The Determinant as an Area Scale Factor

Where the unit square goes

The unit square has corners (0, 0), (1, 0), (1, 1) and (0, 1), and area 1. A matrix sends (1, 0) to its first column and (0, 1) to its second, so the unit square lands on the parallelogram whose sides are the two columns. Its area is the size of the determinant, |ad − bc|.

The matrix (2 0; 0 2) doubles both directions. It sends the unit square to a 2 by 2 square, of area 4, and its determinant is 2 × 2 − 0 × 0 = 4. Because every shape in the plane is made of small squares, and each one is enlarged in the same way, every area is multiplied by 4.

A slanted example

Take M = (2 1; 1 3). Its determinant is 2 × 3 − 1 × 1 = 6 − 1 = 5. Its columns are (2, 1) and (1, 3), so the unit square lands on the parallelogram with corners (0, 0), (2, 1), (3, 4) and (1, 3).

Check that area directly. The parallelogram fits in a 3 by 4 rectangle of area 12. Outside it but inside the rectangle are two right triangles with legs 2 and 1, of area 1 each, two right triangles with legs 1 and 3, of area 1.5 each, and two 1 by 1 squares. So the parallelogram has area 12 − 2 − 3 − 2 = 5, the determinant.

The unit square, of area 1, and its image under (2 1; 1 3), the parallelogram on the columns (2, 1) and (1, 3), of area 5.

Every shape, the same factor

Any shape can be filled with small squares, as closely as you like. M turns each small square into a small parallelogram with 5 times its area, so it multiplies the area of the whole shape by 5.

Check it on a triangle. The triangle with corners (0, 0), (1, 0) and (0, 1) has area $\frac{1}{2}$1/2. M sends it to the triangle with corners (0, 0), (2, 1) and (1, 3), which is half of the parallelogram above, so its area is $\frac{5}{2}$5/2, and $5\times \frac{1}{2}=\frac{5}{2}$5 × 1/2 = 5/2.

Determinant 1: moved, not resized

The quarter turn (0 −1; 1 0) has determinant 0 × 0 − (−1) × 1 = 1. It moves the unit square round the origin without resizing it, so every area stays the same.

A shear changes the shape and still keeps the area. (1 2; 0 1) has determinant 1 × 1 − 2 × 0 = 1. It sends the unit square to a parallelogram with base 1 along the x-axis and height 1, so its area is still 1 × 1 = 1, however far over it leans.

The two handles are the two columns of (a b; 0 1). Moving b shears the square and the area stays a. Moving a stretches it, and the area follows a. At a = 0 the square is flat, and past 0 it is turned over, with a negative determinant.

A negative determinant: a flip

The reflection in the x-axis, (1 0; 0 −1), has determinant 1 × (−1) − 0 × 0 = −1. The unit square lands on the square with corners (0, 0), (1, 0), (1, −1) and (0, −1), which still has area 1. So the area is multiplied by 1, the size of the determinant.

The minus sign says the plane has been turned over. Going round the unit square from (0, 0) to (1, 0) to (1, 1) to (0, 1) is counterclockwise; going round their images, (0, 0) to (1, 0) to (1, −1) to (0, −1), is clockwise. An area is never negative: the size of the determinant gives the area factor, and its sign says whether the order of the corners is reversed.

Determinant 0: no area at all

For B = (2 4; 1 2), ad − bc = 2 × 2 − 4 × 1 = 0. Its columns, (2, 1) and (4, 2), lie along one line, since (4, 2) is twice (2, 1). B sends (x, y) to (2x + 4y, x + 2y), which is (x + 2y) times (2, 1), so every point of the plane lands on the line through the origin and (2, 1), the line $y=\frac{x}{2}$y = x/2.

The unit square is squashed onto the segment from (0, 0) to (6, 3), with no area left, and the area of every shape is multiplied by 0. Different points land on the same point, which is why B has no inverse.

Under (2 4; 1 2) the unit square lands on a segment of the line $y=\frac{x}{2}$y = x/2, from (0, 0) to (6, 3): a shape with no area.

Using the scale factor

The stretch (3 0; 0 2) has determinant 3 × 2 − 0 × 0 = 6. It turns the circle of radius 1, area $\pi$π, into an ellipse that reaches 3 along the x-axis and 2 up the y-axis, and the ellipse has area $6\times \pi =6\pi$6 × π = 6π. That agrees with the formula for the area of an ellipse, $\pi \times 3\times 2$π × 3 × 2.

An enlargement with scale factor k has matrix (k 0; 0 k), whose determinant is $k\times k=k^2$k × k = k². So an enlargement by 3 multiplies areas by 9, as similar shapes do.

The usual mistakes

Adding the two diagonal products. For (2 1; 1 3), 6 + 1 = 7 is not the area factor; it is 6 − 1 = 5.

Using the main diagonal alone. 2 × 3 = 6 leaves out the bc term.

Adding the stretch factors. (3 0; 0 2) multiplies widths by 3 and heights by 2, so areas are multiplied by 3 × 2 = 6, not 3 + 2 = 5.

Giving a negative area. A determinant of −1 keeps every area the same and turns the plane over.

Reading a determinant of 0 as every point going to the origin. The points of the plane land on a line; it is the area that becomes 0.

A flower bed, a stencil and a photo

In the applications below, a flower bed is stretched by (3 0; 0 2) and its new area is found twice, directly and as the old area times the determinant 6. A stencil is reflected in y = x, a matrix with determinant −1, which keeps its area and reverses the order of its corners. And a photo is flattened onto a line by a matrix with determinant 0, which no matrix can undo.