438 CHAPTER 20. THEORY OF DETERMINANTS∗

Letting θ denote the position of n in the ordered list, (k1, · · · ,kn) then using Lemma 20.1.1,det(M) equals

∑(k1,··· ,kn)

(−1)n−θ sgnn−1

(k1, · · · ,kθ−1,

θ

kθ+1, · · · ,n−1kn

)m1k1 · · ·mnkn

Now suppose 20.11. Then if kn ̸= n, the term involving mnkn in the above expression equalszero. Therefore, the only terms which survive are those for which θ = n or in other words,those for which kn = n. Therefore, the above expression reduces to

a ∑(k1,··· ,kn−1)

sgnn−1 (k1, · · ·kn−1)m1k1 · · ·m(n−1)kn−1 = adet(A) .

To get the assertion in the situation of 20.10 use Corollary 20.2.5 and 20.11 to write

det(M) = det(MT )= det

((AT 0∗ a

))= adet

(AT )= adet(A) .

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 definition ofa determinant.

Definition 20.2.12 Let A = (ai j) be an n×n matrix. Then a new matrix called thecofactor matrix, cof(A) is defined by cof(A) = (ci j) where to obtain ci j delete the ith rowand the jth column of A, take the determinant of the (n−1)× (n−1) matrix which results,(This is called the i jth minor of A. ) and then multiply this number by (−1)i+ j. To makethe formulas easier to remember, cof(A)i j will denote the i jth entry of the cofactor matrix.

The following is the main result. Earlier this was given as a definition and the outra-geous totally unjustified assertion was made that the same number would be obtained byexpanding the determinant along any row or column. The following theorem proves thisassertion.

Theorem 20.2.13 Let A be an n×n matrix where n ≥ 2. Then

det(A) =n

∑j=1

ai j cof(A)i j =n

∑i=1

ai j cof(A)i j . (20.12)

The first formula consists of expanding the determinant along the ith row and the secondexpands the determinant along the jth column.

Proof: Let (ai1, · · · ,ain) be the ith row of A. Let B j be the matrix obtained from A byleaving every row the same except the ith row which in B j equals

(0, · · · ,0,ai j,0, · · · ,0) .

Then by Corollary 20.2.6,

det(A) =n

∑j=1

det(B j)

Denote by Ai j the (n−1)× (n−1) matrix obtained by deleting the ith row and the jth col-umn of A. Thus cof(A)i j ≡ (−1)i+ j det

(Ai j). At this point, recall that from Proposition