Apologize, completely understood the question now.
Given: \begin{align*} v_1 &= (1, −1, 2, 0) \\ v_2 &= (1, 0, 1, 1) \\ v_3 &= (1, −2, 3, −1) \\ v_4 &= (3, 1, 2, 4) \end{align*}
Are $v_1, v_2, v_3, v_4$ linearly independent? I was confused with the matrix linearly independent.
How is vectors linearly independently vs matrix linearly independently?
If that is the REF that you obtained, you could have divide the third row by $-4$ and you will get a leading one in that row and the conclusion is the rank is $3$.
However, that is not the right REF. Check the step when you first update $R_3$, you make an arithmetic mistake there. The rank is indeed $2$.
Also notice that $cR_i-R_j$ is not actually a single elementary operations, it consist of $-R_i + R_j$ and $-R_j$.
If you have a zero row in the end, clearly they are not linearly indepedent.
If your original matrix is $A$, working on $A^T$ will help you identify a basis in the original vector set. Also, note that since there is no row swapping operations, the first two rows are linearly indepedent.