Finding the standard basis for $\mathbb{R}^4$ that can be added to the set $\{(1,-4,2,-3),(-3,8,-4,6)\}$ to produce a basis for $\mathbb{R}^4$ Finding the standard basis for $\mathbb{R}^4$ that can be added to the set $\{(1,-4,2,-3),(-3,8,-4,6)\}$ to produce a basis for $\mathbb{R}^4$.

I first check that the two vectors in the set are not linearly dependent, and they are not. So using these vectors to form a matrix, the echelon form shows that there are 2 pivots.
My question is, since elementary operations preserves row equivalent matrices, I can swap the rows containing the pivots, so that the two vectors I am given can be row equivalent to any two combinations of $e_{1},e_{2},e_{3},e_{4}$, and so I just choose the two $e_{i}$'s currently not in the set to produce a basis for $\mathbb{R}^4$? Also, suppose the set was linearly dependent, then I need to choose $3$ of them. Are my understanding correct? 

Yes your understanding is correct.
To apply to the example you gave. You have $u=(1,-4,2,-3), v=(-3,8,-4,6)$. Hence $2u+v=(-1,0,0,0)$ by elementary operation. Finally $(u, v, e_2, e_3)$ is a basis of $\mathbb R^4$
