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.

Edit: The row operations you performed and the row echelon matrix are indeed correct. 
You're correct that since $r = 3 = n$ and hence, the vectors are linearly independent.
Note, since the echelon form of the matrix is now in upper triangular form, we can "read off" the determinant of the matrix in echelon form, which is the same as the determinant of the matrix we started with: why? (Because the only elementary row operations used here are those that did not change the value of the determinant of the original matrix.) 
$$\det(A) = \left|\begin{matrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -1 & 0 & 1 \end{matrix}\right| = \left|\begin{matrix} 1 & 1 & -1 \\ 0 & -1 & 2\\ 0 & 0 & 2 \end{matrix}\right| = (1)(-1)(2) = -2 \neq 0$$
Since the $\det A \neq 0$, we know the rows (and columns) of $A$ are linearly independent.

And you are correct that when $r\lt n$, the vectors represented by the rows are then linearly dependent.
