Can determinants be calculated using tensor notation?

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The discussion centers on calculating the determinant of the product of two 3x3 matrices A and B using tensor notation. The key equation established is that det(C) = det(A) * det(B), where C = AB. The user initially struggles with the connection between the determinant expressions in indicial notation and the standard determinant multiplication rule. They explore the implications of invertibility and the use of elementary matrices to demonstrate the relationship. Ultimately, the user finds clarity in the application of tensor notation to prove the determinant property.
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[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"

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:cry:
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...
 
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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.
 
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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!:smile:
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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