CONTINUUM MECHANICS for ENGINEERS

2.12 Starting with Eq 2.4-11 of the text in the form
det A = ε ijk A i1A j2 A k3
show that by an arbitrary number of interchanges of columns of A ij
we obtain
ε qmn det A = ε ijkA iqA jmA kn
which is Eq 2.4-12. Further, multiply this equation by the appropriate
permutation symbol to derive the formula,
6 det A = ε qmnε ijkA iqA jmA kn
2.13 Let the determinant of the tensor A ij be given by
det A =
Since the interchange of any two rows or any two columns causes a
sign change in the value of the determinant, show that after an arbitrary
number of row and column interchanges
A A A
A A A
A A A
mq mr ms
nq nr ns
pq pr ps
= ε mnpε qrs det A
Now let A ij ≡ δ ij in the above determinant which results in det A = 1
and, upon expansion, yields
ε mnpε qrs = δ mq(δ nrδ ps – δ nsδ pr) – δ mr (δ nqδ ps – δ nsδ pq) + δ ms(δ nqδ pr – δ nrδ pq)
Thus, by setting p = q, establish Eq 2.2-13 in the form
2.14 Show that the square matrices
⎢ [ bij]= −
A A A
A A A
A A A
11 12 13
21 22 23
31 32 33
ε mnqε qrs = δ mrδ ns – δ msδ nr
⎡1
0 0⎤
⎥
⎢
0 1 0
⎥
⎣
⎢0
0 1 ⎦
⎥
and
⎡
[ cij]= ⎢
− −
are both square roots of the identity matrix.
5 2⎤
⎥
⎣ 12 5⎦