# Why do we still teach the determinant formula for cross product? And is it as bad as I think it is?

The cross product is an important vector operation in in any serious multivariable calculus course. In most textbooks that I'm aware of, right after the definition, we always introduce the determinant formula $$\mathbf{u} \times \mathbf{v} = \det \begin{bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{bmatrix}$$ where $$\mathbf{i}$$, $$\mathbf{j}$$, $$\mathbf{k}$$ are the unit vectors in the $$x$$, $$y$$, and $$z$$ directions (in a right-handed coordinate system).

This is pretty standard, and it is how I learned this. The first time I taught this, however, I had to stop myself in the middle of the lecture, because as I was explaining this to students I realized that this makes no sense. I must have realized the nonsensical nature of this equation long ago, but it was not until I had to explain this to someone else that I realized how bad this is.

That was four years ago, and I have taught this course several times since then. Now I still introduce this equation just because it is in the textbook, but I try to be extra careful not to mislead students.

I see very little value of teaching this equation:

• It makes no sense. ([Update] I regret not stating this explicitly and hence misled many answers. Here I'm saying "it makes no sense" in the same way that $$1+1=0$$ makes no sense. To disprove this claim, simply state this equation as a statement without introducing any new terms or redefine any existing terms in a 2nd year college level multivariable calculus course)
• Most students in my class don't know how to compute 3x3 determinants, so for this to make sense, I have to teaching that, which is not trivial. ([Update] This statement is not completely true. I'm sure a good percentage of students have learned determinants, but when they enter my class, they still won't be able to compute one unless I spend a lot of class time to teach them)
• In order to teach this right, we have to teaching mathematics wrong --- any student who feels comfortable writing down this formula either has no idea what a matrix is or can happily ignore the meanings behind mathematical symbols. In either case, he/she will not deserve a degree in mathematics.
• It only marginally makes computation faster (even that is arguable).

Are my views reasonable? If yes, why do almost every textbook include it?

P.S., of course I understand we could perform some serious mathematical yoga and inject meaning into the determinant formula: E.g. by passing to $$\Lambda^*(\mathbb{R}^3)$$. But I consider that to be more of a rationalization after the fact.

[Update] I guess I should have stated a few things explicitly:

• I'm certainly not questioning the value of formal notations/expressions in undergraduate level classes. I teach Gram-Schmidt determinants in senior level numerical analysis, which is also meaningless. I have no issue with that, because most students at this level had some exposure to mathematical nonsense that are useful, and it is thus not hard to make this distinction. My question is specifically about multivariable calculus in first or second year in college. It is my observation that students in this stage have poor understanding for the distinction between purely formal and actual meaningful expressions (probably nonexistence). Well, either that, or I'm teaching in the wrong university.
• This formal expression clearly works and I guess anyone teaching such a course knows exactly what it means. It is nothing but a more compact way of writing down $$\mathbf{u} \times \mathbf{v} = [(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{i}] \, \mathbf{i} + [(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{j}] \, \mathbf{j} + [(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{k}] \, \mathbf{k}$$ This is just how cofactor expansion works. I don't think we have any question there. The question is about whether or not teach this as the primary way of computation in 1st/2nd year level multivariable course.
• Some comments/answers say this is a useful way to memorize the cross product formula. Sure. But is it much better than the alternatives? What about just write down $$(u_1 \mathbf{i} + u_2 \mathbf{j} + u_3 \mathbf{k}) \times (v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k})$$ and just ask students to multiply them out? There is no additional information to memorize in this case. Or what about the "Xyzzy" mnemonic? That's even faster. So we are certainly not comparing with just the definition here.
• I also have small scale data collected from 3 colleges last year showing the difference in how quickly average students can computing cross product using the formal determinantal formula vs just the naive multiplication technique is basically unmeasurably small. (Hopefully this year we can get proper IRB approval and perform a slightly more scientific experiment)
• This answer of mine may be useful to you: math.stackexchange.com/a/606720/34287 – Steven Gubkin Feb 3 at 1:32
• @StevenGubkin, yes that's my favorite justification of the definition cross product. Using triple product, most students can see, intuitively, the definition of the cross product has to be how it is defined. That's what I have been following for the past 3 years when it comes to the geometric justification. I have no issues there, but it is independent from what this question is about. I'm not questioning whether we can justify the coordinates of $\mathbf{u} \times \mathbf{v}$. I'm questioning whether we should ask students to put vectors into matrices and pretend to compute determinant. – user13395 Feb 3 at 1:59
• It is just a mnemonic device. I think it is very useful and encodes many important facts about the cross-product. It builds a memory location in the brain so that one can easily revisit the issue as one learns more mathematics over the years. I appreciate your point that many students have never computed 3x3 determinants. When teaching vector calculus, I turn this on its head and use this to also teach about 3x3 determinants. – user52817 Feb 3 at 2:35
• @user52817, it's not usually presented as a mnemonic device. We write as an equation, don't we? At least that's the case in the 4 textbooks I have on myself. – user13395 Feb 3 at 4:55
• @ssquid, curious to know what you do when calculating a flux integral and you need something like ${\bf r}_u\times{\bf r_v}$. Do you just have the expanded form of the determinant memorized, and plug into that? – user52817 Feb 3 at 18:04


