What the determinant measures: area and orientation
What does the image of the unit square tell you about a 2×2 matrix?
Its area, with a minus sign when the map turns the plane over, is the determinant of the matrix. It can be read off the four entries, and it tells whether the map can be undone.
Before you start · 3 quick checks on columns, area and turning, about 2 min
From the first page on matrices: a 2×2 matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) sends \(e_1 = (1, 0)\) to its first column \(Ae_1 = (a, c)\) and \(e_2 = (0, 1)\) to its second column \(Ae_2 = (b, d)\), and it sends \(x = (x_1, x_2)\) to \(x_1Ae_1 + x_2Ae_2\).
The unit square has corners \(0\), \(e_1\), \(e_1 + e_2\) and \(e_2\); its points are \(x_1e_1 + x_2e_2\) with \(0 \le x_1 \le 1\) and \(0 \le x_2 \le 1\). So \(A\) sends it to the parallelogram with corners \(0\), \(Ae_1\), \(Ae_1 + Ae_2\) and \(Ae_2\). When \(Ae_1\) and \(Ae_2\) lie on one line through 0, this parallelogram is flat: a segment, or the single point 0.
The turn from \(Ae_1\) to \(Ae_2\): when they do not lie on one line through 0, turn the arrow \(Ae_1\) about the origin through the smaller angle (less than a half turn) until it points along \(Ae_2\). That turn is counterclockwise or clockwise. From \(e_1\) to \(e_2\) it is a counterclockwise quarter turn.
Definition. The determinant \(\det A\) is the signed area of the parallelogram that \(A\) sends the unit square to: its area, with a plus sign when the turn from \(Ae_1\) to \(Ae_2\) is counterclockwise, a minus sign when it is clockwise, and 0 when the parallelogram is flat. For the identity matrix, \(\det I = 1\).
Why the turn goes through the smaller angle
Any two arrows can be turned onto each other either way: one way through an angle \(\theta\), the other way through \(360° - \theta\). When the arrows do not lie on one line, one of these angles is less than 180° and the other is more, so turning through the smaller angle picks one direction.
When the two arrows point in opposite directions, both ways are half turns. They then lie on one line through 0, the parallelogram is flat, and its signed area is 0, so no direction is needed.
Build three parallelograms
Drag the two dots, the tips of \(Ae_1\) and \(Ae_2\), or type their coordinates in the boxes below the figure. The area measurements stay hidden until you press Measure: plan the parallelogram from the grid first.
How the tips, the grid and Measure work
The tips sit on a grid of 0.5, between −2 and 2 in both directions, so the whole parallelogram always fits in the figure. In a coordinate box the arrow keys change the value by 0.5, and a typed value moves to the nearest point of the grid.
The background squares have side 1 and area 1. Measure reports the signed area of the shaded parallelogram, its area and the turn from \(Ae_1\) to \(Ae_2\), for the tips as they are when you press it. Once you move a tip, that measurement is withdrawn. A goal is reached only when the signed area equals the goal exactly. After the three goals the numbers follow the tips as you move.
Other regions, and the sign
Two questions first. The explorer below opens when you have answered both.
The explorer opens when you have answered the two questions above.
The explorer adds a disk of radius 0.5 about the point \((-0.5, -0.5)\), of area \(\pi/4\), and draws where your map sends it. The disk is inscribed in the square with corners \(0\) and \(-e_1 - e_2\), so its image lies in the parallelogram turned a half turn about the origin. Here the numbers follow the tips as you move them.
Why the determinant is ad − bc
For every 2×2 matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\),
\[ \det A = ad - bc. \]In other words, the parallelogram with sides \(x = (x_1, x_2)\) and \(y = (y_1, y_2)\) has signed area \(x_1y_2 - x_2y_1\), for all vectors \(x\) and \(y\).
Fix a matrix \(A\). Write \(u = Ae_1 = (a, c)\) and \(v = Ae_2 = (b, d)\) for its columns, \(P\) for the parallelogram with sides \(u\) and \(v\), and \(n = (d, -b)\): the vector \(v\) turned a quarter turn clockwise. The figure shows an example. Build the proof one line at a time; each line has one correct choice.
