Page 26 - Vector Analysis

22 CHAPTER 1. Linear Algebra

Theorem 1.70. Let A be an n ˆ n matrix. Then

Adj(A)A = AAdj(A) = det(A)In.

Proof. Let A = [aij]. By definition of matrix multiplications,

() n () n [ ÿ ź ]
Adj(A)A Adj(A) imamj
ÿ ÿ
ij
= = εk1k2¨¨¨kn aℓkℓ amj

m=1 m=1 (k1,¨¨¨ ,kn)PP(n), km=i 1ďℓďn
ℓ‰m

$n

’ ÿź if i = j,
’
& εk1k2¨¨¨kn aℓkℓ

= (k1,¨¨¨ ,kn)PP(n) ℓ=1

’

’ 0 if i ‰ j.
%

The conclusion then follows from the definition of the determinant. ˝

Corollary 1.71. Let A = [aij] be an n ˆ n matrix, and C = [cij] be the adjoint matrix of

A. Then nn

det(A) = ÿ aijcji = ÿ ajicij @ 1 ď i ď n .

j=1 j=1

Corollary 1.72. Let A be an nˆn matrix and det(A) ‰ 0. Then the matrix Adj(A) is the
det(A)

inverse matrix of A, or equivalently,

Adj(A) = det(A)A´1. (1.2)

1.5.1 Variations of determinants

Let δ be an operator satisfying the “product rule” δ(f g) = f δg + (δf )g. Typically δ will be

differential operators. By the definition of the determinant,

n

δ det(A) = ÿ ź

εk1k2¨¨¨kn δ aℓkℓ

(k1,¨¨¨ ,kn)PP(n) ℓ=1

n[ ]
ÿÿ ź

= εk1k2¨¨¨kn δaiki aℓkℓ

i=1 (k1,¨¨¨ ,kn)PP(n) 1ďℓďn

ℓ‰i

n[ ÿ ź ]
ÿ
= εk1k2¨¨¨kn δaiki aℓkℓ

i,j=1 (k1,¨¨¨ ,kn)PP(n), ki=j 1ďℓďn
ℓ‰i

n (´1)i+j det (A(ˆi, ˆj))δaij

ÿ

= .

i,j=1

Therefore, we obtain the following