Let me rephrase it a bit. Choosing a positively oriented orthonormal basis $$\mathbf{e_1},\mathbf{e_2},\mathbf{e_3}$$ (e.g., the standard basis) and expressing $$\u = (u_1,u_2,u_3)$$, $$\v = (v_1,v_2,v_3)$$ in this basis, we define the cross product as $$\mathbf{u} \times \mathbf{v} = \det \begin{bmatrix} u_1 & v_1 & \mathbf{e_1} \\ u_2 & v_2 & \mathbf{e_2} \\ u_3 & v_3 & \mathbf{e_3} \end{bmatrix}.$$ To make this definition rigorous, we should add one more thing - by the above determinant we actually mean the Laplace expansion with respect to the last column. One could think of it as an exercise in the spirit of this 3Blue1Brown video - in mathematics, one is often led to make sense of some nonsensical notions, and this is a good example of this. There's more than one way of formalizing this definition, and all sensible solutions are probably equivalent.

There are certain properties that are easy to see with this definition. For any $$\w \in \R^3$$ the dot product with $$\u \times \v$$ can be computed as \begin{align*} (\u \times \v) \cdot \w & = \left( \det \begin{bmatrix} u_1 & v_1 & \mathbf{e_1} \\ u_2 & v_2 & \mathbf{e_2} \\ u_3 & v_3 & \mathbf{e_3} \end{bmatrix} \right) \cdot \w \\ & = \det \begin{bmatrix} u_1 & v_1 & \mathbf{e_1} \cdot \w \\ u_2 & v_2 & \mathbf{e_2} \cdot \w \\ u_3 & v_3 & \mathbf{e_3} \cdot \w \end{bmatrix} \\ & = \det \begin{bmatrix} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{bmatrix}. \end{align*} In fact, the above can serve as an alternative definition. With the geometric interpretation of determinant, it can be made into a coordinate-free definition (which I was taught as the basic definition at university). Moreover, some other things are evident:

• $$(\u \times \v) \cdot \u = 0$$, $$(\u \times \v) \cdot \v = 0$$
• $$\u \times \v$$ is bilinear in $$\u,\v$$, and also alternating: $$\v \times \u = - \u \times \v$$,
• if $$\u$$ and $$\v$$ are linearly independent, then $$\u \times \v$$ is nonzero, and $$\u$$, $$\v$$, $$\u \times \v$$ form a positively oriented basis.

Another advantage of this definition is that it easily generalizes. There's nothing special in $$\R^3$$ - a similar formula defines the cross product $$\v_1 \times \ldots \times \v_{n-1}$$ of $$n-1$$ vectors in an $$n$$-dimensional space (with dot product), with the same basic properties. The easiest example of this is in $$\R^2$$: the cross product of $$\u = (u_1,u_2)$$ is $$\det \begin{bmatrix} u_1 & \mathbf{e_1} \\ u_2 & \mathbf{e_2} \end{bmatrix} = (-u_2,u_1).$$ This is exactly the vector $$\u$$ rotated by $$90^\circ$$, as one may expect.

