5.19. Expansion Formulas by Minors
Definition:
Minor
Given a square matrix A of order n  2 , the square matrix Akj 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   aij  of order 3 has 9 minors, one for each element. The minors of
the 1st row are
a
A11   22
 a32
a23 
a33 
a 23
a
A12   2 1

 a31 a33 
a 22
a
A13   2 1

 a31 a32 
The expansion (5.2) can therefore be written as
det A  a11 det a11  a12 det A12  a13 det A13
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 akj  det Akj    
k j
det Akj
(5.33)
Therefore, the expansion of detA by the kth row minors is given by
n
det A     
k j
j 1
(5.34)
akj det Akj
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
0
 1
a
 21 a22
A11   a31 a32


a
 n1 a n 2
0
a23
a33
an 3
0 
a2 n 
a3 n 


ann 
It is obvious that A11 can be transformed into
1 0
0 a
22

0
A11   0 a32


0 a
n2

0 
a2 n 
1
a3 n   

O

ann 
0
a23
a33
an 3
O
A11 
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 A110  det A11 .
Furthermore, since A110 is block diagonal, we have det A110  det A11 .
Hence, (5.33)
is verified for this case.
Next, similar consideration for the case k  1 and j  1 gives
 0
a
 21
A1 j   a31


a
 n1
0 
a2 n 
a3n 


ann 
1
a2 j
a3 j
anj
with det A1 j  det A10j .
 a21
a
31
A1 j  


 a n1

 0
a
 21
A10j   a 3 1


a
 n1
1
0
0
0
0
a n2 
a n 3


ann 
Also
a2 j 1 a2 j 1
a3 j 1
a3 j 1
anj 1
anj 1
a2 n 
a3n 


ann 
We now regard det A10j as a function of the n1 rows of the minor A1 j and write
det A10j  f  A1 j  .
It is obvious that f  A1 j  satisfies Axioms 1 and 2 [see §5.3] for
a determinant function of order n1.

By the uniqueness Theorem 5.6, we have

f  A1 j   f I  n 1 det A1 j
where I n1 is the unit matrix of order n1. Note that the diagonal elements of A1 j
are ak 1, k for k  2,
, j  1 and ak k for k  j  1,
,n .
Setting A1 j  J in
A10j gives
0
1

0 1 0
0 0 0
0
0
O

f I  n 1  det 0
0
1 0 0
0 0 1
0
0
0
0 0 0
1
I  j 1 O
  
j 1
det
O
O
1
O
 det I  j 1
O
O
O
  
j 1
1
O
O
O
O
I n j 
 
det I  n     
j 1
I n j 
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 1st. This introduces an extra factor
verified.
 
k 1
so that (5.33) is
Finally, putting (5.33) into (5.24) gives (5.34). QED.