4.1 The Determinant as Area

The determinant of the $2 \times 2$ matrix with columns $(a, c)$ and $(b, d)$ is the number $\det = ad - bc$ . The definition earns its keep through two facts you can watch. First, $|\det|$ is the area of the image of the unit square — and therefore the factor by which the transformation multiplies the area of any region, since any region is well approximated by a fine mosaic of little squares, each of which is scaled by the same factor. Second, the sign of $\det$ records whether the plane kept its orientation or was flipped over.

The figure below is built to be driven through the interesting event. It starts at $a = 2$ , $b = 1$ , $c = 1$ , $d = 1$ , where $\det = 2 \cdot 1 - 1 \cdot 1 = 1$ : a sheared parallelogram with exactly the area of the gray unit square beneath it.

A flat parallelogram in the xy-plane with its two origin edges drawn in red (first matrix column) and blue (second column), over a faint gray unit square. Four sliders a, b, c, d set the matrix entries: dragging them changes the parallelogram's area, collapses it to a segment when ad - bc reaches zero, and turns it inside out when ad - bc goes negative. — The image parallelogram with red edge $(a, c)$ and blue edge $(b, d)$ , over a faint gray unit square for scale. The starting entries $a = 2$ , $b = 1$ , $c = 1$ , $d = 1$ give determinant $1$ ; dragging $d$ downward drives the determinant through zero and out the other side.

1

Determinant at the starting entries ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The determinant at the figure's starting entries, computed by a computer algebra system: the matrix with columns $(2, 1)$ and $(1, 1)$ has determinant exactly $1$ — a violently sheared parallelogram covering precisely one gray square of area.

Explore the figure

At the defaults, compare the parallelogram with the gray unit square. The shapes differ, but the areas match: $\det = 1$ . A matrix can distort violently while preserving area exactly.
Drag $d$ up to $2$ . Now $\det = 2 \cdot 2 - 1 \cdot 1 = 3$ ; judge by eye that the parallelogram holds about three gray squares.
Drag $d$ slowly down to $0.5$ . Watch the parallelogram thin and die: at $d = 0.5$ the blue edge $(1, 0.5)$ is exactly half the red edge $(2, 1)$ , the columns are dependent, and $\det = 2(0.5) - 1 = 0$ .
Keep going to $d = 0$ . The parallelogram reopens — but flipped. Before the collapse, sweeping from the red edge to the blue edge went one way around; now it goes the other. Here $\det = -1$ : area $1$ again, orientation reversed.

Read the number while you drag

Everything above you judged by eye: three gray squares, about half, flipped. The determinant is a number, though, and the second figure prints it — the same four entries on sliders, in the plane the parallelogram actually lives in, with $\det$ updating as you drag. Use it to check the estimates you just made.

Interactive figureThe determinant is the signed area of the image of the unit squareDrag the 4 sliders below the figure.

XYZ Graph · viewer build 5edf91b

The image of the unit square, drawn flat. The dashed square is where the shape started; the red arrow is the first column

(a, c)

and the blue arrow the second column

(b, d)

. The readout gives

\det

live, together with the area factor and the orientation.

At the starting entries the readout says $\det = 1$ — the number you took on trust in step 1 above. The parallelogram and the dashed unit square have the same area.
Drag $d$ to $2$ and read $\det = 3$ . Compare with your by-eye estimate of "about three squares": how close were you?
Drag $d$ down through $0.5$ slowly and watch the readout rather than the shape. It reaches exactly $0$ at the instant the parallelogram becomes a segment, and the readout says so in words instead of leaving you to judge a flat sliver.
Continue to $d = 0$ . The readout now reports a negative determinant and orientation reversed — the same flip you saw, stated as the sign of a number.

Sign is orientation

Why does the sign change mean a flip? For edge vectors $(a, c)$ and $(b, d)$ , the quantity $ad - bc$ is positive exactly when the second edge lies counterclockwise from the first — the same relative position $\mathbf{e}_2$ holds to $\mathbf{e}_1$ . A positive-determinant matrix moves both basis vectors while preserving that relative arrangement; a negative determinant means the images are in the reversed arrangement, which no amount of rotating or stretching within the plane can produce — the plane must have been turned over. And because your slider moved continuously, the only route from $\det = 1$ to $\det = -1$ passed through $\det = 0$ : you cannot flip a page without, at some instant, seeing it edge-on.

Determinant zero is the same event Chapter 2 called dependence and Chapter 3 called collapse, viewed once more. Dependent columns span a line instead of a plane; the transformation crushes two dimensions of input into one dimension of output; distinct points are forced to share images, so no inverse transformation can exist. A matrix is invertible exactly when its determinant is nonzero — a numerical test for a geometric catastrophe.

At the figure's starting entries the test passes: $\det = 1 \ne 0$ , and the inverse exists concretely — the matrix with rows $(1, -1)$ and $(-1, 2)$ , which carries the sheared parallelogram back onto the unit square. A determinant of exactly $1$ even keeps the inverse's entries whole numbers.

\begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}

The inverse at the starting entries ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The undo matrix, computed entry by entry and checked against the rows stated above. Determinant $1$ is what keeps every entry a whole number: the inverse's entries are the original's, shuffled and signed, divided by the determinant.

Example 1 Locate the collapse exactly

With $a = 2$ , $b = 1$ , $c = 1$ held fixed, find every value of $d$ at which the image parallelogram degenerates to a segment, and compute the area scale factor at $d = 2$ and at $d = 0$ .

The determinant is $\det = ad - bc = 2d - 1$ , which vanishes only at $d = 0.5$ . There the columns are $(2, 1)$ and $(1, 0.5) = 0.5\,(2, 1)$ — dependent, so the image is the segment of the line through $(2, 1)$ , matching the collapse seen on the slider. At $d = 2$ : $\det = 3$ , so every region's area is tripled, orientation preserved. At $d = 0$ : $\det = -1$ , so areas are preserved in size but the plane is flipped; a shape and its image are mirror versions of each other. Note the two settings $d = 2$ and $d = 0$ both give $|\det| \ge 1$ , yet only one of them can be continuously undone without a collapse along the way.

-1 + 2 \cdot d

The determinant as d varies ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The whole slider experiment as one symbolic determinant: with $a = 2$ , $b = 1$ , $c = 1$ held fixed, a computer algebra system expands $\det = 2d - 1$ — the line that vanishes at $d = 0.5$ , reaches $3$ at $d = 2$ , and lands on $-1$ at $d = 0$ .

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.