10. Expansion formulae

We will now collect in (6.1) all those terms which involve a₁₁. We can then write

|A| = a₁₁D₁₁ + R,

where R is the sum of all terms which do not contain a₁₁. In order to find a₁₁D₁₁, we must add all those terms in (6.1) in which p₁ = 1, whence

where P₁ passes through all permutations

P = [1, p₂, ··· , p_n]

with 1 is in the first place. Obviously, there corresponds to every permutation P₁ exactly one permutation of the numbers 2, ··· , n, in fact,

Q₁ = [p₂, ··· , p_n],

and, conversely, there corresponds to every Q₁ exactly one P₁. As P passes through all permutations from 1, 2, ··· , n, in which 1 is in the first place, Q₁ passes through all permutations from 2, ··· , n. Moreover, you realize that

c (P₁) = c (Q₁),

because the number 1 in P₁ does not cause any inversions. Hence there follows from (10.1)

where Q₁ passes through all permutations from 2, ··· , n. However, this is an expression analogous to (6.1) for a determinant of order (n - 1), that is,

The determinant D is called the minor determinant or minor of |A|, in other words, the minor determinant arises after cancellation of the first row and first column of |A|.

So far, we have obtained the result: In |A|, there enters the element a₁₁ - the element at the top on the left hand side - multiplied by the minor determinant , which is obtained by omission of the first row and first column of |A|.

Next, we will investigate what factor multiplies an arbitrary element a_ik in |A|. For this purpose, we interchange first of all the i-th row with all the rows above it and then the k-th column with all the columns to the left of it. This process involves altogether (i - 1) + (k -1) interchanges of rows and columns. After this action, we have a matrix, the first row of which, apart from the sequence of its elements, agrees with the i-th row of A and the first column of which, apart from their sequence, contains the elements of the k-th column of A. After these changes, the element a_ik is now located at the left hand top corner of the matrix. The determinant |A| becomes due to these interchanges

(-1)^i-1+k-1|A| = (-1)^i+k|A|.

The earlier derived result relating to the factor of the element in the top left hand corner of a determinant now yields: In (-1)^i+k|A|, the element a_ik is multiplied by that minor determinant D_ik, which arises from |A| by omission of the i-th row and the k-th column. Hence the element a_ik in |A| is multiplied by

(-1)^i+k|D_ik|= A_ik.

A_ik is called the adjoint minor of a_ik.

Every term of |A| has exactly one term out of the i-th row, whence we can collect always the terms with

a_i1, ··· , a_ik.

As we have just seen, here a_ik is multiplied by A_ik, whence

|A| = a_i1A_i1+ a_i2A_i2+ ··· + a_inA_in. (10.2)

This presentation of |A | is referred to as the expansion with respect to the i-th row.

By the interchangeability of the terms row and column, stated in Theorem 4, you obtain from (10.2) the expansion with respect to the k-th column

|A| = a_1kA_1k+ a_2kA_2k+ ··· + a_nkA_nk. (10.3)

Replacement in (10.2) of the elements a_i1, ··· , a_in. in the i-th row by the elements a_j1, ··· , a_jn on another row yields a determinant in which the j-th row occurs twice and which therefore is zero. Hence

|0 = a_j1A_i1+ a_j2A_i2+ ··· + a_jnA_in. (10.4)

Correspondingly, you find

0 = a_1lA_1k+ a_2lA_2k+ ··· + a_nlA_nk. (10.5)

by combining (10.2) and (10.4), (10.3) and (10.5), respectively.

Determinant Expansion formulae

		|A|	for	i = j
aj1Ai1+ aj2Ai2+ ··· + ajnAin =
		0	"	i j
		|A|	"	k = l
a1lA1k+ a2lA2k+ ··· + anlAnk =
		0	"	k l

The sign (-1)^i+k, which must be taken into consideration in the formation of the adjoint minor of a_ik, is readily determined for every element. In fact, the + and - signs re distributed like the black and white fields of a chess board; at the top on the left hand side is +, whence the sign for every element of the principal diagonal is +.

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