Determinants and Their Applications in Mathematical Physics

48 3. Intermediate Determinant Theory

Proof.

Bij=(−1)

i+j

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

a 11 a 12 ··· a 1 ,j− 1 a 1 ,j+1 ··· a 1 n x 1

a 21 a 22 ··· a 2 ,j− 1 a 2 ,j+1 ··· a 2 n x 2

..........................................................

ai− 1 , 1 ai− 1 , 2 ··· ai− 1 ,j− 1 ai− 1 ,j+1 ··· ai− 1 ,n xi− 1

ai+1,i ai+1, 2 ··· ai+1,j− 1 ai+1,j+1 ··· ai+1,n xi+1

..........................................................

an 1 an 2 ··· an,j− 1 an,j+1 ··· ann xn

y 1 y 2 ··· yj− 1 yj+1 ··· yn z

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ n

.

The expansion is obtained by applying arguments toBijsimilar to those

applied toBin Theorem 3.9. Since the second cofactor is zero whenr=i

ors=jthe double sum contains (n−1)

2

nonzero terms, as expected. It

remains to prove thatBij=Eij.

Transfer the last row ofBijto theith position, which introduces the sign

(−1)
n−i
and transfer the last column to thejth position, which introduces

the sign (−1)
n−j

. The result isEij, which completes the proof.

The Cauchy expansion of an arbitrary determinant focuses attention on

one arbitrarily chosen elementaijand its cofactor.

Theorem 3.11. The Cauchy expansion

A=aijAij+

n

∑

r=1

n

∑

s=1

aisarjAir,sj.

First Proof.The expansion is essentially the same as that given in Theorem

3.10. TransformEijback toAby replacingzbyaij,xrbyarjandysby

ais. The theorem appears after applying the relation

Air,js=−Air,sj. (3.7.2)

Second Proof.It follows from (3.2.3) that

n

∑

r=1

arjAir,sj=(1−δjs)Ais.

Multiply byaisand sum overs:

n

∑

r=1

n

∑

s=1

aisarjAir,sj=

n

∑

s=1

aisAis−

n

∑

s=1

δjsaisAis

=A−aijAij,

which is equivalent to the stated result.

Theorem 3.12. Ifys=1, 1 ≤s≤n, andz=0, then

n

∑

j=1

Bij=0, 1 ≤i≤n.