5.10.2. Theorem 5.9.

(a)      For matrices of order 2, there is 1 and only 1 determinant function , namely,

det( a 11 a 12 a 21 a 22 )=|    a 11 a 12 a 21 a 22    |   = a 11 a 22 a 12 a 21                            (5.12a)

(b)      For matrices of order 2, there is 1 and only 1 determinant function , namely

det( a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 )=|    a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33    |

        = 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    |              (5.12)

Proof of (a)

The easiest way is to show that the formula satisfies all 3 axioms for a determinant function and then invoke the uniqueness theorem.  However, it is more instructive to derive them directly as follows.

Given A=( a 11 a 12 a 21 a 22 ) , we have

        A 1 =( a 11 , a 12 )   = a 11 i+ a 12 j                  A 2 =( a 21 , a 22 )   = a 21 i+ a 22 j

where i=( 1,0 )  and j=( 0,1 )  are the unit coordinate vectors.  Using the linearity in the 1st row, we have

        d( A 1 , A 2 )=d( a 11 i+ a 12 j  ,   A 2 )   = a 11 d( i, A 2 )+ a 12 d( j, A 2 )

Using the linearity in the 2nd row, we have

        d( i, A 2 )=d( i  ,   a 21 i+ a 22 j )   = a 21 d( i,i )+ a 22 d( i  ,  j )   = a 22 d( i  ,  j )

        d( j, A 2 )=d( j  ,   a 21 i+ a 22 j )   = a 21 d( j,i )+ a 22 d( j  ,  j )   = a 21 d( j  ,  i )

where we have used d( i,i )=d( j,j )=0 .  Using d( j,i )=d( i,j ) , we have

        d( A 1 , A 2 )= a 11 a 22 d( i,j )+ a 12 a 21 d( j,i )   =( a 11 a 22 a 12 a 21 )d( i,j )

Using

d( i,j )=|    1 0 0 1    |   =d( I )=1

then completes the proof.

Proof of (b)

This is left as an exercise.

General Formula

Let A be an n ´ n matrix.  Its jth row can be written as

        A j =( a j1 ,, a jn )= k=1 n a jk I k

where I k =( 0,, 1 k ,,0 )  is the kth Cartesian basis vector.  Thus,

        d( A )=d( A 1 ,, A j ,, A n )   =d( k 1 =1 n a 1 k 1 I k 1 ,, k j =1 n a j k j I k j ,, k n =1 n a n k n I k n )

                = k 1 =1 n k j =1 n k n =1 n a 1 k 1 a j k j a n k n d ( I k 1 ,, I k j ,, I k n )

Now,

        d( I k 1 ,, I k j ,, I k n )={    1 1 0              if     { k 1 ,, k j ,, k n } is a even odd not  permutation of { 1,,j,,n }

 

Therefore, d( A )  is a signed sum of product terms of the form a 1 k 1 a j k j a n k n , where the set of indices { k 1 ,, k j ,, k n }  is a permutation of { 1,,j,,n } .  This is usually denoted by

        d( A )= P ( ) P a 1P( 1 ) a jP( j ) a nP( n )

where P stands for permutation.