ALGEBRAIC INVARIANTS
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9. B
By the multiplication * of determinants, we get
a' b'
A' B'
=
a h
A B
cos 6 — sin 9
sin 9 cos 9
= aB — bA,
a' -b'
B' A'
=
a —h
B A
cos 9 sin 9
— sin 9 cos 9
— a A T bB.
The expressions at the right are therefore invariants of the
pair of linear functions l and L under every transformation
of type T. The straight lines represented by / = 0 and L = 0
are parallel if and only if aB — bA — 0; they are perpendicular
if and only if aA+bB = 0. Moreover, the quotient of aB — bA
by aA-\-bB is an invariant having an interpretation; it is the
tangent of one of the angles between the two lines.
As in the first example, A 2 -\-B 2 is an invariant of L. Between
our four invariants of the pair ¿ and L the following identity
holds:
{aA +bB) 2 + (aB -bA) 2 = (a 2 +b 2 ){A 2 +B 2 ).
The equation of any conic is of the form 5 = 0, where
S — ax 2 +2bxy-\-cy 2 -\-2kx-\-2ly- J rm.
Under the transformation T, S becomes a function of x' and
y', in which the part of the second degree
F = a'x' 2 -\-2b'x'y' -\-c'y' 2
is derived solely from the part of 5 of the second degree:
f = ax 2 2bxy+cy 2 .
The coefficient a! of x' 2 is evidently obtained by replacing
x by cos 9 and y by sin 9 in /, while c is obtained by replacing
x by — sin 9 and y by cos 6 in /. It follows at once that
a r -\-c = a-\-c.
Using also the value of b f , we can show that
a! c'—b' 2 =ac — b 2 ,
* We shall always employ the rule which holds also for the multiplication
of matrices; the element in the rth row and 5th column of the product is found
by multiplying the elements of the rth row of the first determinant by the cor
responding elements of the 5th column of the second determinant, and adding the
products.