2.3 Elementary Formulas 9Example.
∣
∣
C 1 C 3 C 4 C 2∣
∣
=−
∣
∣
C 1 C 2 C 4 C 3
∣
∣
=
∣
∣
C 1 C 2 C 3 C 4
∣
∣
.
Applying this property repeatedly,i.
∣
∣C
mCm+1···CnC 1 C 2 ···Cm− 1∣
∣=(−1)(m−1)(n−1)A,1 <m<n.The columns in the determinant on the left are a cyclic permutationof those inA.ii.∣
∣
CnCn− 1 Cn− 2 ···C 2 C 1∣
∣
=(−1)
n(n−1)/ 2
A.d.Any determinant which contains two or more identical columns is zero.
∣
∣
C 1 ···Cj···Cj···Cn∣
∣
=0.
e. If every element in any one column ofAis multiplied by a scalarkandthe resulting determinant is denoted byB, thenB=kA.B=
∣
∣
C 1 C 2 ···(kCj)···Cn∣
∣
=kA.Applying this property repeatedly,|kaij|n=∣
∣
(kC 1 )(kC 2 )(kC 3 )···(kCn)∣
∣
=kn
|aij|n.This formula contrasts with the corresponding matrix formula, namely[kaij]n=k[aij]n.Other formulas of a similar nature include the following:i.|(−1)i+j
aij|n=|aij|n,ii.|iaij|n=|jaij|n=n!|aij|n,iii.|xi+j−r
aij|n=xn(n+1−r)
|aij|n.f.Any determinant in which one column is a scalar multiple of anothercolumn is zero.
∣
∣
C 1 ···Cj···(kCj)···Cn∣
∣
=0.
g.If any one column of a determinant consists of a sum ofmsubcolumns,then the determinant can be expressed as the sum ofmdeterminants,each of which contains one of the subcolumns.
∣
∣
∣
∣
∣C 1 ···
(
m
∑s=1Cjs)
···Cn∣
∣
∣
∣
∣
=
m
∑s=1∣
∣C
1 ···Cjs···Cn∣
∣.
Applying this property repeatedly,
∣
∣
∣
∣
∣(
m
∑s=1C 1 s)
···
(
m
∑s=1Cjs)
···
(
m
∑s=1Cns