R2 13 + 12 = R1-1 (p 11) (5.7) and taking the inner-product with vector z provides (z R2 13) + (z 12) = (z R1-1 p) (z RI-1 11). Applying (4.5) to the first term of both sides of this equation gives (after rearranging terms) R1 p R2-1 z 13 = z RI- 11 + z 12. (5.8) The right hand side of Eq. (5.8) is constant since ai Ri-1 i = dici and z.1i=di are independent of ei. diri Multiplying Eq. (5.7) by R2-1 gives 13 + R2-1 12 = R2-1 RI-1 (p 11) (5.9) and multiplication of Eq. (5.5) by R2-1 R1-1 yields R3 z = R2-1 R1-1 t. (5.10) The inner product of corresponding sides of equations (5.9) and (5.10) produces (13 + R2-1 12).(R3 z) = [R2- R1-1 (p 11)].[R2-1 RI-1 t]. Repeated use of properties (4.4) and (4.5) and reordering simplifies this last equation to ..