Cross product and determinants

Question

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 ?

Answer 1

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.