Multivariable Advanced Calculus

3.5. DETERMINANTS 41Summing over all ordered lists, (r 1 , · · · , r n ) where the r i are distinct, (If the r i are notdistinct, sgn (r 1 , · · · , r n ) = 0 and so there is no contribution to the sum.)∑∑(r 1 ,··· ,r n ) (k 1 ,··· ,k n )n! det (A) =sgn (r 1 , · · · , r n ) sgn (k 1 , · · · , k n ) a r1k 1· · · a rnk n.This proves the corollary. since the formula gives the same number for A as it doesfor A T .Corollary 3.5.7 If two rows or two columns in an n×n matrix, A, are switched, thedeterminant of the resulting matrix equals (−1) times the determinant of the originalmatrix. If A is an n × n matrix in which two rows are equal or two columns are equalthen det (A) = 0. Suppose the i th row of A equals (xa 1 + yb 1 , · · · , xa n + yb n ). Thendet (A) = x det (A 1 ) + y det (A 2 )where the i th row of A 1 is (a 1 , · · · , a n ) and the i th row of A 2 is (b 1 , · · · , b n ) , all otherrows of A 1 and A 2 coinciding with those of A. In other words, det is a linear functionof each row A. The same is true with the word “row” replaced with the word “column”.Proof: By Proposition 3.5.4 when two rows are switched, the determinant of theresulting matrix is (−1) times the determinant of the original matrix. By Corollary3.5.6 the same holds for columns because the columns of the matrix equal the rows ofthe transposed matrix. Thus if A 1 is the matrix obtained from A by switching twocolumns,det (A) = det ( A T ) = − det ( A T )1 = − det (A1 ) .If A has two equal columns or two equal rows, then switching them results in the samematrix. Therefore, det (A) = − det (A) and so det (A) = 0.It remains to verify the last assertion.det (A) ≡∑sgn (k 1 , · · · , k n ) a 1k1 · · · (xa ki + yb ki ) · · · a nkn= x+y(k 1 ,··· ,k n )∑(k 1 ,··· ,k n )∑(k 1 ,··· ,k n )sgn (k 1 , · · · , k n ) a 1k1 · · · a ki · · · a nknsgn (k 1 , · · · , k n ) a 1k1 · · · b ki · · · a nkn≡ x det (A 1 ) + y det (A 2 ) .The same is true of columns because det ( A ) T = det (A) and the rows of A T are thecolumns of A.The following corollary is also of great use.Corollary 3.5.8 Suppose A is an n × n matrix and some column (row) is a linearcombination of r other columns (rows). Then det (A) = 0.Proof: Let A = ( a 1 · · · a n)be the columns of A and suppose the conditionthat one column is a linear combination of r of the others is satisfied. Then by usingCorollary 3.5.7 you may rearrange the columns to have the n th column a linearcombination of the first r columns. Thus a n = ∑ rk=1 c ka k and sodet (A) = det ( a 1 · · · a r · · · a n−1∑ rk=1 c ka k).