Determinant + indicial notation proof

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To prove that the determinant of a second-order tensor equals det[A] = (1/6) ε_{ijk} ε_{pqr} A_{pi} A_{qj} A_{rk}, one should start by analyzing the 3x3 case and understanding the determinant's definition. The determinant for a 3x3 matrix is expressed as a specific sum involving its elements. A suggested approach is to rewrite the determinant using the Levi-Civita symbol, ε_{ijk}, to relate it to the standard determinant formula. By comparing the standard determinant expansion with expressions involving ε_{ijk}, a clearer connection can be established. This method will help in deriving the desired proof effectively.
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Hello, I am supposed to prove that the determinant of a second order tensor (a matrix) is equal to the following:

det[A] = \frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{pi} A_{qj} A_{rk}

anyone have any idea how i would go about this? any method is welcome
 
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One thing which I would do is look at the 3X3 case amd see if you can see where the formula comes from. You also need to tell us the definition of the determinant that you're using.
 
well by det[A] i mean the following:

det[A_{ij}] = the determinant of a 3 by 3 matrix with the first row being A_{11}, A_{12}, A_{13} and the second row would be A_{21}, A_{22}, A_{23} and lastly the third row is A_{31}, A_{32}, A_{33}
 
How would you write that as a sum?
 
det[A] =
A_{11}(A_{23}A_{32}-A_{22}A_{33}) + A_{12}(A_{21}A_{33}-A_{23}A_{31}) + A_{13}(A_{22}A_{31}-A_{21}A_{32})
 
I am assuming that you're aware of the notation, how can you write the above sum as a sum which includes \varepsilon_{ijk}. A good start would be to write out one of the sums, say for example:
<br /> \varepsilon_{ijk}A_{1i}A_{1j}A_{3k}<br />
and see how this compares to your sum.
 
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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