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Linear Algebra, Theory And Applications, 2012a

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3.3. THE MATHEMATICAL THEORY OF DETERMINANTS 87<br />

∑<br />

(k 1 ,··· ,k n )<br />

sgn (k 1 , ··· ,k r , ··· ,k s , ··· ,k n ) a 1k1 ···a rkr ···a sks ···a nkn ,<br />

and renaming the variables, calling k s ,k r and k r ,k s , this equals<br />

= ∑<br />

sgn (k 1 , ··· ,k s , ··· ,k r , ··· ,k n ) a 1k1 ···a rks ···a skr ···a nkn<br />

= ∑<br />

(k 1,··· ,k n)<br />

Consequently,<br />

(k 1 ,··· ,k n )<br />

⎛<br />

− sgn ⎝k 1 , ··· ,<br />

These got switched<br />

{ }} {<br />

k r , ··· ,k s<br />

, ··· ,k n<br />

⎞<br />

⎠ a 1k1 ···a skr ···a rks ···a nkn<br />

= − det (A (1, ··· ,s,··· ,r,··· ,n)) . (3.11)<br />

det (A (1, ··· ,s,··· ,r,··· ,n)) = − det (A (1, ··· ,r,··· ,s,··· ,n)) = − det (A)<br />

Now letting A (1, ··· ,s,··· ,r,··· ,n) play the role of A, and continuing in this way, switching<br />

pairs of numbers,<br />

det (A (r 1 , ··· ,r n )) = (−1) p det (A)<br />

where it took p switches to obtain(r 1 , ··· ,r n )from(1, ··· ,n). By Lemma 3.3.1, this implies<br />

det (A (r 1 , ··· ,r n )) = (−1) p det (A) =sgn(r 1 , ··· ,r n )det(A)<br />

and proves the proposition in the case when there are no repeated numbers in the ordered<br />

list, (r 1 , ··· ,r n ). However, if there is a repeat, say the r th row equals the s th row, then the<br />

reasoning of (3.10) -(3.11) shows that det(A (r 1 , ··· ,r n )) = 0 and also sgn (r 1 , ··· ,r n )=0<br />

sotheformulaholdsinthiscasealso.<br />

Observation 3.3.7 There are n! ordered lists of distinct numbers from {1, ··· ,n} .<br />

To see this, consider n slots placed in order. There are n choices for the first slot. For<br />

each of these choices, there are n − 1 choices for the second. Thus there are n (n − 1) ways<br />

to fill the first two slots. Then for each of these ways there are n−2 choices left for the third<br />

slot. Continuing this way, there are n! ordered lists of distinct numbers from {1, ··· ,n} as<br />

stated in the observation.<br />

3.3.3 A Symmetric Definition<br />

With the above, it is possible to give a more symmetric description of the determinant from<br />

which it will follow that det (A) =det ( A T ) .<br />

Corollary 3.3.8 The following formula for det (A) is valid.<br />

det (A) = 1 n! ·<br />

∑<br />

∑<br />

(r 1 ,··· ,r n ) (k 1 ,··· ,k n )<br />

sgn (r 1 , ··· ,r n ) sgn (k 1 , ··· ,k n ) a r1k 1<br />

···a rnk n<br />

. (3.12)<br />

<strong>And</strong> also det ( A T ) =det(A) where A T is the transpose of A. (Recall that for A T = ( a T ij)<br />

,<br />

a T ij = a ji.)

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