3
$\begingroup$

The determinant of a $3 \times 3$ matrix gives the volume of the parallepiped formed by 3 vectors. With the cross product, say to find the torque, we find the $3 \times 3$ determinant value of displacement vector $\mathbf{r}$ and force $\mathbf{F}$, \begin{align} \boldsymbol{\tau} &= \mathbf{r} \times \mathbf{F} \\ &= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ r_1&r_2&r_3\\ F_1&F_2&F_3\\ \end{vmatrix}, \end{align} and the definition of the cross product yields the area vector.

But the determinant by definition is volume. How is the definition of the determinant changed to find the cross product ?

$\endgroup$
5
  • $\begingroup$ There is a great 3b1b video on determinants. However, this questions seems to be about pure math to me. $\endgroup$ – Jonas Feb 22 at 9:33
  • $\begingroup$ Your equation is incorrect, as it equates a (pseudo)vector to a number. The problem is ill-posed. $\endgroup$ – my2cts Feb 22 at 11:48
  • 2
    $\begingroup$ I’m voting to close this question because the question is ill-posed. Also, determinants are not volumes. Only if their columns or rows are vectors in N dimensional space they are, and then you still have to add on the proper dimension. $\endgroup$ – my2cts Feb 22 at 11:48
  • $\begingroup$ The equation is due to an edit of mine, and is correct as written. If you think it could be improved, by all means improve it! $\endgroup$ – JCW Feb 22 at 12:34
  • 2
    $\begingroup$ Notice that $\bf i$, $\bf j$, $\bf k$ are unit vectors; this may help you to see why something which comes out with physical dimensions (length$^3$) in one case can come out with physical dimensions (length$^2$) in another case. $\endgroup$ – Andrew Steane Feb 22 at 14:05
3
$\begingroup$

When we write a cross-product as a determinant, we are really abusing notation. If you look at the expression, \begin{align} \mathbf{a\times b} &= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ a_1&a_2&a_3\\ b_1&b_2&b_3\\ \end{vmatrix} \\ &= (a_2b_3 - a_3b_2)\mathbf{i} -(a_1b_3 - a_3b_1)\mathbf{j} +(a_1b_2 - a_2b_1)\mathbf{k} \\ &=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}, \end{align} you can see that in each term we have a product of two dimensionful `length' coordinates, giving an interpretation as area.

When we talk about it being numerically the area of a parallogram, we are talking about the norm of this vector, \begin{align} \left\lVert \mathbf{a\times b} \right\rVert &= \sqrt{(a_2b_3 - a_3b_2)^2 + (a_1b_3 - a_3b_1)^2 + (a_1b_2 - a_2b_1)^2}. \end{align} Counting the dimensions is easiest if we just consider \begin{align} &\mathbf{a} = \begin{pmatrix} a_1 \\ 0 \\ 0\end{pmatrix} &\mathbf{b} = \begin{pmatrix} 0 \\ b_2 \\ 0\end{pmatrix}, \end{align} then \begin{align} \left\lVert \mathbf{a\times b} \right\rVert &= \sqrt{(a_1b_2)^2} \\ &= a_1 b_2, \end{align} ie an area.

Why do we understand a determinant as corresponding to volume? Consider instead the scalar triple product, \begin{align} \left(\mathbf{a\times b}\right) \cdot \mathbf{c} &= \left( a_2b_3-a_3b_2 \right) c_1 + \left( a_3b_1-a_1b_3 \right) c_2 + \left( a_1b_2-a_2b_1 \right) c_3 \\ &= \begin{vmatrix} c_1&c_2&c_3\\ a_1&a_2&a_3\\ b_1&b_2&b_3\\ \end{vmatrix}. \end{align}

Then if we count the dimensions, eg by choosing \begin{align} &\mathbf{a} = \begin{pmatrix} a_1 \\ 0 \\ 0\end{pmatrix} &\mathbf{b} &= \begin{pmatrix} 0 \\ b_2 \\ 0\end{pmatrix} &\mathbf{c} = \begin{pmatrix} 0 \\ 0 \\ c_3\end{pmatrix}, \end{align} to simplify the algebra, we get $$ \left(\mathbf{a\times b}\right) \cdot \mathbf{c} = a_1 b_2 c_3, $$ a product of three lengths and hence a volume.

Final comment. Why do we abuse notation to write a cross-product as a determinant? Really, we are interested in the only totally-antisymmetric tensor in three dimensions, called the Levi-Civita, or alternating, symbol $\varepsilon_{ijk}$, defined as $$ \varepsilon_{ijk} = \begin{cases} +1 & \text{if } (i,j,k) \text{ is } (1,2,3), (2,3,1), \text{ or } (3,1,2), \\ -1 & \text{if } (i,j,k) \text{ is } (3,2,1), (1,3,2), \text{ or } (2,1,3), \\ \;\;\,0 & \text{if } i = j, \text{ or } j = k, \text{ or } k = i \end{cases}. $$

Both the cross-product and determinant are properly defined using this, with $$ \mathbf{a}\times\mathbf{b} = \sum_{i,j,k=1}^3 \mathbf{e}_i \;\varepsilon_{ijk} \, a_j b_k $$ and $$ \det A = \begin{vmatrix} A_{11}&A_{12}&A_{13}\\ A_{21}&A_{22}&A_{23}\\ A_{31}&A_{32}&A_{33}\\ \end{vmatrix} = \sum_{i,j,k=1}^3 \varepsilon_{ijk} \, A_{1i} A_{2j} A_{3k} $$ which allows us to use our mnemonic for calculating the latter (an expansion in the minors, the $2\times 2$ submatrices) to remember how to calculate the former.

$\endgroup$
1
  • 2
    $\begingroup$ It is worth noting the cross product operator matrix $$[\boldsymbol{a} \times] \boldsymbol{b} = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix} \pmatrix{b_1 \\ b_2 \\ b_3} $$ $\endgroup$ – JAlex Feb 22 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.