We can prove that if the three equations be satisfied by a system of values of the variable, the same system will also satisfy the **Jacobian** or functional determinant.

For if u, v, w be the polynomials of orders m, n, p respectively, the **Jacobian** is (u 1 v 2 w3), and by Euler's theorem of homogeneous functions xu i +yu 2 +zu 3 = mu xv1 +yv2 +zv3 = /IV xw 1+y w 2+ zw 3 = pw; denoting now the reciprocal determinant by (U 1 V2 W3) we obtain Jx =muUi+nvVi+pwWi; Jy=ï¿½.., Jz=..., and it appears that the vanishing of u, v, and w implies the vanishing of J.

CY The proof being of general application we may state that a system of values which causes the vanishing of k polynomials in k variables causes also the vanishing of the **Jacobian**, and in particular, when the forms are of the same degree, the vanishing also of the differential coefficients of the **Jacobian** in regard to each of the variables.

Since, If F = An, 4) = By, 1 = I (Df A4) Of A?) Ab A"'^1Bz 1=, (F, Mn Ax I Ax 2 Axe Ax1) J The First Transvectant Differs But By A Numerical Factor From The **Jacobian** Or Functional Determinant, Of The Two Forms. We Can Find An Expression For The First Transvectant Of (F, ï¿½) 1 Over Another Form Cp. For (M N)(F,4)), =Nf.4Y Mfy.4), And F,4, F 5.4)= (Axby A Y B X) A X B X 1= (Xy)(F,4))1; (F,Ct)1=F5.D' 7,(Xy)(F4)1.

Solving the equation by the Ordinary Theory Of Linear Partial Differential Equations, We Obtain P Q 1 Independent Solutions, Of Which P Appertain To S2Au = 0, Q To 12 B U =0; The Remaining One Is Ab =Aobl A 1 Bo, The Leading Coefficient Of The **Jacobian** Of The Two Forms. This Constitutes An Algebraically Complete System, And, In Terms Of Its Members, All Seminvariants Can Be Rationally Expressed.

1 Ab' Establishing The Ground Forms Of Degrees Order (I, O; I), (O, I; I), (I, I; O), Viz: The Linear Forms Themselves And Their **Jacobian** J Ab.

1 A2B' Where The Denominator Factors Indicate The Forms Themselves, Their **Jacobian**, The Invariant Of The Quadratic And Their Resultant; Connected, As Shown By The Numerator, By A Syzygy Of Degreesorder (2, 2; 2).

There is no linear covariant, since it is impossible to form a symbolic product which will contain x once and at the same time appertain to a quadratic. (v.) is the **Jacobian**; geometrically it denotes the bisectors of the angles between the lines ax, or, as we may say, the common harmonic conjugates of the lines and the lines x x .