Why determinant is area?

Last updated: 2026-04-12

This article try to explain why the determinant naturally emerges as an area spanned by its vectors.

Determinant in 2-dimensional space

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

Definition (Determinant in R2\mathbb{R}^2)
The determinant det⁡\det of a 2×22 \times 2 matrix are 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*}
\end{definition}
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 following video.
Geometric interpretion of determinant in R2\mathbb{R}^2

Geometric Proof

We can give a simple geometric proof by expressing sin⁡θ\sin \theta via cos⁡(π2−θ)\cos( \frac{\pi}{2} - \theta )
Geometric proof that determinant is area of parallelogram in R2\mathbb{R}^2
We can split proof in several steps:
  • the area of parallelogram is given by ∣Aˉ∣∣Bˉ∣sin⁡θ| \bar{A} | |\bar{B}| \sin \theta
  • rewrite sin⁡θ\sin \theta as cos⁡(π2−θ)\cos( \frac{\pi}{2} - \theta )
  • angle π2−θ\frac{\pi}{2} - \theta is between Bˉ\bar{B} and Cˉ=(−a2,a1)\bar{C} = (-a_2, a_1)
  • now ∣Aˉ∣∣Cˉ∣cos⁡(π2−θ)| \bar{A} | |\bar{C}| \cos( \frac{\pi}{2} - \theta ) is the formula for inner product between Aˉ\bar{A} and Cˉ\bar{C}, which is exactly a1b2−a2b1a_1b_2 - a_2b_1

Second proof(computationally extensive)

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 using equation 1.4.7:
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{a_1^2b_1^2 + a_1^2b_2^2 + a_2^2b_1^2 + a_2^2b_2^2 - a_1^2b_1^2 - 2a_1b_1a_2b_2 - 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∣⏟determinant.□\begin{align*}\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}} = \underbrace{|a_1b_2 - a_2b_1|}_{\text{determinant}}. \quad \Box\end{align*}

Determinant in 3-dimensional space

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

In R3\mathbb{R}^3 the determinant formula gets more complicated.
Definition (Determinant in In R3\mathbb{R}^3 )
The determinant of a 3×33 \times 3 matrix is
det⁡[a1b1c1a2b2c2a3b3c3]=defa1det⁡[b2c2b3c3]−a2det⁡[b1c1b3c3]+a3det⁡[b1c1b2c2]\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} \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}\end{equation*}=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*}
Each entry of the first column of the original matrix serves as the coefficient for the determinant of a 2×22 \times 2 matrix; the first and third (a1a_1 and a3a_3) are positive, the middle one is negative. To remember which 2×22 \times 2 matrix goes with which coefficient, cross out the row and column the coefficient is in; what is left is the matrix you want. To get the 2×22 \times 2 matrix for the coefficient a2a_2,
[a1b1c1a2b2c2a3b3c3]=[b1c1b3c3].\begin{equation*}\begin{bmatrix}\cancel{a_1} & b_1 & c_1 \\\cancel{a_2} & \cancel{b_2} & \cancel{c_2} \\\cancel{a_3} & b_3 & c_3\end{bmatrix} =\begin{bmatrix}b_1 & c_1 \\b_3 & c_3\end{bmatrix}.\end{equation*}

Proof with determinant manipulation

We can prove the determinant formula in R3\mathbb{R}^3 by examining how the determinant decomposes into simpler components. Consider a 3×33 \times 3 matrix whose columns represent three vectors Aˉ\bar{A}, Bˉ\bar{B}, and Cˉ\bar{C}.
Decomposition of determinant in R3\mathbb{R}^3
The original determinant can be decomposed into a sum of 27 determinants, each using single components from vectors Aˉ\bar{A}, Bˉ\bar{B}, and Cˉ\bar{C}. However, only certain combinations of these components contribute to the volume. Specifically:
  • When three components are selected from the same row, they collapse into a single vector, contributing zero volume
  • When two components are selected from the same row, they form a degenerate two-dimensional parallelogram, again contributing zero volume
  • Only combinations where components are selected from different rows can form a proper three-dimensional parallelepiped, thus contributing to the final volume
Final formula for determinant in R3\mathbb{R}^3
The determinant formula in R3\mathbb{R}^3 reduces to just six non-zero terms, each representing a valid three-dimensional volume. When we factor out the coefficients from each determinants and transform determinant to unitary matrix we finally get the final formula.
Decomposition of determinant in R3\mathbb{R}^3
The proof is finished □\Box.

References