Put (aox 3 -}-a l x 2 +a 2 x +a 3) (box' +b1x'+b2) - (aox'3+aix'2+a2x'+a3) (box' + bix + b2) = 0; after division by x-x the three equations are formed aobcx 2 = aobix+aob2 =0, aobix 2 + (aob2+a1b1-a2bo) x +alb2 -a3bo = 0, aob2x 2 +(a02-a3bo)x+a2b2-a3b1 =0 and thence the resultant aobo ao aob2 aob 1 aob2+a1b1-a2bo alb2-a3b0 aob 2 a1b2 - a 3 bo a2b2 - a3b1 which is a symmetrical determinant.
Since (ab) = a l b 2 -a 2 b l, that this may be the case each form must be linear; and if the forms be different (ab) is an invariant (simultaneous) of the two forms, its real expression being aob l -a l b 0.
By simple multiplication (al b l b2 -24a2bib2+ala2b;)xi +(aibz -ala214b2-aia2blb2+a2b2)xlx2 + (aia 2 b2 - 2a l a2b l b2 +a2/4b 2)x2; and transforming to the real form, (aob 2 - 2a1b,+a2bo)xi (aob 3 -a l b 2 - alb,+a3bo)xlx2 + (aib3 - 2a2b2+a3b1)x2, the simultaneous covariant; and now, putting b = a, we obtain twice.
For w = i the form is A i ai+Bib i, which we may write aob l -albo = ao(I) b -(I)abo; the remaining perpetuants, enumerated by z I - 2' have been set forth above.
This can be done by placing at B an equal negative point-charge -q in the place which would be occupied by the optical image of A if PO were a mirror, that is, let -q be placed at B, so that the distance BO is equal to the distance AO, whilst AOB is at right angles to PO.
How would you define aob? Add your definition here.