The determinant in this visualization
A 2 × 2 matrix A = (a b; c d) is a linear map of the plane. It sends î and ĵ to its columns a₁ = (a, c) and a₂ = (b, d), so the unit square becomes the parallelogram with sides a₁ and a₂. The determinant of A is the signed area of that parallelogram.
The picture shows the plane before and after A: the standard grid as faint dashed lines, the grid A makes of it in ink, and the image of the unit square as a yellow parallelogram. The small yellow arc near the origin turns from a₁ to a₂ the shorter way round. The flag is a shape carried along, dashed where it was and solid where A puts it.
The determinant is not only about the unit square. Every square of the grid becomes a copy of the same parallelogram, so a region made of grid squares has its area multiplied by |det A|. A finer grid works the same way. Shapes such as a disk can be filled ever more closely with its small squares, so their areas are multiplied by |det A| as well. With Circle chosen, the unit disk of area π becomes an ellipse of area |det A|·π, or a segment or a point when det A = 0.
The sign tells whether the plane is flipped over. If the shorter turn from a₁ to a₂ is counter-clockwise, as from î to ĵ, A keeps the plane’s orientation and det A > 0. If it is clockwise, A reverses the plane’s orientation: the flag faces the other way, the yellow fill becomes hatching, and det A < 0. The determinant changes continuously with the entries, so it cannot jump from positive to negative. Drag a₂ across the line through a₁, and the parallelogram first flattens into a segment, where det A = 0, and only then flips.
When det A = 0, the columns lie on one line and A flattens the whole plane onto it, or onto the single point O if A is the zero matrix. Different points then land in the same place, nothing can undo that, and so A has no inverse.
Why ad − bc? Turn on Shear construction and push. A shear slides one side of the parallelogram along the line of the other. The base and the height stay the same, and so does the area. When det A ≠ 0 and |a| is at least |c|, a₂ is first pushed along a₁ until it stands upright at (0, h), with h = d − bc/a. Then a₁ is pushed along that upright side until it lies on the x-axis at (a, 0). The result is a rectangle with sides |a| and |h|, and its signed area is a·h = ad − bc. When |c| is larger than |a|, the page builds the rectangle the other way round, with the same signed area. If det A = 0, the parallelogram is flat and no rectangle can be built.
Adding a multiple of one column to the other is exactly such a shear, so it never changes the determinant.
Things to notice
- With Shear construction on, A does not change. The pushed copy is yellow, while A’s own parallelogram stays as a faint dashed outline.
- When det A = 0 but A is not zero, the solid line is where the whole plane lands and the dashed line is what lands on O. They can be the same line, as for A = (1 −1; 1 −1).