got[1 3;0 -4] after some reductions ...which involved ensuring theres a zero beneath the first leading entry..
row2/-4 then gives [ 1 3 ;0 1]
now assigning parameters y=t
then [x ;y]=t[-3;1]
eigenvector then in [-3;1]
Yes, that is correct for "row-reduction" but I don't know what you mean about "finding eigenvectors". You cannot find eigenvectors (or eigenvalues) by row reduction!
In this particular problem, it is easy to see that which is NOT a multiple of ([3, 1]. [-3, 1] is NOT an eigenvector for this matrix!
(Here, the eigenvalues are and . Eigenvectors corresponding to eigenvalue are multiples of and eigenvectors corresponding to eigenvalue are multiples of ).