I'm trying to find the rref(A) where A = \begin{bmatrix}2&1&4&-4&11\\1&2&3&1&8\\1&1&2&-1&6\end{bmatrix}
I used row operations and came out with the following: \begin{bmatrix}1&0&0&-17&13\\0&1&0&8&-6\\0&0&1&-6&8\end{bmatrix}
However the Linear Algebra Toolkit came out with this: \begin{bmatrix}1&0&0&-3&3\\0&1&0&2&1\\0&0&1&0&1\end{bmatrix}
I read in other questions here that it isn't possible to have two different possible rrefs and I'm unclear as to what I've done incorrectly as the rows satisfy the requirements of a rref matrix. Can anyone identify what I've done incorrectly? I can submit my exact row operations if required. Thanks.
EDIT: Here's a list of my operations: $R_1-R_3,R_2-R_3, R_3-R_2, R_1-2R_2, R_3*R_2, R_2-R_3, R_1-2R_2$