Why are determinants in 2x2 matrices and 3x3 matrices computed the way they are?

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Hi,

I'd like to have a little insight about why the determinants of ℝ2x2 and ℝ3x3 matrices are computed that way.

I know how to calculate said determinants in both the cases and I also know what's the meaning behind it thanks to "3blue1brown"'s youtube channel, which states that they are a scaling factor respectively to the area of a surface in ℝ2 and of a volume of a parallelepiped in ℝ3, but what's the connection between, as in the simpler case, the scaling factor of an area in ℝ2 and it being computed as ab - cd ? (Say the matrix is [a c; b d] meaning "a c" is the first row).
What's the geometric intuition or derivation?

I thank everyone who'll partecipate in this thread in advance.
 

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  • #2
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So, I found my answer. My bad for posting this thread without double checking or continuing said youtube series on the essence of linear algebra.

I don't know how to behave in this case, should I keep the thread or should I (or who has the power) delete it?

In the meanwhile, I'm going to attach an image which answers my question in the ℝ2x2's matrices cases.
The ℝ3x3's and general non square matrices remains to be answered, though.
 

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  • #3
Office_Shredder
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I think the general proof is actually a little bit tricky, and typically goes like
1.) There is only one function which satisfies standard determinant properties (adding one row to another doesn't change the answer, scaling a row scales the determinant etc).

2.) The determinant formula satisfies these properties

3.) Computing volume satisfies these properties.

Therefore they are the same function.

It feels like there should be an inductive argument that makes more sense to prove these are the same, but I haven't quite been able to put it together.
 
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Your thread and posts are fine. Its an interesting question and if you didnt ask someone else would have.

The 3blue1brown youtube channel is an excellent source of math insight and its always good to see that someone has discovered it too.
 
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What's the geometric intuition or derivation?
It's not geometric.

The definition of a determinant is the sum of the signed elementary products (which you can look up if you never heard this before.

For a 2x2 the ab-cd method evaluates by the definition:

For a 3x3, this "shortcut" evaluates the determinant by definition:

1632930300392.png

Geometric interpretations are the consequence of the definition, not the other way.
 
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The article even has a pedagogy section suggesting best practices in teaching the concepts. One suggestion was to NOT teach them at the same time because students will tend to conflate and confuse themselves.
 
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Possibly interesting reading
https://www.maa.org/book/export/html/115944 ( The Geometry of Dot and Cross Products - Tevian Dray and Corinne A. Manogue )
THIS IS PURE GOLD. I can't express my gratitude properly, this was what I was looking for and without doubt what I usually look for when I learn anything that is given as analytical definition. I struggle a lot imagining the geometric consequences of the definitions usually.
 
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It's not geometric.

(...)

Geometric interpretations are the consequence of the definition, not the other way.
I mean, I'm definitely not the person who can say anything about it, but I always imagined this kind of math to be invented (where "invented" could be intended as "defined") to describe a particular real life system or procedure, so it felt logical to me that a geometric truth was interpreted by math.

For instance, the cross product between two vector has the geometric representation of an area and I always thought that it was defined that way because it does indeed output in a comprehensible way an area.


I'm not trying to be stubborn, I find this "point of view" you're proposing me fascinating but counterintuitive.
 
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I don't know what else to say. The definition is the definition.
 
  • #11
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For this part of my linear algebra course last year, a matrix was considered as a collection of column vectors ##\mathbf{A} = (\boldsymbol{c}_1 \ \dots \ \boldsymbol{c}_n)##, and the determinant was introduced as a linear map (say, ##\phi##) of these ##n## vectors into the real numbers which satisfied the rules for column operations. Geometrically, ##|\mathbf{A}|## is the ##n##-volume of the parallelepiped with sides ##\boldsymbol{c}_i##.

From that was derived the familiar Laplace expansion, i.e. using the linearity:\begin{align*}
|\mathbf{A}| = \phi(\boldsymbol{c}_1, \dots, \boldsymbol{c}_n) = \displaystyle{\sum_{\alpha, \beta, \dots, \xi}} (\boldsymbol{c}_1)_{\alpha} (\boldsymbol{c}_2)_{\beta} \dots (\boldsymbol{c}_n)_{\xi} \phi(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}, \dots, \boldsymbol{e}_{\xi})
\end{align*}then using the property that the determinant reverses sign under column interchange, if all of ##\alpha, \beta## etc. are distinct then ##\phi(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}, \dots, \boldsymbol{e}_{\xi}) = (-1)^p \phi(\boldsymbol{e}_1, \boldsymbol{e}_2, \dots, \boldsymbol{e}_3)## where ##p## is the number of swaps to take the permutation ##(1,\dots, n)## to the permutation ##(\alpha, \dots, \xi)##. Whilst if two or more of ##\alpha, \beta## etc. are the same, the term is zero. You can also express this with the antisymmetric symbol ##\epsilon##,\begin{align*}
|\mathbf{A}| = \displaystyle{\sum_{\alpha, \beta, \dots, \xi}} (\boldsymbol{c}_1)_{\alpha} (\boldsymbol{c}_2)_{\beta} \dots (\boldsymbol{c}_n)_{\xi} \epsilon_{\alpha \beta \dots \xi} \phi(\boldsymbol{e}_1, \boldsymbol{e}_2, \dots, \boldsymbol{e}_3)
\end{align*}Choosing a normalisation such that ##\phi(\boldsymbol{e}_1, \boldsymbol{e}_2, \dots, \boldsymbol{e}_3) = 1## gives the familiar formula:\begin{align*}
|\mathbf{A}| = \displaystyle{\sum_{\alpha, \beta, \dots, \xi}} \epsilon_{\alpha \beta \dots \xi} (\boldsymbol{c}_1)_{\alpha} (\boldsymbol{c}_2)_{\beta} \dots (\boldsymbol{c}_n)_{\xi}
\end{align*}There are probably some finer points that I missed from my notes, but I think that captures the gist of it.
 
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  • #12
PeroK
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I mean, I'm definitely not the person who can say anything about it, but I always imagined this kind of math to be invented (where "invented" could be intended as "defined") to describe a particular real life system or procedure, so it felt logical to me that a geometric truth was interpreted by math.
The 2x2 and 3x3 determinants come out naturally if you crank out the inverse matrices. It's not too hard to find that:
$$\begin{bmatrix}
A& B\\
C&D
\end{bmatrix}^{-1} =
\frac 1{AD-BC} \begin{bmatrix}
D& -B\\
-C&A
\end{bmatrix}
$$
The 3x3 case takes a bit more work, but it comes out fairly easily. That gets you started with a potential inductive rule for determinants and generating inverse matrices, using the determinant and signed cofactors.
 
  • #13
robphy
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I mean, I'm definitely not the person who can say anything about it, but I always imagined this kind of math to be invented (where "invented" could be intended as "defined") to describe a particular real life system or procedure, so it felt logical to me that a geometric truth was interpreted by math.

For instance, the cross product between two vector has the geometric representation of an area and I always thought that it was defined that way because it does indeed output in a comprehensible way an area.

I'm not trying to be stubborn, I find this "point of view" you're proposing me fascinating but counterintuitive.

I don't know what else to say. The definition is the definition.

Sometimes a useful mathematical structure arises in one field
that turns out to be found in another field,
possibly with a different name, different application, different motivation, and different intuition.
And, it may be for historical reasons the ideas were developed (or matured) in one field before another.
So this is likely the case with the determinant.
(However, in the end, it may be that every structure will be "fundamentally" defined using https://en.wikipedia.org/wiki/Category_theory... e,g, https://ncatlab.org/nlab/show/determinant .)
 

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