• What is this: $\det \begin{bmatrix} u_1 & v_1 & \mathbf{e_1} \\ u_2 & v_2 & \mathbf{e_2} \\ u_3 & v_3 & \mathbf{e_3} \end{bmatrix} \cdot \mathbf{w} = \det \begin{bmatrix} u_1 & v_1 & \mathbf{e_1} \cdot \mathbf{w} \\ u_2 & v_2 & \mathbf{e_2} \cdot \mathbf{w} \\ u_3 & v_3 & \mathbf{e_3} \cdot \mathbf{w} \end{bmatrix}$ – user13395 Feb 3 at 23:27
• The first expression uses the "cheating determinant" from the definition, which is then multiplied (by dot product) with $\mathbf{w}$. I guess I should have put this determinant in paranthesis to avoid possible confusion. The second is already a valid determinant, with numbers in the last column. To see they're equal, one needs to use one's favorite formalization of the "cheating determinant". – Michał Miśkiewicz Feb 3 at 23:31
• @MichałMiśkiewicz, I read it correctly the first time then. See that's exactly my problem with this kind of thing. To make sense a nonsensical thing, you have to accept even more nonsensical things. Don't be surprised that the next day students will write $\det \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} = \det \begin{bmatrix} 1 & 0 & 0 \cdot 2 \\ 0 & 1 & 0 \cdot 3 \\ 0 & 0 & 1 \cdot 4 \end{bmatrix} = \det \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{bmatrix} = 4$. – user13395 Feb 4 at 1:35
• Wow. After reading OP's question, I was actually in the "yeah it's non-sensical; tensor calculus is how you do it"-camp, but you drew me back in. Your 2D example is pretty nice as well. – freddieknets Feb 4 at 20:43
• Taking this a little further, $u\times v = *(u \wedge v)$, and the correct generalization to any dimension and number of arguments is the wedge product. – pavel Feb 5 at 10:53

I think you're correct, strictly speaking, but overreacting. I just checked two calculus books, Rogawski and Strang. Rogawski defines the cross product as the "symbolic" determinant (his words) in question but then defines exactly what he means by this determinant in the same line, so I don't see a problem there. Strang mentions the determinant mnemonic later in his presentation and claims that it's "probably illegal...but it works". Again, that seems reasonable to me (and humorous, as Strang often is).

I figure that anyone who hasn't seen 3x3 determinants before won't know enough to even be confused in the first place, while anyone who has can easily see that it's just a formal determinant that spits out whatever "official" definition they've already seen.

Sure, give up on the mixed-object determinant formula. Instead use tensor arithmetic. Let $$\epsilon_{ijk}$$ be the completely antisymmetric symbol. Also, while we're at it, let's give up on quaternionic notation for unit-vectors and instead use $$\hat{x},\hat{y},\hat{z}$$ or $$\hat{x_i}$$ for $$i=1,2,3$$ then you have an elegant formula: $$\vec{A} \times \vec{B} = \sum_{i,j,k} \epsilon_{ijk} A_iB_j \hat{x_k}$$ Moreover, you then have a completely legit platform to share the formula for the determinant: $$det(A) = \sum_{i_1,i_2,\dots, i_n} \epsilon_{i_1i_2\dots i_n}A_{1i_1}A_{2i_2}\cdots A_{ni_n}$$ I wish this was in standard textbooks. It would have been helpful to me at a certain point in my mathematical youth. Also, the apparent definition of the determinant in terms of Laplace's expansion by minors is truly horrible. Of course, eventually, if we think longer on this we will be driven to wedge products.

On the other hand, the hypocritical formula can be naturally interpreted to simply mean the formula I write here with the dreaded $$\epsilon_{ijk}$$-type notation. Furthermore, the determinant trick extends to higher dimensional orthogonal complements: $$x=\sum_{i_1,\dots, i_n} \epsilon_{i_1i_2\dots i_n}(v_1)_{i_1}(v_2)_{i_2} \cdots (v_{n-1})_{i_{n-1}}\hat{x}_{i_n} \in \{ v_1,v_2, \dots , v_{n-1} \}^{\perp}$$ and $$\| x \| = 1$$. Most students would know what you meant if you expressed the above as (for the sake of brevity I'll just do $$n=4$$) $$\text{det} \left[ \begin{array}{c} \hat{x}_1 \ \hat{x}_2 \ \hat{x}_3 \ \hat{x}_4 \\ \vec{A} \\ \vec{B} \\ \vec{C} \end{array}\right]$$ where $$\hat{x}_1 = \langle 1,0,0,0 \rangle$$ and $$\vec{A},\vec{B},\vec{C} \in \mathbb{R}^4$$. For example, $$\vec{A} = \langle 0,1,0,0 \rangle$$, $$\vec{B} = \langle 0,0,1,0,\rangle$$ and $$\vec{C} = \langle 0,0,0,1 \rangle$$ gives us: $$\text{det} \left[ \begin{array}{cccc} \hat{x}_1 & \hat{x}_2 & \hat{x}_3 & \hat{x}_4 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array}\right] = \hat{x}_1.$$

