Multivariable Advanced Calculus

3.5. DETERMINANTS 43Proof: Denote M by (m ij ) . Thus in the first case, m nn = a and m ni = 0 if i ≠ nwhile in the second case, m nn = a and m in = 0 if i ≠ n. From the definition of thedeterminant,det (M) ≡∑sgn n (k 1 , · · · , k n ) m 1k1 · · · m nkn(k 1 ,··· ,k n )Letting θ denote the position of n in the ordered list, (k 1 , · · · , k n ) then using the earlierconventions used to prove Lemma 3.5.1, det (M) equals∑(k 1 ,··· ,k n ))(−1) n−θ θsgn n−1(k 1 , · · · , k θ−1 , k θ+1 , · · · , n−1k n m 1k1 · · · m nknNow suppose 3.31. Then if k n ≠ n, the term involving m nkn in the above expressionequals zero. Therefore, the only terms which survive are those for which θ = n or inother words, those for which k n = n. Therefore, the above expression reduces to∑a sgn n−1 (k 1 , · · · k n−1 ) m 1k1 · · · m (n−1)kn−1 = a det (A) .(k 1,··· ,k n−1)To get the assertion in the situation of 3.30 use Corollary 3.5.6 and 3.31 to writedet (M) = det ( M T ) (( )) AT0= det= a det ( A T ) = a det (A) .∗ aThis proves the lemma. In terms of the theory of determinants, arguably the most important idea is that ofLaplace expansion along a row or a column. This will follow from the above definitionof a determinant.Definition 3.5.12 Let A = (a ij ) be an n×n matrix. Then a new matrix calledthe cofactor matrix, cof (A) is defined by cof (A) = (c ij ) where to obtain c ij delete thei th row and the j th column of A, take the determinant of the (n − 1) × (n − 1) matrixwhich results, (This is called the ij th minor of A. ) and then multiply this number by(−1) i+j . To make the formulas easier to remember, cof (A) ijwill denote the ij th entryof the cofactor matrix.The following is the main result.Theorem 3.5.13 Let A be an n × n matrix where n ≥ 2. Thendet (A) =n∑a ij cof (A) ij=j=1n∑a ij cof (A) ij. (3.32)The first formula consists of expanding the determinant along the i th row and the secondexpands the determinant along the j th column.Proof: Let (a i1 , · · · , a in ) be the i th row of A. Let B j be the matrix obtained from Aby leaving every row the same except the i th row which in B j equals (0, · · · , 0, a ij , 0, · · · , 0) .Then by Corollary 3.5.7,n∑det (A) = det (B j )j=1Denote by A ij the (n − 1) × (n − 1) matrix obtained by deleting the i th row and thej th column of A. Thus cof (A) ij≡ (−1) i+j det ( A ij) . At this point, recall that fromi=1