Linear Independence by row space. 
Let $(1,1,-1), (2,1,0)$ and $(-1,0,1)$ be vectors, show if they are independent.

I wrote each vector on the rows of the matrix $A$.
$A=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -1 & 0 & 1 \end{pmatrix}$
Then I put $A$ on an echelon form.
$A=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -1 & 0 & 1 \end{pmatrix}\overset{L2-2L1}  {\rightarrow}\begin{pmatrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ -1 & 0 & 1 \end{pmatrix}\overset{L3+L1} {\rightarrow}\begin{pmatrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ 0 & 1 & 0 \end{pmatrix}\overset{L3+L2}{\rightarrow}\begin{pmatrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ 0 & 0 & 2 \end{pmatrix}$
So, $r(A)=3$. If $\space r(A)=3$ that is equal to the number of rows$(n)$ of $A$, then I can conclued that the $3$ vectors are independent.
If $r(A)<n$ , then the set of the $3$ vectors would be dependent. Is this right? Thanks.

Yes, this is correct.
You could also just compute the determinant of $A$ and conclude that the rows (and columns too) are linear independent.
