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.