## 5.19. Expansion Formulas by Minors

### Definition:  Minor

Given a square matrix A of order n2 , the square matrix A kj  of order n1  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 1st 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 n2 , 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 j1  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 2j1 a 2j+1 a 2n a 31 a 3j1 a 3j+1 a 3n a n1 a nj1 a nj+1 a nn )

We now regard det A 1j 0  as a function of the n-1 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 n-1.  By the uniqueness Theorem 5.6, we have

f( A 1j )=f( I ( n1 ) )det A 1j

where I ( n1 )  is the unit matrix of order n-1.  Note that the diagonal elements of A 1j  are a k+1,k  for k=2,,j1  and a kk  for k=j+1,,n .  Setting A 1j =J  in A 1j 0  gives

f( I ( n1 ) )=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 ( j1 ) O O O O I ( nj )    |

= ( ) j1 det|    I ( j1 ) O O O 1 O O O I ( nj )    |   = ( ) j1 det( I ( n ) )= ( ) j1

which verifies (5.35) for the case k=1  and j1 .

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 ( ) k1  so that (5.33) is verified.  Finally, putting (5.33) into (5.24) gives (5.34).  QED.