Elementary Abstract Algebra- Examples and Applications, 2019a

338 CHAPTER 8 SIGMA NOTATION
Exercise 8.6.4. There is a formula for the determinant of a n × n matrix
in terms of an n-index Levi-Civita symbol. Guess what the formula should
be (you don’t need to prove it).
♦
Based on our definition of the Levi-Civita symbol ɛ ijk in terms of the
sign of the permutation ( 123
ijk)
, we can also write the formula for a 3 × 3
determinant as:
detA =
∑
permutations φ
sign(φ) · a 1φ(1) a 2φ(2) a 3φ(3) .
Exercise 8.6.5. Use this formula to prove that the determinant of any 3×3
square matrix A is equal to the determinant of its transpose. That is,
detA =detA T
(*Hint*)
♦
An important concept to keep in mind when dealing with these Levi-
Civita symbols is what they mean based on when indices are equal or unequal,
and how that relates to permutations. To see how this works, let’s
look at a proof to show that if any two rows in a 3 × 3 matrix are equal, the
determinant is 0. Based on our definition we start out with:
detA = ∑ i,j,k
ɛ ijk a 1i a 2j a 3k
We want to show what happens when any two rows are equal, so let’s do
one case where row 1 equals row 2. In that case a 2j = a 1j . That means we
can rewrite our determinant as:
detA = ∑ i,j,k
ɛ ijk a 1i a 1j a 3k
Now the letters i, j, k are just “dummy indices” or placeholders, so we can
replace them with any letters we want. So we can replace i with j and
vice-versa without changing the value:
detA = ∑ ɛ jik a 1j a 1i a 3k
j,i,k
Now remember what we discussed earlier, if you interchange two indices
(that is, an odd permutation) of ɛ ijk , you get its negative, so ɛ jik = −ɛ ijk .