Page 25 - Vector Analysis

§1.5 Determinants 21

1. ki = j;

2. for each τ P t1, ¨ ¨ ¨ , i ´ 1u and ι P ti, i + 1, ¨ ¨ ¨ , n ´ 1u,

" στ if στ ă j , " σι if σι ă j ,
στ + 1 if στ ě j , σι + 1 if σι ě j .
kτ = and kι+1 =

We now determine the sign of εσ1σ2¨¨¨σn´1 and εk1k2¨¨¨kn. Note that if a process of pair inter-
changes of the permutation (σ1, σ2, ¨ ¨ ¨ , σn´1) leads to (1, 2, ¨ ¨ ¨ , n ´ 1), then similar process
of pair interchanges of the permutation (k1, k2, ¨ ¨ ¨ , ki´1, j, ki+1, ¨ ¨ ¨ , kn), by leaving the i-th
slot fixed, leads to the permutation of degree n

$ (1, 2, ¨ ¨ ¨ , j ´ 1, j + 1, ¨ ¨ ¨ , i ´ 1, j, i, ¨ ¨ ¨ , n) if i ą j,
’
&

(1, 2, ¨ ¨ ¨ , i ´ 1, j, i, ¨ ¨ ¨ , j ´ 1, j + 1, ¨ ¨ ¨ , n) if i ă j,

’ (1, 2, ¨ ¨ ¨ , n) if j = i.
%

For the case that i ‰ j, another |i ´ j|-times of pair interchanges leads to (1, 2, ¨ ¨ ¨ , n). To
be more precise, suppose that i ą j. We first interchange the (i ´ 2)-th and the (i ´ 1)-th
components, and then interchange that (i ´ 3)-th and the (i ´ 2)-th components, and so on.

After (i ´ j)-times of pair interchanges, we reach (1, 2, ¨ ¨ ¨ , n). Symbolically,

ò

(1, 2, ¨ ¨ ¨ , j ´ 1, j + 1, ¨ ¨ ¨ , i ´ 1, j , i, ¨ ¨ ¨ , n)

Ó τ(i ´ 2, i ´ 1)

ò

(1, 2, ¨ ¨ ¨ , j ´ 1, j + 1, ¨ ¨ ¨ , i ´ 2, j , i ´ 1, ¨ ¨ ¨ , n)

Ó τ(i ´ 3, i ´ 2)

ò

(1, 2, ¨ ¨ ¨ , j ´ 1, j + 1, ¨ ¨ ¨ , i ´ 3, j , i ´ 2 ¨ ¨ ¨ , n)

Ó...
Ó
(1, 2, ¨ ¨ ¨ , n).

Similar argument applies to the case i ă j; thus

εσ1σ2¨¨¨σn´1 = (´1)|i´j|εk1k2¨¨¨kn = (´1)i+j εk1k2¨¨¨kn .

As a consequence,

det (A(ˆi, ˆj)) = n´1

ÿ ź

εσ1σ2¨¨¨σn´1 bτ στ

(σ1,σ2,¨¨¨ ,σn´1)PP(n´1) τ =1

= (´1)i+j ÿ ź

εk1k2¨¨¨kn aℓkℓ . ˝

(k1,¨¨¨ ,kn)PP(n), ki=j 1ďℓďn
ℓ‰i