( 253 ) 
by putting this matrix equal to naught, we let the space pz = 0 
satisfy the given condition. Here, however, an obstacle seems to 
present itself. For the five equations obtained by putting the deter- 
minants comprised in the matrix equal to naught, furnish in general 
two respectively independent relations, which cannot be the case 
here. However, as is immediately evident after development, each of 
those five determinants consists really of the form 
(pit Pot Ps + Pa) Ps + Pe P3 Pat P1 P3 Pa + Pi Pa Pa + P1 Pa Ps 
every time multiplied by another linear form, and we find the 
wanted equation of the enveloped surface by putting this common 
factor equal to naught. 
The same obstacle seems to appear when we make use of the 
following method to determine the equation of the enveloped space. 
If pr= 0, gr = 0 represent an arbitrary plane, it intersects the 
four lines aj, ag, a3, a, under the conditions 
P1 P3 Pa P2 Pa Ps P3 Pe Ps Pa P1 P5 
Mi 93 15 ll Q2 9495 | = 0, 93 9275 | = 0, 91491 |= 0 
ee ET: | Ee Dl eae 
and by eliminating gj, 92, q3 We arrive at the equation 
D3 — P5 Oi a (Pies) (Pips) 95 
OF: Pa Pos 0 5 (ha Ps) Ia + (petpa) gs A 
Leh ne a Lam bi (potps)9s| 
—(patps), Oe Oene (P1—Ps) Va + (91+ pa) 93 
which also furnishes two equations, as it must hold good for all 
: 74, î : 2 
values of the quotient —. We recognise in these two equations 
75 
immediately those which are obtained by omitting from the matrix 
found above respectively the last column and the last but one. 
By the way we notice that the second method can prove in a 
