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Exercise 1637Linear algebra › DeterminantsL1Added 3 Nov 2026

The determinant as a volume

#determinant#volume#parallelepiped

Statement

Let u,v,w∈R3u,v,w\in\mathbb{R}^3. Show that ∣det⁡(u,v,w)∣|\det(u,v,w)| is the volume of the parallelepiped spanned by u,v,wu,v,w.

Hint

The volume is ∥u∥⋅(area of the base)⋅(height)\|u\|\cdot\text{(area of the base)}\cdot\text{(height)}. For 33 vectors, it is ∣u⋅(v×w)∣|u\cdot(v\times w)|.

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Solution

Triple product. The scalar triple product u⋅(v×w)u\cdot(v\times w) equals det⁡(u,v,w)\det(u,v,w) (with u,v,wu,v,w as columns). Indeed, the cross product v×wv\times w has norm equal to the area of the parallelogram spanned by v,wv,w, and is orthogonal to the plane of v,wv,w. So u⋅(v×w)u\cdot(v\times w) is the volume of the parallelepiped spanned by u,v,wu,v,w.

Determinant. The scalar triple product can be written as a determinant:
u⋅(v×w)=∣u1v1w1u2v2w2u3v3w3∣=det⁡(u,v,w).u\cdot(v\times w) = \begin{vmatrix} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{vmatrix} = \det(u,v,w).

Conclusion. ∣det⁡(u,v,w)∣|\det(u,v,w)| is the volume of the parallelepiped.

Higher dimension. In Rn\mathbb{R}^n, the same holds: ∣det⁡(u1,…,un)∣|\det(u_1,\ldots,u_n)| is the nn-dimensional volume of the parallelotope spanned by the vectors.