# Tensor notation-Determinants

[SOLVED] Tensor notation-Determinants

I'm trying to learn the basics of Tensor calculus using a free online book (Introduction to Tensor Calculus and Continuum Mechanics), and I got stuck on this question (Part 2 in book, after non-math introduction).

link to part 2, questions (mine #19) at end:
http://www.math.odu.edu/~jhh/part2.PDF" [Broken]

## Homework Statement

Let A and B denote 3x3 matrices with elements $$A_{ij}$$ and $$B_{ij}$$
respectively. Show that if C = AB is a matrix product, then det(C) = det(A)*det(B)
where det = determinant.

Hint: use the result from example 1.1-9

## Homework Equations

det(A) = $$e_{ijk}A_{1i}A_{2j}A_{3k}$$

## The Attempt at a Solution

The matrix multiplication of A*B =C, in indical notation, is
$$C_{ij}=A_{im}B_{mj}$$ (I think) where the first subscript in A,B and C is the row number and the second subscript the column of the matrix.

Then, plugging into 'relevant equation' above, we get
det(C) = $$e_{ijk}C_{1i}C_{2j}C_{3k}$$
det(C) = $$e_{ijk}(A_{1m}B_{mi})(A_{2n}B_{nj})(A_{3x}B_{xk})$$

Then, I compare this to just multiplying det(A)*det(B)
det(A) * det(B) = $$(e_{ijk}A_{1i}A_{2j}A_{3k})(e_{rst}B_{1r}B_{2s}B_{3t})$$

However, from here, I can't seem to make the connection between the two I would try to expand it, but I don't think this is the way to do it, since it would get rid of the advantage of indical notation. Help would be appreciated! Thxs!

PS: on a side note, whenever you preview a post, the template is pasted again to the end of the post, which is annoying. Maybe someone can fix this...

Last edited by a moderator:

I believe you need to prove this in two cases. If B is not invertible, then det(B)=0. There is a non-trivial solution. Thus AB is not invertible.

Else, suppose A and B are invertible/nonsingular, then there are elementary matrices, A=Ej, E_j-1,.... E_1
det(A)=det(E_j)*det(E_j-1)....*(E_1)

Now, det(AB)= det(E_j)*det(E_j-1)...(E_1) * det(B).
Then A and B can be expressed as a product of elementary matrices.

Edit:Since C=AB, then det(C)=det(AB). Hope that helps.

Last edited:
Hurkyl
Staff Emeritus
Gold Member
Maybe I'm reading the intent incorrectly, but I think the problem specifically wants him to prove it using indicial notation.

Oh sorry.

Let C=AB. By indicial notation, C_ij=A_ik B_kj

Since A is 3x3 matrix, then det(A)= A_11(A_22...) - A_12(A21..) + A_13(A_21...) by method of solving for determinants where n>2
Then det(A)= epsilon_ijk A_i1 A_j2 A_k3 = epsilon_ijk A_1i A_2j A_3k
Therefore, epsilon_rst det(A) = epsilon_ijk A_ir A_js A_kt

We know that det(C)=det(AB)=epsilon_ijk C_i1 C_j2 C_k3
=...= epsilon_rst B_r1 B_s2 B_t3 det(A)

I can't believe i didn't see it! Thanks guys! 