What follows
Let \(R\) be a region whose area can be found by filling it ever more closely with small squares, such as a polygon or a disk. Then \(A\) sends \(R\) to a region of area \(|\det A| \cdot \operatorname{area}(R)\).
Reason: the square with corner \(p\) and side \(h\) consists of the points \(p + hs\), with \(s\) in the unit square, and \(A\) sends them to \(Ap + hAs\). That is the image of the unit square, shrunk by the factor \(h\) and moved to \(Ap\), so its area is \(|\det A|\,h^2\), and every small square grows by the same factor \(|\det A|\). When \(\det A \ne 0\), \(A\) can be undone (next box), so it sends different points to different points, and the images of squares that do not overlap do not overlap either. When \(\det A = 0\), everything lands on a line or a point, of area 0.
For a 2×2 matrix \(A\) with columns \(u\) and \(v\), these four statements are equivalent:
- \(\det A = 0\);
- \(u\) and \(v\) lie on one line through 0 (one of them is a multiple of the other);
- \(A\) sends the whole plane into one line through 0;
- \(A\) is not invertible: no matrix undoes it.
(1) and (2) say the same thing: the parallelogram with sides \(u\) and \(v\) has area 0 exactly when it is flat. (2) gives (3), since every \(Ax = x_1u + x_2v\) lies on that line. (3) gives (4): \(u\) and \(v\) then lie on one line, so \(su + tv = 0\) for some numbers \(s, t\) that are not both 0, and \(A\) sends the point \((s, t)\) and the point 0 to the same place; no map can send that place back to both. Finally, when \(\det A \ne 0\) the matrix
\[ A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]undoes \(A\): multiplying out gives \(A^{-1}A = AA^{-1} = I\). So (4) gives (1).
For all 2×2 matrices \(A\) and \(B\), \(\det(AB) = \det A \cdot \det B\), and when \(A\) is invertible, \(\det(A^{-1}) = 1/\det A\).
Reason: write \(D(x, y) = x_1y_2 - x_2y_1\), the signed area of the parallelogram with sides \(x\) and \(y\). For all vectors \(p\) and \(q\),
\[ D(Ap, Aq) = \det A \cdot D(p, q). \]Indeed \(Ap = p_1u + p_2v\) and \(Aq = q_1u + q_2v\); expanding \(D(Ap, Aq)\), the terms with \(p_1q_1\) and with \(p_2q_2\) cancel, and what is left is \((p_1q_2 - p_2q_1)(ad - bc)\). By matrix multiplication as composition, the columns of \(AB\) are \(A\) times the columns \(b_1\) and \(b_2\) of \(B\), so
\[ \begin{aligned} \det(AB) &= D(Ab_1, Ab_2) \\ &= \det A \cdot D(b_1, b_2) \\ &= \det A \cdot \det B. \end{aligned} \]In pictures: \(B\) multiplies areas by \(|\det B|\), then \(A\) multiplies them by \(|\det A|\), and each map that turns the plane over flips the sign once. For an invertible \(A\), \(\det A \cdot \det(A^{-1}) = \det(AA^{-1}) = \det I = 1\).
Problems
Each problem has hints and a full solution. Answers are graded exactly: type whole numbers, exact decimals such as 0.5, or fractions such as 1/2, without thousands separators. Where a problem asks for two numbers, separate them with a comma.
Next in linear algebra: Column space, null space, and rank (in preparation).
Further reading
- G. Strang, Introduction to Linear Algebra, 5th ed. (Wellesley-Cambridge Press, 2016), chapter 5 (Determinants).
- D. C. Lay, S. R. Lay and J. J. McDonald, Linear Algebra and Its Applications, 5th ed. (Pearson, 2016), chapter 3 (Determinants), section 3.3 on volume and linear transformations.
- S. Axler, Linear Algebra Done Right, 4th ed. (Springer, 2024), chapter 9 (Multilinear Algebra and Determinants).