Page 23 - Vector Analysis

§1.5 Determinants 19

thus after applying (n ´ 1)-times elementary row operations of the third type (adding
some multiple of certain row to another row) on A we reach a matrix whose j-th row
is a zero (row) vector. Thereofre, for some elementary matrices E1, ¨ ¨ ¨ , En´1 we have

det(En´1 ¨ ¨ ¨ E1A) = 0

which implies that det(A) = 0. ˝

Corollary 1.65. Let A be an nˆn matrix. Then the determinant of A and AT, the transpose
of A, are the same; that is,

det(A) = det(AT).

Proof. If A is not invertible, then AT is not invertible either because of Theorem 1.48.
Therefore, det(A) = 0 = det(AT).

Now suppose that A is invertible. Then Theorem 1.55 implies that

A = EkEk´1 ¨ ¨ ¨ E2E1

for some elementary matrices E1, ¨ ¨ ¨ , Ek. Since all ETj ’s are also elementary matrices, by
Proposition 1.62 we conclude that

det(AT) = det(ET1 ¨ ¨ ¨ EkT) = det(E1T) ¨ ¨ ¨ det(EkT) ˝
= det(ETk ) ¨ ¨ ¨ det(ET1 )

= det(Ek) ¨ ¨ ¨ det(E1) = det(Ek ¨ ¨ ¨ E1) = det(A) .

Corollary 1.66. Let A, B be n ˆ n matrices. Then det(AB) = det(A) det(B).

Proof. If A is not invertible, then AB is not invertible either; thus in this case det(A) det(B) =
0 = det(AB).

Now suppose that A is invertible. Then Theorem 1.55 implies that

A = EkEk´1 ¨ ¨ ¨ E2E1

for some elementary matrices E1, ¨ ¨ ¨ , Ek. As a consequence, Proposition 1.62 implies that

det(AB) = det(Ek ¨ ¨ ¨ E1B) = det(Ek) det(Ek´1 ¨ ¨ ¨ E1B) ˝
= ¨ ¨ ¨ = det(Ek) ¨ ¨ ¨ det(E1) det(B)
= det(Ek ¨ ¨ ¨ E1) det(B) = det(A) det(B) .