28.1. THE DETERMINANT 531

28.1.6 the determinant of A is zero if and only if the determinant of the matrix B, which hasthis special column placed in the last position, equals zero. Thus an = ∑

rk=1 ckak and so

det(B) = det(

a1 · · · ar · · · an−1 ∑rk=1 ckak

).

By Corollary 28.1.6

det(B) =r

∑k=1

ck det(

a1 · · · ar · · · an−1 ak

)= 0.

because there are two equal columns. The case for rows follows from the fact that det(A) =det(AT). ■

28.1.6 The Determinant Of A ProductRecall the following definition of matrix multiplication.

Definition 28.1.9 If A and B are n×n matrices, A = (ai j) and B = (bi j), AB = (ci j) where

ci j ≡n

∑k=1

aikbk j.

One of the most important rules about determinants is that the determinant of a productequals the product of the determinants.

Theorem 28.1.10 Let A and B be n×n matrices. Then

det(AB) = det(A)det(B) .

Proof: Let ci j be the i jth entry of AB. Then by Proposition 28.1.3,

det(AB) =

∑(k1,··· ,kn)

sgn(k1, · · · ,kn)c1k1 · · ·cnkn

= ∑(k1,··· ,kn)

sgn(k1, · · · ,kn)

(∑r1

a1r1br1k1

)· · ·

(∑rn

anrnbrnkn

)= ∑

(r1··· ,rn)∑

(k1,··· ,kn)

sgn(k1, · · · ,kn)br1k1 · · ·brnkn (a1r1 · · ·anrn)

= ∑(r1··· ,rn)

sgn(r1 · · ·rn)a1r1 · · ·anrn det(B) = det(A)det(B) . ■

28.1.7 Cofactor ExpansionsLemma 28.1.11 Suppose a matrix is of the form

M =

(A ∗0 a

)(28.10)