## 5.11. The Product Formula for Determinants

The product of an *m* ´ *n*
matrix
A=(
a
ij
)
and an *n*
´ *p* matrix
B=(
b
ij
)
is defined as the *m* ´ *p*
matrix
C=(
c
ij
)
with components [see §4.15]

c
ij
=
∑
k=1
n
a
ik
b
kj
(5.13)

### Lemma 5.10.

Given an *m* ´ *n*
matrix
A=(
a
ij
)
and an *n*
´ *p* matrix
B=(
b
ij
)
,
we have

(
AB
)
i
=
A
i
B
(5.14)

which gives the *i*th row of *AB* as the
product of the *i*th row of *A* with *B*.

#### Proof

Using *A*_{i}
(*A*^{i}) to denote the *i*th row (column) of *A*, we can write eq(5.13) as the dot product

c
ij
=
A
i
⋅
B
j

The *i*th
row of *C* is therefore

C
i
=(
c
i1
,
c
i2
,⋯,
c
ip
)
=(
A
i
⋅
B
1
,
A
i
⋅
B
2
, ⋯,
A
i
⋅
B
p
)

=
A
i
⋅(
B
1
,
B
2
,⋯,
B
p
)
=
A
i
B
QED.

### Theorem 5.11.
Product Formula for Determinants.

For any two *n* ´ *n*
matrices *A* and *B*, we have

det(
AB
)=(
detA
)(
detB
)
(5.15)

#### Proof

By Lemma 5.10, we have

det(
AB
)=d[
(
AB
)
1
,
(
AB
)
2
, ⋯,
(
AB
)
n
]
=d(
A
1
B,
A
2
B,⋯,
A
n
B
)

For a fixed *B*, we define the function

f(
A
)=det(
AB
)

In terms of the rows of *A*, this becomes

f(
A
1
,⋯,
A
n
)=d(
A
1
B,
A
2
B,⋯,
A
n
B
)

It is obvious that *f* satisfies Axioms 1 and 2 of a determinant function. Furthermore,

f(
I
)=d(
B
)

so that by the uniqueness Theorem 5.9, we
have

f(
A
)=d(
A
)f(
I
)=d(
A
)d(
B
)
QED.