Geometric / Visual Interpretation of Cramer's Rule

1. Setup and intuitive statement

Let $v_{1}, v_{2}, v_{3} \in R^{3}$ be three linearly independent vectors (columns of a matrix). They form the edges of a parallelepiped; the signed volume of that parallelepiped equals the determinant $det [v_{1} v_{2} v_{3}] .$
Given a vector $w \in R^{3}$ , we ask: what are the coordinates $x, y, z$ such that $w = x v_{1} + y v_{2} + z v_{3} ?$

Cramer's rule tells us that, e.g. the coordinate $z$ is

z = \frac{det [v_{1} v_{2} w]}{det [v_{1} v_{2} v_{3}]} .

2. Why inner products fail for a non-orthonormal basis

If $v_{1}, v_{2}, v_{3}$ were orthonormal, you could find the coordinates of $w$ by taking dot products (projections), e.g. $z = w \cdot v_{3}$ . But when the basis is not orthonormal, projection mixes different coordinates. We need a dual object that "annihilates" (gives zero on) the other basis directions and measures height relative to a chosen plane. Determinants and exterior algebra provide that object.

Two cleaner ways to see it follow.

3. 1-form viewpoint

We can take a 1‑form $ω : R^{3} \to R$ such that
$ω (v_{1}) = 0, ω (v_{2}) = 0, ω (v_{3}) \neq 0.$
It can be defined as $w = v_{1} ⌟ v_{2} ⌟ Ω$ , being $Ω$ a volume form.
Applying $ω$ to the decomposition $w = x v_{1} + y v_{2} + z v_{3}$ annihilates the $v_{1}, v_{2}$ parts, leaving $ω (w) = z \cdot ω (v_{3})$ . This leads directly to Cramer's formula.

4. Geometric algebra viewpoint (concrete for $R^{3}$ )

Given $w = x v_{1} + y v_{2} + z v_{3}$ we can multiply both sides by the bivector $v_{1} \land v_{2}$ . Then

v_{1} \land v_{2} \land w = x v_{1} \land v_{2} \land v_{1} + y v_{1} \land v_{2} \land v_{2} + z v_{1} \land v_{2} \land v_{3}

and therefore

v_{1} \land v_{2} \land w = z v_{1} \land v_{2} \land v_{3}

Geometric interpretation: the determinant in the numerator is the signed volume of the parallelepiped formed by $v_{1}, v_{2}, w$ while the denominator is the signed volume formed by $v_{1}, v_{2}, v_{3}$ . If we view $v_{1}$ and $v_{2}$ as fixing the base (a parallelogram) then the determinant equals (area of base) × (signed height). Because the base is the same for both determinants, the ratio of determinants reduces to the ratio of the heights of $w$ and $v_{3}$ above that base — and that ratio is precisely the coordinate $z$ .