Prove Determinant Using the Triple Scalar Product

In summary: Can you take it from there?In summary, the determinant of a matrix A can be expressed as the scalar triple product of its row vectors. Using the property of antisymmetry of the Levi-Civita symbol, we can rearrange the terms to obtain the desired form for proving det A using the triple scalar product. This approach utilizes the geometrical interpretation of both the scalar triple product and the determinant as representing volumes.
  • #1
ThirdEyeBlind
12
0

Homework Statement


I'm supposed to prove [itex]det A = \frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{ip} A_{jq} A_{kr} [/itex] using the triple scalar product.

Homework Equations


[itex] \frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{ip} A_{jq} A_{ kr} [/itex]
[itex] (\vec u \times \vec v) \cdot \vec w = u_i v_j w_k \epsilon_{ijk} [/itex]

The Attempt at a Solution


I have written out and understand that [itex]det A = \epsilon_{ijk} A_{1i} A_{2j} A_{3k} [/itex] but I don't understand where the triple scalar product comes into play.

I see the similarity between the shorthand notations of the triple scalar product and the determinant but don't see how I can relate the two. I figure I must be missing some geometric/mathematical interpretation that can help me.

I've used Einstein summation notation in a couple previous problems but this is really the first involved problem I have to do and am a bit lost as to what to do.
 
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  • #2
ThirdEyeBlind said:

Homework Statement


I'm supposed to prove [itex]det A = \frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{ip} A_{jq} A_{kr} [/itex] using the triple scalar product.

Homework Equations


[itex] \frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{ip} A_{jq} A_{ kr} [/itex]
[itex] (\vec u \times \vec v) \cdot \vec w = u_i v_j w_k \epsilon_{ijk} [/itex]

The Attempt at a Solution


I have written out and understand that [itex]det A = \epsilon_{ijk} A_{1i} A_{2j} A_{3k} [/itex] but I don't understand where the triple scalar product comes into play.

I see the similarity between the shorthand notations of the triple scalar product and the determinant but don't see how I can relate the two. I figure I must be missing some geometric/mathematical interpretation that can help me.

I've used Einstein summation notation in a couple previous problems but this is really the first involved problem I have to do and am a bit lost as to what to do.

Both the scalar triple product and the determinant (in 3 dimensions) have the geometrical interpretation of volume. ##(\vec u \times \vec v) \cdot \vec w## is the volume of the prism formed by vectors u, v, and w. They are probably expecting you to make the identification

$$ \det A \quad = (\vec{A}_1 \times \vec{A}_2) \cdot \vec{A}_3 ,$$

where, for example, ##\vec{A}_1## is the first row in matrix A. The determinant of A is the volume spanned by its three row vectors.

That gets you up to the point where you are, namely ##\det A = \epsilon_{ijk} A_{1i} A_{2j} A_{3k}##. To get the rest of it, note that

$$\det A = \epsilon_{ijk} A_{1i} A_{2j} A_{3k} = -\epsilon_{ijk} A_{2i} A_{1j} A_{3k} = \epsilon_{213}\epsilon_{ijk} A_{2i} A_{1j} A_{3k}$$

for example.
 

1. What is the triple scalar product?

The triple scalar product, also known as the scalar triple product or box product, is a mathematical operation that takes three vectors and produces a scalar quantity. It is used to calculate the volume of a parallelepiped formed by the three vectors.

2. How is the triple scalar product used to prove determinant?

The triple scalar product can be used to prove the determinant of a 3x3 matrix by taking the three columns of the matrix as vectors and calculating their triple scalar product. If the result is equal to the determinant of the matrix, then it can be considered as a valid proof.

3. What is the purpose of using the triple scalar product to prove determinant?

The triple scalar product provides a geometric interpretation of the determinant of a matrix. It allows us to visualize the volume of the parallelepiped formed by the three vectors and understand how the determinant is related to the scaling factor of the transformation represented by the matrix.

4. Can the triple scalar product be used to prove determinant for matrices of any size?

No, the triple scalar product can only be used to prove determinant for 3x3 matrices. For matrices of higher dimensions, other methods such as cofactor expansion or row reduction must be used.

5. Are there any limitations to using the triple scalar product to prove determinant?

One limitation is that the triple scalar product only works for square matrices. Additionally, it can only be used to prove determinant for 3x3 matrices that have a non-zero volume. If the three vectors are coplanar, the triple scalar product will result in a value of zero, making it an invalid proof.

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