Are linear algebraic groups rigid? The underlying variety of a linear elgebraic group (say, over an algebraically closed field) is affine, so doesn't have nontrivial (infinitesimal) deformations. I'm curious to know whether  it's possible to deform the group structure on a fixed variety (that admits at least a structure of an algebraic group). 
Edit: I should have added, though when I wrote the question I mistakenly didn't expect reductivity made any difference, that I was mostly interested in reductive groups (in which case the comment of user54268 shows my question was not that naive after all..).
 A: How about this: for each $t \in F$, define a group structure on $F^3$ by 
$$(a,b,c) \cdot (d,e,f)=(a+d,b+e,c+f+tae).$$ When $t=0$ this is just $F^3$ with coordinate-wise addition, while for $t \neq 0$ it is $3$ by $3$ unipotent matrices. 
A: A most excellent example of "non-rigidity" beyond the reductive case (depending on how loose one wants to be about the meaning of "rigidity") 
is given in 5.2--5.10 of Exp. XIX of SGA3: a smooth affine group scheme $G$ over $k[t]$ for any field $k$ of characteristic 0 such that $G|_{t \ne 0}$ is a form of ${\rm{PGL}}_2$ (reductive!) but the fiber $G_0$ is solvable with 2 geometric connected components.  By definition, 
$G$ is the automorphism scheme of the Lie algebra over $k[t]$ whose underlying $k[t]$-module is free on a basis $\{X,Y,H\}$ that satisfy
$$[H,X]=X,\,\,\,[H,Y]=-Y,\,\,\,[X,Y]=2tH.$$
(It is clear that $G|_{t\ne 0}$ becomes ${\rm{Aut}}_{\mathfrak{sl}_2} = {\rm{PGL}}_2$ over the degree-2 finite etale cover defined by $\sqrt{t}$, but proving that $G$ is $k[t]$-smooth requires some cleverness.)
