132 CHAPTER 7. DETERMINANTS MATHEMATICAL THEORY∗
= ∑(k1,··· ,kn)
sgn(k1, · · · ,kn)ar1k1 · · ·arnkn (7.5)
= det(A(r1, · · · ,rn)) . (7.6)
Proof: Let (1, · · · ,n) = (1, · · · ,r, · · ·s, · · · ,n) so r < s.
det(A(1, · · · ,r, · · · ,s, · · · ,n)) = (7.7)
∑(k1,··· ,kn)
sgn(k1, · · · ,kr, · · · ,ks, · · · ,kn)a1k1 · · ·arkr · · ·asks · · ·ankn ,
and renaming the variables, calling ks,kr and kr, ks, this equals
= ∑(k1,··· ,kn)
sgn(k1, · · · ,ks, · · · ,kr, · · · ,kn)a1k1 · · ·arks · · ·askr · · ·ankn
= ∑(k1,··· ,kn)
−sgn
k1, · · · ,These got switched︷ ︸︸ ︷
kr, · · · ,ks , · · · ,kn
a1k1 · · ·askr · · ·arks · · ·ankn
=−det(A(1, · · · ,s, · · · ,r, · · · ,n)) . (7.8)
Consequently,det(A(1, · · · ,s, · · · ,r, · · · ,n)) =
−det(A(1, · · · ,r, · · · ,s, · · · ,n)) =−det(A)
Now letting A(1, · · · ,s, · · · ,r, · · · ,n) play the role of A, and continuing in this way, switch-ing pairs of numbers,
det(A(r1, · · · ,rn)) = (−1)p det(A)
where it took p switches to obtain(r1, · · · ,rn) from (1, · · · ,n). By Lemma 7.0.1, this implies
det(A(r1, · · · ,rn)) = (−1)p det(A) = sgn(r1, · · · ,rn)det(A)
and proves the proposition in the case when there are no repeated numbers in the orderedlist, (r1, · · · ,rn). However, if there is a repeat, say the rth row equals the sth row, then thereasoning of 7.7 -7.8 shows that detA(r1, · · · ,rn) = 0 and also sgn(r1, · · · ,rn) = 0 so theformula holds in this case also. ■
Observation 7.1.4 There are n! ordered lists of distinct numbers from {1, · · · ,n} .
To see this, consider n slots placed in order. There are n choices for the first slot. Foreach of these choices, there are n−1 choices for the second. Thus there are n(n−1) waysto fill the first two slots. Then for each of these ways there are n−2 choices left for the thirdslot. Continuing this way, there are n! ordered lists of distinct numbers from {1, · · · ,n} asstated in the observation.