integral linear combinations of $e_1,\ldots ,e_n$ give $\Bbb{Z}^n$. This is the case if and only if the associated matrix $A$ is in the group $GL_n(\Bbb{Z})$. Note that $\det(A)=\pm 1$ then.