Plane and solid analytic geometry, by William F. Osgood and William C. Graustein.

366

ANALYTIC GEOMETRY

sign, according as i j k I presents an even or an odd number of
inversions from the normal order 1 2 3 4. Thus the product
b,2da3c4 would be taken as plus, since, when the factors are
arranged in the order of the letters, viz. - a3b2c4d1, the number of inversions in the subscripts 3 2 4 1 is even, namely 4.
The product a4b2c3d1 would be taken as minus, since the number of inversions in 4 2 3 1 is odd, namely 5.
We can now give a complete definition of the determinant of
the fourth order.
DEFINITION. Form all the products of elements of (1) which
contain just one factor from each row and one factor from each
column of (1); to each product aibjckdl prefix a plus sign or a
minus sign, according as the number of inversions of i j k I from
the normal order 1 2 3 4 is even or odd. The sum of the
products, thus signed, is the determinant.
Determinants of the fifth, sixth, and higher orders are
similarly defined. Let the student think through the definition
for a five-rowed determinant, and let him show, also, that in the
case of two-and three-rowed determinants the definition yields
precisely the expressions which were defined as these determinants in ~ 2.
The signed products which make up a determinant are known
as the terms of the determinant. Thus, + a3b2c4dl and
- a4b2c3d are terms of I a b c d.
EXERCISES
1. What is the number of inversions of each of the following
orders, from the normal order?
(a) 3 1 4 2; (c) 2 5 3 1 4; (e) 3 1 6 4 5 2;
(b) 2 4 3 1; (d) 4 3 5 2 1; (f) 6 5 4 3 21.
2. Write out all the terms of I a b c d 1. To how many products have you prefixed plus signs? To how many, minus signs?
3. How many terms has a determinant of the fifth order?
Prove your answer.