• Has there been any success in teaching tensors in multivariable calculus (a second year college level course)? – user13395 Feb 3 at 14:40
• @ssquidd Beauty is in the eye of the beholder. – James S. Cook Feb 3 at 18:46
• I enjoyed reading this answer and upvoted. It reminds me of another approach that is also "not nonsensical." Identify the ${\bf R}^3$ of undergraduate calculus with the Lie algebra $so(3)$. The cross product is then the Lie bracket. We can use 3x3 skew-symmetric matrices to represent $so(3)$. The cross product then corresponds to the commutator of matrices. – user52817 Feb 3 at 22:03
• I think you should add short definitions to what antisymmeteric symbol and quarternion notation is to keep the answer self contained – Buraian Feb 4 at 9:24
• I think the physics department would be very happy to have you teach tensors early! I remember one of my physics professors claiming that Einstein deserved a Nobel just for his summation convention, even if he'd never invented anything else. That's wild exaggeration for rhetorical effect, but frankly it's much more natural to teach electromagnetism in the language of differential forms, where thanks to Hodge one can write all four of Maxwell's equations as $d*F = 4\pi * J$. – Ryan C Feb 4 at 10:39

I had a lecturer who thought the same as you, so suggested we just learnt the definition of the cross product (which he derived along the same lines as James S Cook's answer). Nobody liked his lectures because he pitched them way to high for our first-year brains.

The point is, it's a very useful mnemonic device, and when you teach it make that clear, but make sure you teach it, because nobody will like you if you don't.

• Kind of feels like you are threatening OP, not sure if that is what we want to do. I do agree that it's a very usefull mnemonic device. – user15043 Feb 5 at 20:11

this makes no sense

But it does. A determinant exists for any $$n\times n$$ matrix for which the Leibniz formula$$\det A:=\sum_{\sigma\in S_n}\epsilon_\sigma\prod_{i=1}^nA_{i\sigma_i}$$is fully antisymmetric. This is why, for example, quaternionic matrices lack determinants: their elements don't commute. But if we apply it to the case at hand, we don't have that problem (as long as the entries of $$u,\,v$$ are C numbers), because vectors commute with scalars. The Gram-Schmidt determinants you've also taught admit the same defence.

If you object to this, perhaps you have a narrower definition of when determinants are defined. From such a perspective, my definition is not "meaningless", it's a different meaning. But we typically go with the most general workable definition in mathematics, as it pays dividends. Nor is being comfortable writing it evidence of misunderstanding or ignoring certain things. At worst, it's an abuse of notation, which is not an insult; it's a reference to utility.

It only marginally makes computation faster... But is it much better than the alternatives [for memorization]?

The main reason to learn this expression is to understand it, not to compute with it or have a mnemonic. Its geometric significance has been discussed in multiple answers already, so I won't repeat those points. What I'd add is it facilitates a sanity check of such details as the cross product's antisymmetry, the scalar triple product's full antisymmetry, and each being multilinear.

• So what does "matrix" mean? – user13395 Feb 5 at 17:09
• @ssquidd In this case, a function in $T^S$ with $S:=\{1,\,2,\,3\}^2,\,T:=\Bbb R\cup\Bbb R^3$, endowed with the usual matrix arithmetic rules; whether the determinant is defined depends on where vectors occur. – J.G. Feb 5 at 17:13
• Comments are not for extended discussion; this conversation has been moved to chat. – Chris Cunningham Feb 6 at 6:41

I agree with the OP: the determinant is only a mnemonic device, but (a) students may not know how to evaluate a determinant, so it's useless as a mnemonic, and (b) it is not a "legal" determinant and so mathematicallly misleading.

I would rather start with $$|a \times b| = |a| |b| \sin \theta$$ as a contrast to $$a \cdot b = |a| |b| \cos \theta$$, and derive the coordinate formula (if you feel a derivation is needed) by expanding $$a$$ to $$a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$$ and similarly for $$b$$. Writing out the terms is a bit of a mess, but after simplifications such as $$\hat{i} \times \hat{i} = 0$$ and $$\hat{i} \times \hat{j} = \hat{k}$$, one eventually reaches the coordinate formula. That this requires distributivity and anticommutativity is a opportunity to explain those properties.

This derivation could be a homework assignment rather than part of the lecture.

