## 5.9. Multilinearity of Determinants

### Theorem 5.7.
Additivity In Each Row.

Given an *n* ´ *n*
matrix
A=(
A
1
,⋯,
A
n
)
and any *n*-D
vector *V*. Let *B*
and *C* be matrices that differ from *A* only in the *k*th row, namely,

A=(
⋯,
A
k
,⋯
)
B=(
⋯,V,⋯
)
C=(
⋯,
A
k
+V,⋯
)

Then,

detC=detA+detB
(5.10)

or,

d(
⋯,
A
k
+V,⋯
)=d(
⋯,
A
k
,⋯
)+d(
⋯,V,⋯
)
(5.10a)

which is known as the **additivity in the ***k*th row.

#### Proof

Let

f(
A
)=detC−detB
=d(
⋯,
A
k
+V,⋯
)−d(
⋯,V,⋯
)
(a)

where the right hand side of the last
equality is assumed to be independent of *V*. We shall prove the theorem by showing that
f(
A
)
satisfies all 3 axioms for a determinant function
so that by the uniqueness theorem,
f(
A
)=detA
.

For operations not involving *A*_{k}, the proof that
f(
A
)
satisfies axioms 1 and 2 follows trivially
from the fact that det*C* and det*B* are determinant functions. For operations involving *A*_{k}, the proof for axiom 1 is as follows

f(
⋯,c
A
k
,⋯
)=d(
⋯,c
A
k
+V,⋯
)−d(
⋯,V,⋯
)

=cd(
⋯,
A
k
+
V
c
,⋯
)−cd(
⋯,
V
c
,⋯
)

=c[
d(
⋯,
A
k
+
V
c
,⋯
)−d(
⋯,
V
c
,⋯
)
]

=c f(
⋯,
A
k
,⋯
)

where the last equality made use of the
assumption that det*C* - det*B* is independent of *V*. For axiom 2, we have

f(
⋯,
A
k
+
A
j
,⋯,
A
j
,⋯
)=d(
⋯,
A
k
+
A
j
+V,⋯,
A
j
,⋯
)−d(
⋯,V,⋯,
A
j
,⋯
)

=d(
⋯,
A
k
+V,⋯,
A
j
,⋯
)−d(
⋯,V,⋯,
A
j
,⋯
)

=f(
⋯,
A
k
,⋯,
A
j
,⋯
)

and

f(
⋯,
A
k
,⋯,
A
k
+
A
j
,⋯
)=d(
⋯,
A
k
+V,⋯,
A
k
+
A
j
,⋯
)−d(
⋯,V,⋯,
A
k
+
A
j
,⋯
)

Applying the theorem itself to the right
hand side gives

f(
⋯,
A
k
,⋯,
A
k
+
A
j
,⋯
)=d(
⋯,
A
k
+V,⋯,
A
j
,⋯
)−d(
⋯,
A
k
+V,⋯,
A
k
,⋯
)

−d(
⋯,V,⋯,
A
j
,⋯
)+d(
⋯,V,⋯,
A
k
,⋯
)

=f(
⋯,
A
k
,⋯,
A
j
,⋯
)

where we’ve used

d(
⋯,
A
k
+V,⋯,
A
k
,⋯
)=d(
⋯,
A
k
+V−
A
k
,⋯,
A
k
,⋯
)
=d(
⋯,V,⋯,
A
k
,⋯
)

Finally, we need to prove *f* satisfies axiom 3, namely,
f(
I
)=1
. Setting
A=I
in eq(a), we have

f(
I
)=d(
I
1
,⋯,
I
k
+V,⋯,
I
n
)−d(
I
1
,⋯,V,⋯,
I
n
)

Let
V=(
v
1
,⋯,
v
n
)
. If
k=1
,
both *C* and *B* are upper triangular. By
Theorem 5.5, we have

detC=1+
v
1
and
detB=
v
1
(b)

so that

f(
I
)=1+
v
1
−
v
1
=1
(c)

and the theorem holds. For
k=2
,
we have

C=(
1
0
0
⋯
0
v
1
1+
v
2
v
3
⋯
v
n
0
0
1
⋯
0
⋮
⋮
⋮
⋱
0
0
0
0
⋯
1
)
B=(
1
0
0
⋯
0
v
1
v
2
v
3
⋯
v
n
0
0
1
⋯
0
⋮
⋮
⋮
⋱
0
0
0
0
⋯
1
)

To make each of these upper triangular, we
need only add
−
v
1
×
1^{st} row to the 2^{nd}
row. The operation does not affect the
diagonal elements so that we again recover eq(b) and hence eq(c). Furthermore, a little reflection will show
that the same holds for all *k* so that
the theorem is proved.

### Multilinearity

By repeated applications of Theorem 5.7 and
Axiom 1 to the same row, we can generalize (5.10a) to

d(
⋯,
∑
i=1
p
t
i
U
i
,⋯
)=
∑
i=1
p
t
i
d(
⋯,
U
i
,⋯
)

which expresses the **linearity in each row**. Since
this applies to every row, the determinant function is said to be **multilinear function **in its rows.