432 PROFESSOR EVERETT ON A METHOD OF REDUCING 
and the sum of these four products (since (S,)’ + (8,)?=1) is 6 A,. Hence the value 
of A, can be found. 
Again, adding the quantities which stand opposite to each other in columns 
III. and IV. we have the sums 
oune 4B, §, 4B, S, 2B; 
and if we multiply these respectively by the factors, 
| 0 Se ce 1 
we obtain the products, 
0 4B, (S,)? 4 B, (S,)* 2B, 
The sum of these products is 6 B,; hence B, can be found. 
Adding the terms opposite each other in columns I. and II. we find 
Vv VI. 
v,+%,=2 A, +2 A, Sy+2B,8, | v,+%, =2A,—2 A, 8,—2 B, 8, 
v,+0,=2 A,+2A,8,4+2B,8, | v,+%,=2A,—2 A, 8,—2 B, 8, 
Up +%,=2 A,-2A,S,4+2B,8, | v,+v,,=2 A,+2 A, 8,—2B, S, 
The sum of all the terms in V. and VI. is 12 A,, which is in fact the sum of the 
12 values of v. 
Subtracting the quantities in VI. from those opposite to them in V., we have 
the remainders,— 
4A, Adh US. ASR Soe —4A,8,+4B, S,; 
Multiply these remainders respectively by 
1 S, =8, 
and omitting the two terms 4 B,S,S8, and —4 B, 8, S,, which destroy one 
another, we have the products— 
4 A, 4 A, (8,)? 4 A, (S,)? 
in! 
whose sum (since Si=5) is 6 A,. Hence A, can be found. 
Again, if the above remainders be multiplied respectively by 
0 Ss S, 
the products (omitting terms which destroy each other) are 
0 4 B, (S8,) 4 B, (S8,)? 
and since ee) the sum of these products is 6 B,. Hence B, can be found. 