• By $a \times b = |a| |b| \sin \theta$ you probably meant $|a \times b| = |a| |b| \sin \theta$? To make it a full characterization, one also needs to define the direction of $a \times b$ (orthogonal to both $a$ and $b$) and the sign (chosen so that $a$, $b$, $a \times b$ is positively oriented). These surely pop up in the derivation you mentioned. – Michał Miśkiewicz Feb 4 at 11:25
• I would say that "students may not know how to evaluate a determinant, so it's useless as a mnemonic" is too strong a statement. Usually students in a specific class will either all presumably know how to evaluate the determinant because it's a prerequisite or not. If we couldn't assume prior knowledge in any our teaching because some students don't have it (despite being a pre-req) we'd have to stop teaching any university mathematics.. – user2705196 Feb 4 at 21:43
• @user2705196: You have a point. But I relied on the OP's: "Most students in my class don't know how to compute 3x3 determinants." – Joseph O'Rourke Feb 6 at 0:49

As a preliminary, it's important to realize the incredibly low level of most students who are learning this stuff. Low means a low intellectual level and previous exposure to low-intellectual-level instruction, with low expectations for intellectual understanding. Low means that students who have previously been exposed to $$\hat{\textbf{i}}$$, $$\hat{\textbf{j}}$$, and $$\hat{\textbf{k}}$$ find it an insuperable barrier to adapt to $$\hat{\textbf{x}}$$, $$\hat{\textbf{y}}$$, and $$\hat{\textbf{z}}$$. Low means that after encountering the dot and cross products twice (in first-semester physics and multivariable calculus), they still can't tell you under what circumstances the dot product is zero, or under what circumstances the cross product is zero.

Yes, the presentation using a determinant is awful and should be nuked from orbit.

A good way to go is the following. First we introduce the dot product and build intuition for it through interesting applications, such as computing work, or finding the angle between two vectors. Then we point out that the dot product is rotationally invariant, so that although we may compute it using components that depend on our choice of a coordinate system, the result doesn't actually depend on how we rotate the axes. This is easy to see because the dot product can be defined as $$\textbf{A}\cdot\textbf{B}=|\textbf{A}||\textbf{B}|\cos\theta$$, where $$\theta$$ is the angle between the vectors.

Next we provide some motivation for why we would want to multiply two vectors to get a vector. In a physics class, this could come in the context of angular momentum. We would like this definition of multiplication to be rotationally invariant, like the dot product. Based on these considerations, we clearly must have $$\hat{\textbf{x}}\times\hat{\textbf{x}}=0$$, because otherwise there is no rotationally invariant way to decide what direction to pick for the result. By similar arguments, we determine that $$\hat{\textbf{x}}\times\hat{\textbf{y}}$$ must lie along the $$z$$ axis. We fix it to be $$\hat{\textbf{z}}$$, which is arbitrary up to a nonzero multiplicative constant. Choosing the sign of this constant is arbitrary, and amounts to choosing a right-hand rule. We practice the right-hand rule on some examples.

Finally, we build up a 3x3 multiplication table for cross products of the unit vectors in a particular xyz coordinate system. Any cross product we want to evaluate, given the two vectors in component form, can then be evaluated by using the distributive property (bilinearity).

• Yeah, what I have been following in the last two years of teaching this roughly follow this scheme, but not as detailed. What you have here has a nice progression of conceptual jumps. It should be very reasonable for students at this level. One thing not clear is why does rotational invariance forces $\hat{\mathbf{x}} \times \hat{\mathbf{x}} = \mathbf{0}$? – user13395 Feb 4 at 1:48
• @ssquidd Take two vectors $u$, $v$ of the same length, denote $w := u \times v$, and denote by $\ell$ the bisector of the angle between $u$ and $v$. Rotating $u,v,w$ around $\ell$ by $180^\circ$, we get new vectors $u, v', w'$. Note that $u'=v$, $v'=u$ and $w'=-w$, so if we expect $u' \times v' = w'$ (i.e., rotational invariance), then $v \times u = - u \times v$. Applying this to $u=v$, we obtain $u \times u = 0$. – Michał Miśkiewicz Feb 4 at 11:32
• @MichałMiśkiewicz, why don't you just use anti-commutativity $\mathbf{v} \times \mathbf{v} = - \mathbf{v} \times \mathbf{v}$ to convince students that it has to be $\mathbf{0}$? This is your last step anyway. – user13395 Feb 4 at 21:56
• @ssquidd Perhaps because first you need to motivate why anticommutativity is a reasonable property? The smarter students will probably wonder, “Isn’t there a more straightforward operation with nicer properties?” and you should prove there isn’t, not without breaking more essential properties. – Roman Odaisky Feb 6 at 20:39

Just go with the Cartesian formula.

