#L05 · Linear Algebra & Manifolds · Lv.3View in 3D Mind-Map

Determinant & Matrix Inverse

Seki Kowa, Leibniz & Cauchy · 1683 CE

det⁡(A)=ad−bc,A−1=1ad−bc(d−b−ca)\det(A) = ad - bc,\quad A^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
AA−1=I,det⁡(AB)=det⁡(A)det⁡(B)AA^{-1} = I,\quad \det(AB) = \det(A)\det(B)

Measures the signed area/volume scaling factor of a linear transformation via det(A), and reverses the transformation via A⁻¹ whenever det(A) ≠ 0.

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Entry a
Entry b
Entry c
Entry d
det⁡(A)=(3)(4)−(1)(2)=10.00\det(A) = (3)(4) - (1)(2) = 10.00
행렬식 det(A) = ad-bc
10.0000
역행렬 존재 여부
가역 (Invertible)
A⁻¹의 (1,1) 성분 d/det
0.400
col₁col₂

Prerequisite Path (13 STEP)

Full Path Analyzer

Click steps in order to follow the 3D neon path from basics to target.

STEP 1#A02Polynomial Expansion & Factoring
STEP 2#A03Laws of Exponents & Radicals
STEP 3#G01Pythagorean Theorem
STEP 4#G03Pi, Radian Measure & Circular Sector
STEP 5#L01Vector Operations, L1/L2 Norms & Normalization
STEP 6#A01Linear Equation & Balance Axiom
STEP 7#G022D/3D Euclidean Distance Formula
STEP 8#G05Law of Sines & Law of Cosines
STEP 9#G04Unit Circle Trigonometry & Identities
STEP 10#L02Dot Product & Cosine Similarity
STEP 11#L033D Cross Product & Surface Normal Vector
STEP 12#L04Matrix Multiplication & Linear Transformation
STEP 13#L05Determinant & Matrix Inverse

Geometric & Intuitive Insight

가로·세로가 1인 단위 정사각형에 행렬 A를 곱하면 두 열벡터 (a, c)와 (b, d)가 만드는 평행사변형으로 변형되는데, 그 넓이가 정확히 ad - bc 입니다! 넓이가 0이 되어 선분으로 찌그러지면 다시 정사각형으로 복구(역행렬)할 수 없습니다.

Step-by-Step Derivation

  1. Step 1.1단계: 2×2 행렬 A의 두 열벡터 (a, c, 0)과 (b, d, 0)의 외적 z성분(평행사변형 부호 넓이)은 ad - bc 입니다.
  2. Step 2.2단계: A · [[d, -b], [-c, a]] 를 행렬 곱셈하면 대각선은 ad - bc, 비대각선은 0이 됩니다.
  3. Step 3.3단계: 따라서 ad - bc ≠ 0 일 때 양변을 (ad - bc)로 나누면 AA⁻¹ = I 를 만족하는 역행렬이 유도됩니다.

AI & Coding Real-World Use

3D mouse raycasting unprojection (MVP⁻¹) in games and Jacobian determinants in Normalizing Flow AI

#3D Mouse Raycasting#Jacobian Determinant#Normalizing Flows
const det2 = (a:number, b:number, c:number, d:number) => a * d - b * c;

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