4.3 The Determinant as Volume

Section 4.1 established what $|\det|$ means for a $2 \times 2$ matrix: the area of the parallelogram built on its columns, and therefore the factor by which the transformation scales every area. Nothing in that argument was two-dimensional. Chop any region into a fine mosaic of tiny cubes, note that a linear map takes each cube to the same slanted box scaled by the same factor, and the same conclusion follows one dimension up:

$|\det A| = \text{the volume of the parallelepiped built on the columns of } A$

The parallelepiped is the solid Section 0.3 already built: the three-dimensional analogue of a parallelogram, swept out by three edge vectors leaving a common corner, with each pair of opposite faces parallel. There it was a device for adding — three vectors laid head to tail arrive at the far corner, which the long diagonal reaches directly. Here the same solid is asked a different question, about the space it encloses rather than the corner it reaches, and the determinant is the answer. Take the columns from the last section as its edges:

$\mathbf{u} = (2, 1, 0), \qquad \mathbf{v} = (1, 3, 1), \qquad \mathbf{w} = (0, 1, g)$

Their determinant is $5g - 2$ , so the box's volume is $|5g - 2|$ and the slider drives it through the interesting event.

Three semi-transparent parallelogram faces meeting at the origin, forming half of a slanted box - a parallelepiped - with a red, a blue and a green segment along the three edges that leave the origin, and three amber edges meeting at the far corner, where the sum of the three edge vectors lands. A slider tilts the green edge. As it comes down the box flattens and the far corner descends; at one particular setting all three edges lie in a single plane and the box has no thickness at all, and pushing further reopens it with its corner on the other side, so the frame of edges has turned inside out. — The three faces of the parallelepiped that meet at the origin, with the red, blue and green edges $\mathbf{u}$ , $\mathbf{v}$ and $\mathbf{w}$ along them and, in amber, the three edges that meet at the far corner $\mathbf{u} + \mathbf{v} + \mathbf{w}$ . The slider $g$ tilts the green edge, and with it the whole solid.

Explore the figure

At the default $g = 2$ the box is comfortably solid. Its volume is $|5(2) - 2| = 8$ , the determinant computed by cofactor expansion in the last section — now visible as an amount of space rather than an amount of arithmetic.
Drag $g$ down towards $0.4$ . The box flattens: the green edge is folding down towards the plane of the other two, and the far corner is descending with it. The volume $|5g - 2|$ is shrinking at a steady rate, because the determinant is linear in $g$ .
At $g = 0.4$ the solid has no thickness at all. All three edges lie in one plane, the box is a flat sheet, and the volume is exactly zero — which is the determinant vanishing, and also the columns being dependent, and also the matrix losing its inverse. Three descriptions, one event.
Continue to $g = 0$ . The box reopens, and $\det = -2$ , so it has volume $2$ — but it is not the same box scaled down. Compare the arrangement of the three edges now with their arrangement at $g = 2$ : the frame has been turned inside out.

What the sign means with three vectors

In the plane, a negative determinant meant the image had been flipped over. In space the corresponding idea is handedness. Point the fingers of your right hand along $\mathbf{u}$ and curl them towards $\mathbf{v}$ ; your thumb points to one side of their common plane. If $\mathbf{w}$ points to that side, the triple is right-handed and the determinant is positive; if $\mathbf{w}$ points the other way, it is left-handed and the determinant is negative.

Handedness cannot be changed by any amount of rotating or stretching, exactly as a left glove cannot be rotated into a right one. So a continuous journey from a positive determinant to a negative one must pass through zero — the box must flatten completely before it can reopen the other way, which is what step 4 made you watch.

-2

The flipped frame ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The determinant at $g = 0$ , where exploration step 4 stops. The magnitude $2$ is the box's volume, and the minus sign says the three edges, in the order written, form a left-handed frame. The sign belongs to the ordered triple, not to the solid: listing the same three edges in a different order flips it without moving the box at all. What cannot be rotated onto one another are two frames of opposite handedness.

0

Exactly flat ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The determinant at $g = 2/5$ : zero, on the nose. Three vectors in space are coplanar exactly when the box they build has no volume, and that is the whole content of a vanishing $3 \times 3$ determinant. Every other statement in the Invertible Matrix Theorem fails at this same slider setting.

Volume scaling, not just one box

The reading that matters most is the one about every region, not the one special box. If $T(\mathbf{x}) = A\mathbf{x}$ then $T$ takes the unit cube — volume $1$ — to the parallelepiped on the columns of $A$ , of volume $|\det A|$ . Since any solid can be approximated by a mosaic of small cubes, and each is scaled by that same factor, $T$ multiplies the volume of any region by $|\det A|$ .

Two consequences worth stating. First, determinants multiply, and now you can see why they must: doing $A$ and then $B$ scales volume by $|\det A|$ and then by $|\det B|$ , so the composite scales it by the product, and the signs compose in the same way because handedness either survives each step or is reversed by it. Second, a matrix with $\det A = 0$ destroys volume everywhere at once — every solid is flattened into something with no thickness. Usually that is a sheet; if the matrix is more degenerate still it can be a line or even the single point $\mathbf{0}$ , as the zero matrix shows. Either way no inverse can exist, because no transformation manufactures volume from something flat.

Example 1 Two boxes, one volume, opposite frames

Compute the volume of the parallelepiped on $\mathbf{u} = (2, 1, 0)$ , $\mathbf{v} = (1, 3, 1)$ , $\mathbf{w} = (0, 1, g)$ at $g = 0.8$ , and at the value of $g$ that gives the same volume with the opposite handedness. Confirm your answer against the slider.

Show solution

At $g = 0.8$ : $\det = 5(0.8) - 2 = 2$ , so the volume is $2$ and the frame is right-handed.

For the same volume with the opposite sign, solve $5g - 2 = -2$ , giving $5g = 0$ and $g = 0$ . So at $g = 0$ the box again has volume $2$ , this time left-handed — matching the $-2$ computed above, and matching exploration step 4.

The two settings $g = 0.8$ and $g = 0$ sit either side of $g = 0.4$ , which is exactly where the box is flat, as they must: to reach one from the other by sliding, the determinant has to travel from $+2$ to $-2$ , and a linear function cannot do that without passing through zero. Notice also that the two boxes are genuinely different solids — the green edge has length $\sqrt{41}/5$ in one and $1$ in the other — but handedness is not what separates them. It belongs to the ordered frame of edges, not to the box: read the same box from its far corner, along $-\mathbf{u}$ , $-\mathbf{v}$ and $-\mathbf{w}$ , and the frame reverses without the box moving at all.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Chapters 1–6 follow the topic sequence of Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL). Chapter 0's scope (its topics, their order, and the idea each figure shows) follows the Vectors chapter of Ximera's Linear Algebra: An Interactive Introduction. The prose, worked examples and figures were written for this book, which is not affiliated with or endorsed by the authors of either work. License: CC-BY-NC-SA-4.0.