5.19. Expansion Formulas by Minors
Definition: Minor
Given a square matrix A of order
n≥2
,
the square matrix
A
kj
of order
n−1
obtained by deleting the kth row and jth column of
A is called the k, j minor of A.
Example
A matrix
A=(
a
ij
)
of order 3 has 9 minors, one for each
element. The minors of the 1^{st}
row are
A
11
=(
a
22
a
23
a
32
a
33
)
A
12
=(
a
21
a
23
a
31
a
33
)
A
13
=(
a
21
a
22
a
31
a
32
)
The expansion (5.2) can therefore be
written as
detA=
a
11
det
a
11
−
a
12
det
A
12
+
a
13
det
A
13
Theorem 5.19:
Expansion by kth Row Minors.
For a square matrix A of order
n≥2
,
the cofactors and minors are related by
cof
a
kj
=det
A
′
kj
=
(
−
)
k+j
det
A
kj
(5.33)
Therefore, the expansion of detA by the kth row minors is given by
detA=
∑
j=1
n
(
−
)
k+j
a
kj
det
A
kj
(5.34)
Proof
We shall demonstrate the validity of (5.33)
by a series of progressively more general cases. To begin, consider the case
k=j=1
where
A
′
11
=(
1
0
0
⋯
0
a
21
a
22
a
23
⋯
a
2n
a
31
a
32
a
33
⋯
a
3n
⋮
⋮
⋮
⋱
⋮
a
n1
a
n2
a
n3
⋯
a
nn
)
It is obvious that
A
′
11
can be transformed into
A
11
0
=(
1
0
0
⋯
0
0
a
22
a
23
⋯
a
2n
0
a
32
a
33
⋯
a
3n
⋮
⋮
⋮
⋱
⋮
0
a
n2
a
n3
⋯
a
nn
)
=(
1
O
O
A
11
)
by adding
−
a
j1
×(row 1)
to the jth
row for each
j=2,⋯,n
. Since these are elementary row operations of
type (3) [see § 5.5], we have
det
A
11
0
=det
A
′
11
. Furthermore, since
A
11
0
is block diagonal, we have
det
A
11
0
=det
A
11
. Hence, (5.33) is verified for this case.
Next, similar consideration for the case
k=1
and
j≠1
gives
A
′
1j
=(
0
⋯
1
⋯
0
a
21
⋯
a
2j
⋯
a
2n
a
31
⋯
a
3j
⋯
a
3n
⋮
⋮
⋮
⋱
⋮
a
n1
⋯
a
nj
⋯
a
nn
)
Þ
A
1j
0
=(
0
⋯
1
⋯
0
a
21
⋯
0
⋯
a
2n
a
31
⋯
0
⋯
a
3n
⋮
⋮
⋮
⋱
⋮
a
n1
⋯
0
⋯
a
nn
)
with
det
A
′
1j
=det
A
1j
0
. Also
A
1j
=(
a
21
⋯
a
2j−1
a
2j+1
⋯
a
2n
a
31
⋯
a
3j−1
a
3j+1
⋯
a
3n
⋮
⋱
⋮
⋮
⋱
⋮
a
n1
⋯
a
nj−1
a
nj+1
⋯
a
nn
)
We now regard
det
A
1j
0
as a function of the n1 rows of the minor
A
1j
and write
det
A
1j
0
=f(
A
1j
)
. It is obvious that
f(
A
1j
)
satisfies Axioms 1 and 2 [see §5.3] for a
determinant function of order n1. By the uniqueness Theorem
5.6, we have
f(
A
1j
)=f(
I
(
n−1
)
)det
A
1j
where
I
(
n−1
)
is the unit matrix of order n1. Note that the diagonal
elements of
A
1j
are
a
k+1, k
for
k=2,⋯,j−1
and
a
k k
for
k=j+1,⋯,n
. Setting
A
1j
=J
in
A
1j
0
gives
f(
I
(
n−1
)
)=det
0
⋯
0
1
0
⋯
0
1
⋯
0
0
0
⋯
0
⋮
⋱
⋮
⋮
⋮
⋱
⋮
0
⋯
1
0
0
⋯
0
0
⋯
0
0
1
⋯
0
⋮
⋱
⋮
⋮
⋮
⋱
⋮
0
⋯
0
0
0
⋯
1

=det
O
1
O
I
(
j−1
)
O
O
O
O
I
(
n−j
)

=
(
−
)
j−1
det
I
(
j−1
)
O
O
O
1
O
O
O
I
(
n−j
)

=
(
−
)
j−1
det(
I
(
n
)
)=
(
−
)
j−1
which verifies (5.35) for the case
k=1
and
j≠1
.
Now, the general case of arbitrary k can be obtained from the case
k=1
by moving the kth row to the 1^{st}.
This introduces an extra factor
(
−
)
k−1
so that (5.33) is verified. Finally, putting (5.33) into (5.24) gives
(5.34). QED.