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Solution:
a. If n = 2, there are only two possibilities for permuting these two
numbers, giving the following: (1, 2) and (2, 1). In the first permu-
tation, no transposition was necessary; that is, the multiplying
factor in Eq. (8.5) is 1. In the second term, one transposition is
needed; that is, the multiplying factor in Eq. (8.5) is –1, giving for
the determinant the value:

∆= MM − M M (8.6)
11 22 12 21
b. If n = 3, there are only six permutations for the sequence (1, 2, 3):
namely, (1, 2, 3), (2, 3, 1), and (3, 1, 2), each of which is an even
permutation and (3, 2, 1), (2, 1, 3), and (1, 3, 2), which are odd
permutations, thereby giving for the determinant the value:

∆= MM M + M M M + M M M
11 22 33 12 23 31 13 21 32
(8.7)
− MM M( + M M M + M M M )
13 22 31 12 21 33 11 23 32
MATLAB Representation
Compute the determinant and the inverse of the matrices M and N, as keyed
below:

M=[1 3 5; 7 11 13; 17 19 23];
detM=det(M)
invM=inv(M)

gives

detM=
-84
invM=
-0.0714 -0.3095 -0.1905
-0.7143 -0.7381 -0.2619
-0.6429 -0.3810 -0.1190

On the other hand, entering:

N=[2 4 6; 3 5 7; 5 9 13];
detN=det(N)
invN=inv(N)

© 2001 by CRC Press LLC