The Determinant as a Volume Scale Factor

Where the unit cube goes

The unit cube has one corner at the origin and its three edges from that corner along i, j and k, so its volume is 1. A 3 × 3 matrix sends i, j and k to its three columns, so it sends the unit cube to the solid whose three edges from the origin are the columns. That solid is a parallelepiped, a box whose faces are parallelograms.

The volume of that solid is the size of the determinant of the matrix. Since the unit cube has volume 1, the determinant is the factor by which the matrix multiplies volume.

A stretch in each direction

The matrix A = (2 0 0; 0 3 0; 0 0 4) sends i to 2i, j to 3j and k to 4k. It sends (x, y, z) to (2x, 3y, 4z): it stretches x by 2, y by 3 and z by 4, each direction on its own.

The unit cube becomes a box 2 by 3 by 4. Count the unit cubes inside it: each layer holds 2 × 3 = 6 of them, and there are 4 layers, so 24 unit cubes fit inside. The volume has gone from 1 to 24.

A sends the unit cube to a 2 by 3 by 4 box, which holds 24 unit cubes.

The determinant is the scale factor

Find det A by expanding along the first row. Only its first entry is not 0, so det A = 2 × (3 × 4 − 0 × 0) = 2 × 12 = 24, the same number as the new volume. For any diagonal matrix the determinant is the product of the diagonal entries.

Every solid, not only the cube, has its volume multiplied by 24. Any solid can be filled with small cubes as closely as you like, and A turns each small cube into a small box with 24 times its volume. A solid of volume 5 becomes a solid of volume 5 × 24 = 120.

A sphere of radius 1 has volume $\frac{4\pi }{3}$4π/3. A turns it into an ellipsoid that reaches 2 along the x-axis, 3 along the y-axis and 4 along the z-axis, with volume $24\times \frac{4\pi }{3}=32\pi$24 × 4π/3 = 32π. That agrees with the formula for the volume of an ellipsoid, $\frac{4\pi }{3}\times 2\times 3\times 4$4π/3 × 2 × 3 × 4.

Slanted solids

The matrix S = (1 0 1; 0 1 0; 0 0 1) leaves i and j alone and sends k to (1, 0, 1). It sends (x, y, z) to (x + z, y, z): each layer of the cube slides along x by its own height, like a stack of cards pushed over. The unit cube leans into a slanted solid with the same unit square as its base and the same height 1, so its volume is still 1. Its determinant agrees: S is zero below the diagonal, so det S = 1 × 1 × 1 = 1.

Now take N = (2 1 0; 1 2 0; 0 0 3). Expand along the third row, where only the last entry is not 0: det N = 3 × (2 × 2 − 1 × 1) = 3 × 3 = 9. The block (2 1; 1 2), with determinant 3, multiplies areas in the floor by 3, and the 3 in the corner multiplies heights by 3, so volumes are multiplied by 3 × 3 = 9.

The enlargement (k 0 0; 0 k 0; 0 0 k) has determinant $k\times k\times k=k^3$k × k × k = k³. An enlargement by 2 multiplies every volume by 8, as similar solids do.

S seen from the side, with x across the page and z up the page: each layer slides along by its height. The base stays on the floor and the height stays 1, so the volume stays 1.

A negative determinant

The reflection in the plane z = 0, M = (1 0 0; 0 1 0; 0 0 −1), has determinant 1 × 1 × (−1) = −1. The unit cube lands on the cube below the floor, which still has volume 1, so every volume is multiplied by 1, the size of the determinant.

The minus sign says space has been turned inside out. Point the thumb, first finger and middle finger of your right hand along i, j and k. Their images under M are i, j and −k, and only a left hand can point along those three. A reflection turns a right hand into a left hand, and no turn can do that.

A turn has determinant 1. The quarter turn about the z-axis, (0 −1 0; 1 0 0; 0 0 1), expanded along its third row, has determinant 1 × (0 × 0 − (−1) × 1) = 1: it moves every solid without resizing it or turning it inside out.

Determinant 0: no volume at all

Take B = (1 2 3; 2 4 6; 1 0 1). Its second row is twice its first. Expand along the first row: det B = 1 × (4 × 1 − 6 × 0) − 2 × (2 × 1 − 6 × 1) + 3 × (2 × 0 − 4 × 1) = 4 + 8 − 12 = 0.

B sends (x, y, z) to (x + 2y + 3z, 2x + 4y + 6z, x + z). The second coordinate of every image is twice the first, so every image lies in the plane y = 2x. The unit cube is flattened onto part of that plane, with no volume left, and every volume is multiplied by 0.

Different points land on the same point. B sends (1, 1, −1) to (1 + 2 − 3, 2 + 4 − 6, 1 − 1) = (0, 0, 0), and every multiple of (1, 1, −1) lands on the origin too. That is why B has no inverse.

Row 2 of B is twice row 1, so the second coordinate of every image is twice the first, and det B = 0.

The usual mistakes

Adding the stretches. A stretches by 2, 3 and 4, and the volume is multiplied by 2 × 3 × 4 = 24, not 2 + 3 + 4 = 9.

Cubing the largest stretch, as if A were an enlargement by 4. Each direction has its own factor, and the determinant combines all three.

Giving a negative volume. A determinant of −1 keeps every volume and turns space inside out.

Reading a determinant of 0 as sending every point to the origin. B flattens space onto a plane; it is the volume that becomes 0.

A model shed

In the application below, a model shed is stretched by 3 along its length and by 2 in its depth and its height. The determinant, 12, gives the new volume straight from the old, and the new length, depth and height check it.