The first time I saw a cross product was in computer code and I always felt like the typical way it's coded is the easiest way to compute $$\mathbf{c} = \textbf{a} \times \mathbf{b}$$:

$$c_x = a_y b_z - a_z b_y$$

$$c_y = a_z b_x - a_x b_z$$

$$c_z = a_x b_y - a_y b_x$$

The pattern is obvious. Each component of $$\mathbf{c}$$ is the difference between the products of the other two different-direction components of $$\mathbf{a}$$ and $$\mathbf{b}$$. The only tricky part is which product gets the minus sign. You just have to remember that the component of $$a$$ (or the first multiplicand) in the first term is the next letter in the xyz series from the component of the cross product. $$c_x$$ has $$a_y$$ in the first term. $$c_y$$ has $$a_z$$ in the first term. For $$c_z$$, we hit the end of series and start back at $$a_x$$.

No need to think about matrices or tensors.

You can also pretty easily see the properties of the cross product from this form. It's pretty clear that $$\textbf{a} \times \mathbf{a} = 0$$, since $$a_i a_j = a_j a_i$$. It's also pretty clear that $$\textbf{b} \times \mathbf{a} = -\textbf{a} \times \mathbf{b}$$.

• This is the "Xyzzy" mnemonic then. – user13395 Feb 5 at 21:05

The question of whether formal relationships "make sense" is an old one in mathematics. People asked the question about whether negative numbers "made sense" (how can you start with three pebbles and take away six of them?) or complex numbers (what number when squared gives minus 1?). But we found that putting them in algebraic expressions as if they were sensible numbers proved useful, and so eventually they got accepted.

A matrix may be initially defined as a grid of numbers - integer, real, complex - and that makes sense. We have some intuitive idea of what it means. But formally we can put other objects inside a matrix, and so long as we have some operations analogous to addition and multiplication (i.e. the objects are elements of a ring or something similar), we can reproduce a lot of the matrix theory we've already built up for reals in the new arena. So integers modulo k, or quaternions, or matrices could just as easily be used. If the new entity doesn't have quite the same properties as the reals or integers, not all the theory gets reproduced. Matrices of integers often don't have inverses when the corresponding matrix of reals does. Real and complex matrices have significant differences in the theories that arise, so you do have to be careful with this sort of generalisation. It is important to emphasise to students that this usage isn't standard, and you can't apply any real-number matrix theory you know without going back to basics. But that doesn't make it invalid, and it is frequently the case that what are initially merely formal relationships and parallels can reflect deeper relationships. (As with the progression from natural numbers to integers to rationals to reals to complex numbers...)

I'm guessing that the reason you consider the formula to "make no sense" is that it mixes vectors and scalars in the same algebra, and while the subset of operations needed for a determinant are allowed (addition of vectors, multiplication of a vector by a scalar), there are other operations that are not (multiplication of two vectors, addition of a vector to a scalar). There is in fact a more general algebra where these operations are allowed: Geometric Algebra, (aka Clifford algebra) does allow these operations. It generalises vector algebra to incorporate scalars, vectors, bivectors, pseudovectors, spinors, complex numbers, and quaternions. You can add a vector to a scalar, in the same sort of way as adding a real number to an imaginary one, and the multiplication of vectors in Geometric Algebra incorporates both the dot product and cross product of vector algebra into a single, usually invertible operation.

If instead of defining matrices over the reals, you define them over the multivectors of geometric algebra, the determinant now makes algebraic sense. Whether it helps intuitively is another matter. Geometric Algebra has a far simpler way of expressing cross products: as the dual to the antisymmetric part of the product. In 3D geometric algebra $$ab=(ab+ba)/2+(ab-ba)/2=a\cdot b+a\wedge b=a\cdot b+I(a \times b)$$ where $$I$$ is the unit trivector and acts like an imaginary unit. This dual is only needed to turn the valid-in-all-dimensions bivector $$(ab-ba)/2=a\wedge b$$ (which vector algebra can't cope with) into a pseudovector - something looking a lot like a vector but which has different transformation properties under reflection.

