589 
of Edinburgh, Session 1885 - 86 . 
The set of equations being 
ax -t-by + cx +d =0 
a'x + b'y + dx +d’ =0 
a"x + b"y + c"x + d" = 0 
we know that the numerator of the values of x, y , z, and the com- 
mon denominator are 
b 
c 
d 
a 
c 
d 
a 
b 
d 
a 
b 
c 
V 
d 
d! 
+ 
a' 
d 
d! 
- 
a' 
V 
d! 
+ 
a' 
b' 
d 
b" 
e" 
d" 
a" 
c" 
d" 
a" 
b" 
d" 
3 
a" 
b" 
c" 
They are therefore the coefficients of x , y , z, t in the determinant 
abed 
a! V e' d! 
a" b" c" d" 
x y z t 
, or A say. 
Thus the problem of solving the set of equations is transformed into 
finding the development of this determinant. In doing so let us 
use \xyz\ to stand for the determinant of which x , y , z is the last 
row, and whose other rows are the two rows immediately above 
x , y, z in A : similarly let [ zt J stand for the determinant of which 
z, t is the last row, and its other row the row c", d" immediately 
above z, t in A ; and so on in all possible cases, including even 
\xyzt\, which of course is A itself. 
Then clearly we have 
\xyzt\ = a[yzt\ - b\_xzt\ + c\xyt\ - d\xyz\ .... (1) 
Developing in the same way the four determinants here on the right 
side, we have as our next step 
\xyzt] = a(b'[zt] - c f [yt\ + 
- b{a'\zf\ - c'[xf\ + d'lxz]) 
+ c [a'[yt] - b'\xf[ + d!\xyf) 
~ d(a\yz\ - b'\xz\ + c'[xy \) , 
= (ab' - a'b)\zt\ - (ad - a! d)[yt ] + (ad! - a'd\yz~\ 
+ (bd - b'c)\xt\ - (bd' - b'd)\xz!\ + (ed! - dd)\xy ] . 
