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Linear Algebra
Step 11 of 22

The determinant

Understand first: linear transformations

A transformation moves everything. Some spread space out, some crush it together. It would be useful to have a single number for how much.

Watch the shaded square. It starts as the unit square on ı^\hat{\imath} and ȷ^\hat{\jmath}, area 1. Under the transformation it becomes a parallelogram, and its area rises and falls.

Computing it, and checking it

For a 2×22 \times 2 matrix the formula is

det⁡[abcd]=ad−bc\det \begin{bmatrix} a & b \ c & d \end{bmatrix} = ad - bc

Take [3002]\begin{bmatrix} 3 & 0 \ 0 & 2 \end{bmatrix}: the determinant is 3(2)−0(0)=63(2) - 0(0) = 6. Check it against the picture. This stretches 3 wide and 2 tall, so the unit square becomes a 3×23 \times 2 rectangle — area 6. The formula agrees with what you can see.

The number applies to everything, not just that one square. If the unit square sextuples, so does any shape you draw, because the grid it sits on was scaled uniformly.

Zero is the interesting case

Take [1224]\begin{bmatrix} 1 & 2 \ 2 & 4 \end{bmatrix}: 1(4)−2(2)=01(4) - 2(2) = 0.

Area zero means the parallelogram has no area at all — it has been flattened onto a line. Every point in the plane now lands on that single line, and a whole dimension is gone.

That collapse is irreversible, and it is the source of most of the trouble ahead: once many inputs land on the same output, nothing can tell them apart again.

Negative means flipped

det⁡[−1001]=−1\det \begin{bmatrix} -1 & 0 \ 0 & 1 \end{bmatrix} = -1.

Areas do not go negative. The sign is reporting orientation: space has been turned over, like a sheet of paper flipped face down. ȷ^\hat{\jmath} used to sit counter-clockwise from ı^\hat{\imath}, and now it sits clockwise. The magnitude still gives the area factor; the sign tells you the mirror was involved.

In three dimensions the same number measures volume instead — worth seeing before we chase down what that collapse actually destroys.