“…If α 2 = 0 and β 1 ≠ 0 and α 1 = β 2 then choosing x = 1 2α1 a , y = α1+β1 2α1 a 2 + 1 2 b, z = a 2 + b gives A = span{x, y, z} with the nonzero products given in(3). On the other hand, if α 2 = 0 and β 1 ≠ 0 andα 1 ≠ β 2 then choosing x = 1 β2 a − 1 α1β2 a 2 , y = β1 β2−α1 a 2 + b, z = 1β gives A = span{x, y, z} with the nonzero products given in(2) with α = α1 β2 ∈ C {0, 1}. If α 2 ≠ 0 and α 1 +β 2 = 0 then choosing x = 1 (α2β1−β2α1) 1 2 a− α1 (α2β1−β2α1) 3 2 a 2 − α2 (α2β1−β2α1) 3 2 b, y = [(α 2 β 1 −β 2 α 1 ) 1 2 −α 1 ]a 2 −α 2 b, z = [(α 2 β 1 −β 2 α 1 ) 1 2 +α 1 ]a 2 +α 2 b gives A = span{x, y, z} with the nonzero products given in (2) with α = −1.…”