The determinant formula probably reflects some deeper geometric relationship in Geometric Algebra, but I agree that exploring something like that is not suitable for a beginner/introductory class. But as for "not making sense", there are plenty of things in vector algebra that don't make sense. Why can dot products be done in any dimension but cross products only work in three dimensions? (Or non-uniquely and without some of the most useful properties, seven.) Why are there two different products, both of them producing objects of a different type? (Dot product produces scalars, cross product produces pseudovectors.) Why can't you invert vector multiplication, as with real, rational, complex, and matrix multiplication, to get a 'vector division' operation? What is a pseudovector anyway? What do you get if you add a vector to a pseudovector? Why are the products non-associative? [$$(a\cdot b)\cdot c$$ is undefined, $$(a\times b)\times c\neq a\times (b\times c)$$.] You have entities to represent directed length (i.e. vectors) but not directed areas or volumes. You have entities to represent translations but not the other isometries like rotations, reflections, screws, etc. Translation is the limit of rotation as the centre-of-rotation is moved towards infinity, so why aren't they represented by the same type of object? Angular velocity - treated as a vector - can't be integrated directly to yield angular displacement. Angular velocities add like vectors, but angular displacements (i.e. rotations) don't. It's messy and ugly. But as you say, most teachers don't notice, as vector algebra is now the conventional way it has 'always' been taught.

• Completely agree. But that's not what I meant. I regret not stating the context. I'm not saying it "doesn't make sense" as a mathematician. Pretty much everything makes sense in mathematics. A significant percentage of Ph.D. theses are about taking a false statement and create a new context in which it will be true: That includes my own thesis. Here I'm just stating this as an instructor multivariable calculus. I'm saying that equation "makes no sense" in the same way "$1+1 = 0$ makes no sense in a calculus classroom. Sure, it makes sense in a different classroom, but not this one. – user13395 Feb 5 at 19:30
• Exactly. This is not a question about mathematics, it's about teaching mathematics in a very specific classroom setting. That's why I'm asking in matheducator.se not math.se. Should've made that clear in the OP. In numerical analysis (4th year), I have zero issue teaching Gram-Schmidt determinant. It's the same nonsense, but it would be appropriate in students' senior year. – user13395 Feb 5 at 21:17
• OK. Then the question is whether it is better to not present it (because it introduces topics they're not ready for), or to present it but point out that it belongs to a more advanced theory that they haven't covered, and they should treat it with caution. I wouldn't say it 'makes no sense'. I'm generally in favour of making students aware of the limits of their knowledge, it provides a hook into later studies when you revisit it and motivates the best students to explore. But it can also confuse the less able students. That's a trade-off you can only decide based on your class's ability. – Nullius in Verba Feb 5 at 21:28
• Agree with the general point. Say, for the complex roots you mentioned or logarithm of negative numbers, I think any math educator deserving a job probably would not say "they make no sense", because there is something beautiful beckoning from just beyond. Back to this case though, the earliest appearance of the determinantal formula was from Cayley's paper and it is nothing more than a creative and compact way of writing down the coordinate of cross product using triple product. So I'm not sure if it can be put into the same category as complex roots of imaginary logarithm. – user13395 Feb 5 at 21:42
• Geometric Algebra is certainly a beautifiul theory beckoning from just beyond (David Hestenes Oersted lecture on the subject is worth reading), and I'd be very surprised if there wasn't something more elegant and general lurking behind Cayley's formula. Determinants and wedge-products are closely connected. But I'll not argue with what you say in this particular context. – Nullius in Verba Feb 5 at 22:06

In my opinion $$2 \times 2$$ matrices, particularly $$2 \times 2$$ real matrices are fairly easy to understand as mathematical objects, even to folks who haven't been exposed to linear algebra. $$2 \times 2$$ real matrices can be motivated by linear functions in $$\mathbb{R}^2$$, (equivalently) as scaled rotations of the plane that fix the origin, as a convenient way to represent complex numbers, or not motivated at all.

The determinant formula for $$2 \times 2$$ matrices specifically is compact, easy to remember, and can be justified if necessary by its geometric interpretation as the signed area of a parallelogram.

$$\left| \begin{matrix} x & y \\ z & w \end{matrix} \right| = xw - yz$$

I will introduce the nonstandard notion of a cyclic index. If $$x$$ is a vector in $$\mathbb{R}^3$$, then $$x_{(0)}$$ is $$x_{3}$$ and is $$x_{(100)}$$ is $$x_1$$ and $$x_{(2)}$$ is trivially $$x_2$$. The componentwise entries of the cross product can be written as follows.

$$s = a \times b$$

$$s_{(i+0)} = \left| \begin{matrix} a_{(i+1)} & b_{(i+1)} \\ a_{(i+2)} & b_{(i+2)} \end{matrix} \right|$$

By defining the $$i$$th component of the result vector $$s$$ in terms of the next dimension mod $$3$$ ($$i+1$$ symbolically) and the next next dimension mod $$3$$ ($$i+2$$ symbolically), we can remove the need to remember whether the determinant associated with $$\hat{i}$$, $$\hat{j}$$ or $$\hat{k}$$ is supposed to be negated or not.

