Page 44 - Numerical Methods for Chemical Engineering

The determinant 33

N
If det(A) = 0; i.e., if A is nonsingular, then Ax = b has a unique solution for all b ∈ .
Otherwise, if det(A) = 0; i.e., if A is singular, then no unique solution to Ax = b exists
(either there are no solutions or an infinite number of them).
A fuller discussion of matrix determinants is provided in the supplemental material in
the accompanying website, but here we summarize the main results. The determinant of a
matrix is computed from its matrix elements by the formula

N N N

a
det(A) = ... ε i 1 ,i 2 ,...,i N i 1 ,1 a i 2 ,2 ··· a i N ,N (1.168)
i 1 =1 i 2 =1 i N =1
takes the values
a i k ,k is the element of A in row i k , column k and ε i 1 ,i 2 ,...,i N

0, if any two of {i 1 , i 2 ,..., i N } are equal
= 1, if (i 1 , i 2 ,..., i N ) is an even permuation of (1, 2,..., N) (1.169)
ε i 1 ,i 2 ,...,i N

−1, if (i 1 , i 2 ,..., i N ) is an odd permuation of (1, 2,..., N)
By an even or odd permutation of (1, 2, . . . , N) we mean the following: if no two of
{i 1 , i 2 ,..., i N } are equal, then we can rearrange the ordered set ( i 1 , i 2 ,..., i N ) into the
ordered set (1, 2,..., N) by some sequence of exchanges. Consider the set (i 1 , i 2 ,..., i N ) =
(3, 2, 4, 1) which we can reorder into (1, 2, 3, 4) by any of the following sequences, where
at each step we underline the two members that are exchanged,

(3, 2, 4, 1) → (3, 2, 1, 4) → (3, 1, 2, 4) → (1, 3, 2, 4) → (1, 2, 3, 4)
(3, 2, 4, 1) → (1, 2, 4, 3) → (1, 2, 3, 4)
(3, 2, 4, 1) → (4, 2, 3, 1) → (2, 4, 3, 1) → (1, 4, 3, 2) → (1, 4, 2, 3)
→ (1, 4, 3, 2) → (1, 2, 3, 4)

Some sequences reorder (3, 2, 4, 1) to (1, 2, 3, 4) in fewer steps than others, but for (3, 2, 4,
1), any possible sequence of simple exchanges that yield (1, 2, 3, 4) must comprise an even
number of steps; (3, 2, 4, 1) is therefore an even permutation. Similarly, because (3, 2, 1, 4)
can be reordered by the three exchanges

(3, 2, 1, 4) → (3, 1, 2, 4) → (1, 3, 2, 4) → (1, 2, 3, 4)

it is an odd permutation. The even or odd nature of a permutation is called its parity.
We can determine whether an ordered set ( i 1 , i 2 ,..., i N ) has even or odd parity
by the following procedure: for each m = 1, 2,..., N, let α m be the number of inte-
gers in the set {i m+1 , i m+2 ,..., i N } that are smaller than i m . We sum these α m to
obtain
N

v = α m (1.170)
m=1

If v = 0, 2, 4, 6,..., (i 1 , i 2 ,..., i N ) is even. If v = 1, 3, 5, 7,..., it is odd.