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Why determinant is area?

Geometric interpretation of determinant in R2\mathbb{R}^2
In R2\mathbb{R}^2, the determinant gives the signed area of the parallelogram spanned by two vectors. In R3\mathbb{R}^3, it gives the signed volume of the parallelepiped spanned by three vectors.

The foolowing lemma is partially from [Hubbard2015] p. 75

Determinant in R2\mathbb{R}^2
The determinant det⁡\det of a 2×22 \times 2 matrix is given by
det[a1b1a2b2]=defa1b2−a2b1\begin{equation*}\text{det}\begin{bmatrix}a_1 & b_1 \\a_2 & b_2\end{bmatrix} \stackrel{\text{def}}{=} a_1b_2 - a_2b_1\end{equation*}
If we think of the determinant as a function of the vectors Aˉ=(a1,a2)\bar{A} = (a_1, a_2) and Bˉ=(b1,b2)\bar{B} = (b_1, b_2) in R2\mathbb{R}^2, it has a geometric interpretation, illustrated by the following video.
Geometric proof that determinant is area of parallelogram in R2\mathbb{R}^2
We can give a simple geometric proof by rewriting sin⁡θ\sin \theta as cos⁡(π2−θ)\cos\left(\frac{\pi}{2}-\theta\right). The area of the parallelogram is
∣Aˉ∣ ∣Bˉ∣sin⁡θ=∣Aˉ∣ ∣Bˉ∣cos⁡(π2−θ).\begin{equation*}|\bar{A}|\,|\bar{B}| \sin \theta = |\bar{A}|\,|\bar{B}| \cos\left(\frac{\pi}{2}-\theta\right).\end{equation*}
Now let
Cˉ=(−a2,a1).\begin{equation*}\bar{C} = (-a_2,a_1).\end{equation*}
Then the angle between Bˉ\bar{B} and Cˉ\bar{C} is π2−θ\frac{\pi}{2}-\theta, and also ∣Cˉ∣=∣Aˉ∣|\bar{C}|=|\bar{A}|. Therefore
∣Aˉ∣ ∣Bˉ∣cos⁡(π2−θ)=∣Bˉ∣ ∣Cˉ∣cos⁡(π2−θ)=∣Bˉ⋅Cˉ∣=∣a1b2−a2b1∣.\begin{equation*}|\bar{A}|\,|\bar{B}| \cos\left(\frac{\pi}{2}-\theta\right) = |\bar{B}|\,|\bar{C}| \cos\left(\frac{\pi}{2}-\theta\right) = |\bar{B}\cdot\bar{C}| = |a_1b_2-a_2b_1|.\end{equation*}
We can also prove the determinant formula by directly computing sin⁡θ\sin \theta. The area of the parallelogram is height times base. We will choose as base ∣Bˉ∣=b12+b22|\bar{B}| = \sqrt{b_1^2 + b_2^2}. If θ\theta is the angle between Aˉ\bar{A} and Bˉ\bar{B}, the height hh of the parallelogram is
h=sin⁡θ∣Aˉ∣=sin⁡θa12+a22.\begin{equation*}h = \sin \theta |\bar{A}| = \sin \theta \sqrt{a_1^2 + a_2^2}.\end{equation*}
To compute sin⁡θ\sin \theta we first compute cos⁡θ\cos \theta:
cos⁡θ=Aˉ⋅Bˉ∣Aˉ∣∣Bˉ∣=a1b1+a2b2a12+a22b12+b22.\begin{equation*}\cos \theta = \frac{\bar{A} \cdot \bar{B}}{|\bar{A}||\bar{B}|} = \frac{a_1b_1 + a_2b_2}{\sqrt{a_1^2 + a_2^2}\sqrt{b_1^2 + b_2^2}}.\end{equation*}
We then get sin⁡θ\sin \theta as follows:
sin⁡θ=1−cos⁡2θ=(a12+a22)(b12+b22)−(a1b1+a2b2)2(a12+a22)(b12+b22)==a12b12+a12b22+a22b12+a22b22−a12b12−2a1b1a2b2−a22b22(a12+a22)(b12+b22)=(a1b2−a2b1)2(a12+a22)(b12+b22).\begin{align*}\sin \theta = \sqrt{1 - \cos^2 \theta} &= \sqrt{\frac{(a_1^2 + a_2^2)(b_1^2 + b_2^2) - (a_1b_1 + a_2b_2)^2}{(a_1^2 + a_2^2)(b_1^2 + b_2^2)}} = \\&= \sqrt{\frac{\cancel{a_1^2b_1^2} + a_1^2b_2^2 + a_2^2b_1^2 + \cancel{a_2^2b_2^2} - \cancel{a_1^2b_1^2} - 2a_1b_1a_2b_2 - \cancel{a_2^2b_2^2}}{(a_1^2 + a_2^2)(b_1^2 + b_2^2)}} = \sqrt{\frac{(a_1b_2 - a_2b_1)^2}{(a_1^2 + a_2^2)(b_1^2 + b_2^2)}}.\end{align*}
Using this value for sin⁡θ\sin \theta in the equation for the area of a parallelogram gives
Area=∣Bˉ∣⏟base∣Aˉ∣sin⁡θ⏟height=b12+b22⏟basea12+a22(a1b2−a2b1)2(a12+a22)(b12+b22)⏟height=∣a1b2−a2b1∣.\begin{equation*}\text{Area} = \underbrace{|\bar{B}|}_{\text{base}} \underbrace{|\bar{A}|\sin \theta}_{\text{height}} = \underbrace{\sqrt{b_1^2 + b_2^2}}_{\text{base}} \underbrace{\sqrt{a_1^2 + a_2^2} \sqrt{\frac{(a_1b_2 - a_2b_1)^2}{(a_1^2 + a_2^2)(b_1^2 + b_2^2)}}}_{\text{height}} = |a_1b_2 - a_2b_1|.\end{equation*}
So the area is the absolute value of the determinant. The sign of a1b2−a2b1a_1b_2-a_2b_1 records the orientation of the ordered pair (Aˉ,Bˉ)(\bar{A},\bar{B}). □\Box