It surprises me that nobody has brought up the rule of Sarrus yet:

• For $$\mathbf{u} \times \mathbf{v}$$, make the table $$\begin{array}{|ccc|cc} \mathbf{i}& \mathbf{j}& \mathbf{k}& \mathbf{i}& \mathbf{j} \\ u_1 & u_2 & u_3 & u_1 & u_2 \\ v_1 & v_2 & v_3 & v_1 & v_2 \end{array}$$ , then sum up products of diagonals (with signs, as usual).

(If we want Wikipedia to rather link to the cross product, there is also Sarrus's scheme, but what is presented there is different enough to not be immediately recognisable as the same as the determinant rule; I suspect they may have been more similar, but don't care enough to go examine the page history.)

Using the rule of Sarrus has a number of advantages.

1. It is clearly a memnonic device, rather than something pretending to be a formula, so there is no problem with the rigor.

2. It is usable even for students who have not learnt to compute determinants in a previous course (and also for those who did learn that, but have now forgotten the details).

• Depending on what is covered in your multivariate calculus course, you may of course need to do determinants anyway to perform change of variables in multiple integrals, but that is far more likely to get deferred to a more advanced course than the cross product is. Even when you cover change of variables in multiple integrals, it tends to be dominated by the standard cartesian/polar/spherical changes where people anyway look up the factor in a table.
3. For those students who care, you can explain that the reason the textbook instead writes that formula with the determinant is that $$3 \times 3$$ determinants can be calculated using the exact same rule, which have tempted some authors to bend the interpretations so that they could present the cross product as a determinant. Then point out that a determinant which could be applied to a matrix mixing scalar and vector entries must in general also cope with products of two vectors — in particular it must be known how to evaluate such products before one starts using it to evaluate that determinant…

• It's of course not impossible to define such a determinant. The easiest approach is probably to say that the matrix elements are quaternions, as it is perfectly possible to define $$\det\colon \mathbb{H}^{n \times n} \longrightarrow \mathbb{H}$$ (there's even a method for approximating the permanent which makes use of such determinants), but then care is needed in the definition and algebraic manipulation — do you know which of your favourite determinant properties were derived under the assumption that element multiplication is commutative? Probably not. Even something as basic as $$\begin{vmatrix} 0 & \mathbf{i} \\ \mathbf{j} & 0 \end{vmatrix}$$ ends up having a different value if elements are multiplied left-to-right than it does if they're multiplied top-to-bottom.

It might feel as if not using a determinant in the definition weakens the link to the triple product formula

$$\mathbf{u} \mathbin{\pmb{\cdot}} (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}$$

but getting to this from the nonsense determinant formula for the cross product $$\mathbf{v} \times \mathbf{w}$$ still requires doing and undoing cofactor expandsion along the first row, so the proof length remains the same for a definition using the rule of Sarrus.

• Mind you, taking the triple product formula as definition of the cross product provides easy routes not only to getting explicit expressions for the elements of the cross product (just let $$\mathbf{u}$$ range over the vectors in the standard basis), but also for identifying $$\Vert \mathbf{v} \times \mathbf{w} \Vert$$ as the area of the parallelogram spanned by $$\mathbf{v}$$ and $$\mathbf{w}$$ (which in turn leads to the $$\left\Vert \mathbf{v} \right\Vert \left\Vert \mathbf{w} \right\Vert \sin\theta$$ formula for $$\Vert \mathbf{v} \times \mathbf{w} \Vert$$) as a consequence of that determinant being the (signed) volume of the parallelepiped spanned by $$\mathbf{u}$$, $$\mathbf{v}$$, and $$\mathbf{w}$$.

• It's probably also worth pointing out that while this definition caters for the cross product as the $$n=3$$ instance of the $$\mathbb{R}^n$$-valued $$(n{-}1)$$-ary product of vectors in $$\mathbb{R}^n$$, it does not so much cater for the cross product as being the $$n=3$$ exterior product of two $$1$$-forms or bracket in $$\mathfrak{so}(n)$$.

Still, even if being notationally rigorous in the definition of the cross product, teaching a multivariate calculus course probably means that you will also have to address the even larger degree of symbolic interpretation that is needed to write the curl as $$\nabla \times \mathbf{F}$$.