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Why determinant is area?

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*}
In R3\mathbb{R}^3, it gives the signed volume of the parallelepiped spanned by three vectors.
Determinant in R3\mathbb{R}^3
The determinant of a 3×33 \times 3 matrix is
det⁡[a1b1c1a2b2c2a3b3c3]=def=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(b_2c_3 - b_3c_2) - a_2(b_1c_3 - b_3c_1) + a_3(b_1c_2 - b_2c_1).\end{align*}

Determinant in 2-dimensional space

The following lemma is partially from Hubbard and Hubbard [HubbardHubbard2015], p. 75.
If we think of the determinant as a function of the vectors A=(a1,a2)A = (a_1, a_2) and B=(b1,b2)B = (b_1, b_2) in R2\mathbb{R}^2, then we can prove that it is an area of the parallelogram spanned by AA and BB. We provide the two proof of this fact.

Geometric Proof

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
∣A∣ ∣B∣sin⁡θ=∣A∣ ∣B∣cos⁡(π2−θ).\begin{equation*}|A|\,|B| \sin \theta = |A|\,|B| \cos\left(\frac{\pi}{2}-\theta\right).\end{equation*}
Now let
C=(−a2,a1).\begin{equation*}C = (-a_2,a_1).\end{equation*}
Then the angle between BB and CC is π2−θ\frac{\pi}{2}-\theta, and also ∣C∣=∣A∣|C|=|A|. Therefore
∣A∣ ∣B∣cos⁡(π2−θ)=∣B∣ ∣C∣cos⁡(π2−θ)=∣B⋅C∣=∣a1b2−a2b1∣.\begin{equation*}|A|\,|B| \cos\left(\frac{\pi}{2}-\theta\right) = |B|\,|C| \cos\left(\frac{\pi}{2}-\theta\right) = |B\cdot C| = |a_1b_2-a_2b_1|.\end{equation*}

Second proof

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 ∣B∣=b12+b22|B| = \sqrt{b_1^2 + b_2^2}. If θ\theta is the angle between AA and BB, the height hh of the parallelogram is
h=sin⁡θ∣A∣=sin⁡θa12+a22.\begin{equation*}h = \sin \theta |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=A⋅B∣A∣∣B∣=a1b1+a2b2a12+a22b12+b22.\begin{equation*}\cos \theta = \frac{A \cdot B}{|A||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=∣B∣⏟base∣A∣sin⁡θ⏟height=b12+b22⏟basea12+a22(a1b2−a2b1)2(a12+a22)(b12+b22)⏟height=∣a1b2−a2b1∣.\begin{equation*}\text{Area} = \underbrace{|B|}_{\text{base}} \underbrace{|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 (A,B)(A,B). □\Box

Determinant in 3-dimensional space

The following lemma is partially from Hubbard and Hubbard [HubbardHubbard2015], p. 79. Animations are inspired by Sanderson [Sanderson2016].
Let
A=[a1a2a3],B=[b1b2b3],C=[c1c2c3].\begin{equation*}A=\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}, \qquad B=\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}, \qquad C=\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}.\end{equation*}
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*}
We now write the determinant as a sum of simpler determinants.
Decomposition of determinant in R3\mathbb{R}^3
We first split each column into its coordinate pieces:
A=[a100]+[0a20]+[00a3],B=[b100]+[0b20]+[00b3],C=[c100]+[0c20]+[00c3].\begin{equation*}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 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 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.
For instance, linearity in the first column gives
det⁡[A B C]=det⁡[a1b1c10b2c20b3c3]+det⁡[0b1c1a2b2c20b3c3]+det⁡[0b1c10b2c2a3b3c3].\begin{equation*}\det[A\ B\ 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,
det⁡[A B C]=∑i=13∑j=13∑k=13det⁡[aiei  bjej  ckek]=∑i=13∑j=13∑k=13aibjckdet⁡[ei ej ek].\begin{equation*}\det[A\ B\ 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*}
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⁡[A B C]=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[A\ B\ 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*}
Decomposition of determinant in R3\mathbb{R}^3
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

References

  • [HubbardHubbard2015]Hubbard, John H., and Barbara Burke Hubbard. Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach. Fifth edition. Matrix Editions, 2015.
  • [Sanderson2016]Sanderson, Grant. ``The determinant.'' 3Blue1Brown, 2016. Video.