The foolowing lemma is partially from [Hubbard2015] p. 79. Animations are inspired by ttps://www.youtube.com/watch?v=HZDvpuJfYU8&t=176s

Determinant in R3\mathbb{R}^3
The determinant of a 3×33 \times 3 matrix is
det⁡[a1b1c1a2b2c2a3b3c3]=defa1det⁡[b2c2b3c3]−a2det⁡[b1c1b3c3]+a3det⁡[b1c1b2c2]==a1(b2c3−b3c2)−a2(b1c3−b3c1)+a3(b1c2−b2c1).\begin{align*}\det \begin{bmatrix}a_1 & b_1 & c_1 \\a_2 & b_2 & c_2 \\a_3 & b_3 & c_3\end{bmatrix} \stackrel{\text{def}}{=}&a_1 \det\begin{bmatrix}b_2 & c_2 \\b_3 & c_3\end{bmatrix} -a_2 \det\begin{bmatrix}b_1 & c_1 \\b_3 & c_3\end{bmatrix} +a_3 \det\begin{bmatrix}b_1 & c_1 \\b_2 & c_2\end{bmatrix} = \\=& a_1(b_2c_3 - b_3c_2) - a_2(b_1c_3 - b_3c_1) + a_3(b_1c_2 - b_2c_1).\end{align*}
The absolute value of this determinant is the volume of the parallelepiped spanned by Aˉ\bar{A}, Bˉ\bar{B}, and Cˉ\bar{C}, while the sign records the orientation of the ordered triple (Aˉ,Bˉ,Cˉ)(\bar{A},\bar{B},\bar{C}).
We now write the determinant as a sum of simpler determinants. Let
Aˉ=[a1a2a3],Bˉ=[b1b2b3],Cˉ=[c1c2c3].\begin{equation*}\bar{A}=\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix},\qquad\bar{B}=\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix},\qquad\bar{C}=\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}.\end{equation*}
We first split each column into its coordinate pieces:
Aˉ=[a100]+[0a20]+[00a3],Bˉ=[b100]+[0b20]+[00b3],Cˉ=[c100]+[0c20]+[00c3].\begin{equation*}\bar{A}=\begin{bmatrix} a_1 \\ 0 \\ 0 \end{bmatrix}+\begin{bmatrix} 0 \\ a_2 \\ 0 \end{bmatrix}+\begin{bmatrix} 0 \\ 0 \\ a_3 \end{bmatrix},\qquad\bar{B}=\begin{bmatrix} b_1 \\ 0 \\ 0 \end{bmatrix}+\begin{bmatrix} 0 \\ b_2 \\ 0 \end{bmatrix}+\begin{bmatrix} 0 \\ 0 \\ b_3 \end{bmatrix},\qquad\bar{C}=\begin{bmatrix} c_1 \\ 0 \\ 0 \end{bmatrix}+\begin{bmatrix} 0 \\ c_2 \\ 0 \end{bmatrix}+\begin{bmatrix} 0 \\ 0 \\ c_3 \end{bmatrix}.\end{equation*}
Since the determinant is linear in each column, we can expand one column at a time.
Decomposition of determinant in R3\mathbb{R}^3
For instance, linearity in the first column gives
det⁡[Aˉ Bˉ Cˉ]=det⁡[a1b1c10b2c20b3c3]+det⁡[0b1c1a2b2c20b3c3]+det⁡[0b1c10b2c2a3b3c3].\begin{equation*}\det[\bar{A}\ \bar{B}\ \bar{C}]=\det\begin{bmatrix}a_1 & b_1 & c_1 \\0 & b_2 & c_2 \\0 & b_3 & c_3\end{bmatrix}+\det\begin{bmatrix}0 & b_1 & c_1 \\a_2 & b_2 & c_2 \\0 & b_3 & c_3\end{bmatrix}+\det\begin{bmatrix}0 & b_1 & c_1 \\0 & b_2 & c_2 \\a_3 & b_3 & c_3\end{bmatrix}.\end{equation*}
