Biomedical Engineering Reference
In-Depth Information
The first of these relations is obtained by setting the indices p and k equal in
(A.110) and then expanding the determinant. The second is obtained from the first
by setting the indices n and j equal in first. The third is obtained from the second by
setting the indices i and m equal in second.
Example A.8.1
Derive the first of (A.111) from (A.110).
Solution: The first of (A.111) is obtained from (A.110) by setting the indices p and k
equal in (A.110) and then expanding the determinant:
d im
d in
d ik
e ijk e mnk ¼
d jm
d jn
d jk
;
d km
d kn
3
one finds that
e ijk e mnk ¼
3
d im d jn d im d jk d kn
3
d in d jm þ d in d km d jk þ d ik d jm d kn d ik d km d jn :
Carrying out the indicated summation over the index k in the expression above,
e ijk e mnk ¼
3
d im d jn d im d jn
3
d in d jm þ d in d jm þ d in d jm d im d jn :
This is the desired result, the first of (A.111).
Example A.8.2
Prove that Det( AB ) ¼ Det A Det B .
Solution: Replacing A in (A.107) by C and selecting the values of mnp to be 123,
then (A.107) becomes
Det C
¼
e ijk C i1 C j2 C k3 ¼
e ijk C 1i C 2j C 3k :
Now C is replaced by the product A
B using
C i1 ¼
A im B m1 ;
C j2 ¼
A jn B n2 ;
C k3 ¼
A kp B p3 ;
thus
Det A
B
¼
e ijk A im B m1 A jn B n2 A kp B p3 ;
or Det A
B
¼ð
e ijk A im A jn A kp Þ
B m1 B n2 B p3 ;
where the order of the terms in the second sum has been rearranged from the first.
Comparison of the first four rearranged terms from the second sum with the right-
hand side of (A.107) shows that the first four terms in the sum on the right may
be replaced by e mnp Det A ; thus applying the first equation of this solution again with
C replaced by B , the desired result is obtained:
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