### 5.3.2. Theorem 5.1.

If some row of *A* is the zero vector, then
detA=0
.

#### Proof

Let
A
k
=O
. Since
t
A
k
=O=
A
k
,
we have, with
t=−1
,

d(
⋯,−
A
k
,⋯
)=d(
⋯,
A
k
,⋯
)

However, by axiom 1, we have

d(
⋯,−
A
k
,⋯
)=−d(
⋯,
A
k
,⋯
)

Hence,

d(
⋯,
A
k
,⋯
)=−d(
⋯,
A
k
,⋯
)=0

QED.