2.3 Elementary Formulas 9
Example.
∣
∣
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 permutation
of 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 scalarkand
the 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)
∣
∣
=k
n
|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.|x
i+j−r
aij|n=x
n(n+1−r)
|aij|n.
f.Any determinant in which one column is a scalar multiple of another
column 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=1
Cjs
)
···Cn
∣
∣
∣
∣
∣
=
m
∑
s=1
∣
∣C
1 ···Cjs···Cn
∣
∣.
Applying this property repeatedly,
∣
∣
∣
∣
∣
(
m
∑
s=1
C 1 s
)
···
(
m
∑
s=1
Cjs
)
···
(
m
∑
s=1
Cns