Expanding each of these three determinants in the second column, and then in the third, produces 3⋅3⋅3=273\cdot 3\cdot 3=27 terms. In compact form,
det⁡[Aˉ Bˉ Cˉ]=∑i=13∑j=13∑k=13det⁡[aiei  bjej  ckek]=∑i=13∑j=13∑k=13aibjckdet⁡[ei ej ek].\begin{equation*}\det[\bar{A}\ \bar{B}\ \bar{C}]=\sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \det[a_i e_i\ \ b_j e_j\ \ c_k e_k]=\sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 a_i b_j c_k \det[e_i\ e_j\ e_k].\end{equation*}
Now we determine which of these 27 terms are nonzero.
  • If two of the indices i,j,ki,j,k are equal, then two columns of the determinant are equal, so that term is 00.
  • Therefore a term can survive only when i,j,ki,j,k are all different.
  • The only such triples are the six permutations of (1,2,3)(1,2,3).
So the expansion reduces to the six surviving terms
det⁡[Aˉ Bˉ Cˉ]=a1b2c3det⁡[e1 e2 e3]+a1b3c2det⁡[e1 e3 e2]+a2b1c3det⁡[e2 e1 e3]+a2b3c1det⁡[e2 e3 e1]+a3b1c2det⁡[e3 e1 e2]+a3b2c1det⁡[e3 e2 e1].\begin{align*}\det[\bar{A}\ \bar{B}\ \bar{C}]&=a_1b_2c_3\det[e_1\ e_2\ e_3]+a_1b_3c_2\det[e_1\ e_3\ e_2] \\&\quad+a_2b_1c_3\det[e_2\ e_1\ e_3]+a_2b_3c_1\det[e_2\ e_3\ e_1] \\&\quad+a_3b_1c_2\det[e_3\ e_1\ e_2]+a_3b_2c_1\det[e_3\ e_2\ e_1].\end{align*}
Each remaining determinant is the determinant of a permutation matrix. It equals +1+1 for an even permutation and −1-1 for an odd permutation. Hence
det⁡[e1 e2 e3]=1,det⁡[e1 e3 e2]=−1,det⁡[e2 e1 e3]=−1,\begin{equation*}\det[e_1\ e_2\ e_3]=1,\qquad\det[e_1\ e_3\ e_2]=-1,\qquad\det[e_2\ e_1\ e_3]=-1,\end{equation*}det⁡[e2 e3 e1]=1,det⁡[e3 e1 e2]=1,det⁡[e3 e2 e1]=−1.\begin{equation*}\det[e_2\ e_3\ e_1]=1,\qquad\det[e_3\ e_1\ e_2]=1,\qquad\det[e_3\ e_2\ e_1]=-1.\end{equation*}
Substituting these signs gives
det⁡[a1b1c1a2b2c2a3b3c3]=a1b2c3−a1b3c2−a2b1c3+a2b3c1+a3b1c2−a3b2c1.\begin{equation*}\det\begin{bmatrix}a_1 & b_1 & c_1 \\a_2 & b_2 & c_2 \\a_3 & b_3 & c_3\end{bmatrix}=a_1b_2c_3-a_1b_3c_2-a_2b_1c_3+a_2b_3c_1+a_3b_1c_2-a_3b_2c_1.\end{equation*}
Grouping the terms with the same coefficient from the first column, we recover the cofactor formula
=a1(b2c3−b3c2)−a2(b1c3−b3c1)+a3(b1c2−b2c1).\begin{equation*}= a_1(b_2c_3 - b_3c_2) - a_2(b_1c_3 - b_3c_1) + a_3(b_1c_2 - b_2c_1).\end{equation*}
This is exactly the determinant formula stated above. □\Box

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