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The left-hand member of this equation is what is called the determinant of the given system of equations.

We proceed to investigate the law of its formation.
3. First, suppose we have two equations,

ax, + ax=0,

6,26, + box, = 0. Multiply the first by be, the second by ag, and subtract,

a, b, - a,b, = 0. Hence

a, a, b, b,

= a,b, - ab. We may remark in passing that we shall obtain the same result by eliminating 2; , between the equations

a,1, +5,1% = 0,
azł, + 6,1, = 0.

and we get

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A like theorem will be proved to be true for all determinants.

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Multiply these equations in order by the arbitrary multipliers ,, ,, ng, and add them together. Let the two ratios à,:19:be determined by the conditions that the coefficients of

a, and dg in the resulting equation shall each be zero, i.e. let

a,1, + bon + Con07
a,, + bag + Cody =

o

(A). The resulting equation is then reduced to

(a,1, + 6,7, +13) *c, = 0, which requires that

a, + b^4 + 0,1 = 0............ (B).

........

Multiply the first of equations (A) by dg, the second by Qg, and subtract, we then get

(a,b, - azba) 18+ (0,2; – caz) 2g = 0,

re
Cydz – caz agbz azbe

or

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6,6b,c,'

by symmetry....(C).

Hence, dividing each term of (B) by the corresponding member of (C) we get

az (6,63 – boca) + b, (caz – czan) + (a,ba abs),

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6,, bg,

3

It will be seen that the above process is really equivalent to that of eliminating ,,, 1, between the equations (A) and (B). Hence

а, а, аз 2,, b, c,

ba ag, bg, C,

C, CE, Cg ag, B2, C,
5. Next, let us have the four equations

ayx, + zx2 + + QqC = 0,
6,2, + bx, + boks + b2c, = 0,
CX, + code + cgXg + celle = 0,

dx, +dc,+ dgx, +d, x, = 0. To effect the elimination, multiply the equations in order by day lg, lg, lq, add them, and equate the coefficients of Xz, 2, , severally to zero.

We shalĩ then have
az1, +6,12 + cały + d ) = 0)
agle + 6372 +973 += 0

(A'),
a 14 +6,12 +673 +0
which equations involve as a consequence

a,, + 6,22 +0,13 +0,20 = 0 .......... (B'). To determine the three ratios 1:12::1, multiply equations (A') in order by Mig, Migo Ma, add, and equate to zero the coefficients of rug, dy We thus get CM + Calz + cap = 0%

(C'). dou + das + du =) Also (aglig + Agh's + adja) 2, + (bells + bgMz + beva) 1, = 0, whence

λ,

bells + balta + ball - (azhet agtstagen) Now, treating equations (C') as equations (A) were treated, we see that

M

M р.
cad, cd, cd, - cod. cdo - cade'

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These equations may be more conveniently written in the following equivalent forms:

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Eliminating by means of these equations ? 1g, 1.3, 1., from equation B', we get, as the result of the elimination of 24, Xg, Xg, wc, between the four given equations,

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And since the above process is equivalent to the elimination of 1, 1, 1 g, n, between the equations (A') and (B'), we see that

a, a, as a4 a, b, c, d,
bq, bg, bs, be ag, bg, Ca, d.
Cq, Cz, Co, CA ag, bg, Cg, d,
d , dz, dz, da ag, ba, ca de

6. The law of formation will be sufficiently obvious from the above investigations. If we have n lines and columns, it may be similarly proved that

az, az, az...An Qy, by, Cz...k
bq, bz, bz...ba da, bz, C...kg
C4, C2, C3...Ch Ag, bz, C...k,
;
:
:

:
, , kg...kn an, bx, Cg...kr

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