70 CHAPTER 5. SOME IMPORTANT LINEAR ALGEBRA

Proof: Let (1, · · · ,n) = (1, · · · ,r, · · ·s, · · · ,n) so r < s.

det(A(1, · · · ,r, · · · ,s, · · · ,n)) = (5.4.10)

∑(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)) . (5.4.11)

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 5.3.1, 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 5.4.10 -5.4.11 shows that detA(r1, · · · ,rn) = 0 and also sgn(r1, · · · ,rn) = 0 sothe formula holds in this case also.

Observation 5.4.4 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.