GraphMath

Cramer's rule

Determinants as coordinates of the solution

How can a square full-rank system be solved by replacing one column at a time?

Cramer's rule expresses each component of the solution to Ax = b as a ratio of determinants. This chapter derives the rule from determinant of a matrix product and then illustrates it geometrically in 3D, showing how the sign of each coordinate is tied to signed perpendicular components relative to the opposite face.

Key ideas

Cramer's rule turns the unknown coordinates of the solution vector into determinant ratios.

  • For a square matrix A with |A| ≠ 0, each solution component xₖ is determined uniquely
  • The auxiliary matrix Xₖ is built so that its determinant is exactly xₖ
  • Multiplying by A replaces the k-th column by b, producing the numerator matrix Aₖ
  • The determinant identity |A Xₖ| = |A||Xₖ| gives |Aₖ| = |A|xₖ
  • So xₖ = |Aₖ| / |A|, where Aₖ is obtained by replacing column k of A with b

The 3D illustration then interprets the sign of each numerator geometrically through the orientation of b relative to the face formed by the other columns.

Why does replacing one column of A with b reveal one coordinate xₖ?

Because the auxiliary matrix Xₖ is constructed so that its determinant is exactly xₖ. Once A multiplies Xₖ, the k-th column becomes b, so the determinant of that new matrix isolates xₖ through the product rule for determinants.

Related chapters

Examples with visualizations

Cramer's rule in 3D

Animated determinant ratio and sign interpretation for x₁, x₂ and x₃

Animated 3D illustration of Cramer's rule showing the determinant ratios for x₁, x₂ and x₃ and the sign of each coordinate

Chapter contents

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Topic Pages
Cramer's rule 1–2
Cramer's rule: 3D illustration 3–4

Why is Cramer's rule not usually the preferred computation method?

Because it requires a separate determinant computation for each coordinate, which becomes inefficient for larger systems. Its value is mainly conceptual: it shows how determinants encode the coordinates of the unique solution in the square full-rank case.

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