Let's suppose that $ \lambda $ is an eigenvalue of a matrix A. You want to find an eigenvector corresponding to $ \lambda $. To do that you need to solve Ax = $ \lambda $x, which is same as (A- $ \lambda $I) x = 0.
If you solve the 2nd equation (A - $ \lambda $ I) x =0, swapping rows doesn't change your answer.
If you solve the 1 st equation, Ax = $ \lambda $ x, swapping rows changes your answer.
Here's the reason. Let P be a permutation matrix swapping rows 1 and 2.
If you multiply A by P from the left , P will swap the 1 st and 2nd rows of A.
Note that Ax = $ \lambda $x < => PAx = $ \lambda $Px .
This means that if you swap rows of A, rows of x will be swapped too.
However, (A-$ \lambda $I)x = 0 <=> P (A - $ \lambda $I) x = P 0 <=> [P (A - $ \lambda $I)] x = 0 .
This doesn't affect your answer. So, it depends on what equation you use when you swap rows.
