What the determinant means
A matrix moves every point of the plane at once. Is there one number that says how much it stretches or shrinks the area of every shape, and whether it flips shapes over?
Where you are 12 in Matrices
Why this exists
A Linear transformation T(\mathbf{x}) = A\mathbf{x} of the plane moves every point at once. Tshepo prints a tailor’s pattern pieces on stretchy fabric, and a press distorts the fabric by such a map: a collar piece, a round patch and a sleeve all change shape, and Dineo, who buys the cloth, wants to know how much fabric each piece covers afterward, curved pieces included.
Computing a determinant stated that the columns of A span a parallelogram of area \lvert \det A \rvert, proved there for one drawn case. Here it is proved in general and read as a fact about the map: every region, square or curved, has its area multiplied by \lvert \det A \rvert. The sign says whether the map flips the plane over, and \det A = 0 says it flattens the plane, so it cannot be undone.
Definition
The signed area S(\mathbf{u}, \mathbf{v}) of an ordered pair of vectors in \mathbb{R}^2 is the area of the parallelogram they span, with a sign. Stand at the origin facing along \mathbf{u}. If \mathbf{v} points to your left (the counterclockwise side of the line through \mathbf{u}), the sign is +; to your right, -. If \mathbf{u} and \mathbf{v} lie on one line through the origin (a zero vector included), S(\mathbf{u}, \mathbf{v}) = 0. This is the same sign as the turn rule of Computing a determinant: \mathbf{v} is on the left of \mathbf{u} exactly when the shorter turn from \mathbf{u} to \mathbf{v} is counterclockwise.
For a 2 \times 2 matrix A with columns \mathbf{a}_1, \mathbf{a}_2, the map \mathbf{x} \mapsto A\mathbf{x} sends \mathbf{e}_1, \mathbf{e}_2 to \mathbf{a}_1, \mathbf{a}_2. The determinant is the signed area scale factor of that map: the unit square, of signed area S(\mathbf{e}_1, \mathbf{e}_2) = 1, goes to the parallelogram of signed area S(\mathbf{a}_1, \mathbf{a}_2) = \det A. The map keeps orientation when \det A > 0 and reverses orientation when \det A < 0. For a 3 \times 3 matrix the same words hold with volume in place of area.
Intuition
A linear map sends equally spaced grid lines to equally spaced parallel lines, so the grid of small squares becomes a grid of identical parallelograms. Each small square is a shifted, shrunken copy of the unit square, and its image is the same shift and shrink of the big parallelogram. So every small square has its area multiplied by the same factor, 5 for the matrix in the figure.
A curved piece is almost exactly covered by small squares, each scaled by 5, so the disk of area \pi becomes an ellipse of area 5\pi.
Results
Theorem 1 (signed area). For \mathbf{u} = (a, c) and \mathbf{v} = (b, d), S(\mathbf{u}, \mathbf{v}) = ad - bc, the determinant of the matrix with columns \mathbf{u}, \mathbf{v}.
Theorem 2 (every parallelogram). For every 2 \times 2 matrix A and every pair, S(A\mathbf{u}, A\mathbf{v}) = \det A \cdot S(\mathbf{u}, \mathbf{v}).
Theorem 3 (every region). For a region R with an area, \operatorname{area}(A(R)) = \lvert \det A \rvert \operatorname{area}(R).
Theorem 4 (flattening). \det A = 0 exactly when A sends the whole plane onto a line through the origin or onto \mathbf{0}. Then some \mathbf{x} \neq \mathbf{0} has A\mathbf{x} = \mathbf{0}, and the map cannot be undone.
Theorem 5 (products). \det(AB) = \det A \det B.
In \mathbb{R}^3, \lvert \det A \rvert is the volume of the box spanned by the columns, every solid has its volume multiplied by \lvert \det A \rvert, \det A = 0 exactly when space is flattened into a plane, a line or a point, and \det(AB) = \det A \det B. These are stated here without proof; Properties of determinants: row operations, products, transposes proves the product rule for every size.
Why it is true
Three rules for signed area. Each comes from the picture.
- Swap: S(\mathbf{v}, \mathbf{u}) = -S(\mathbf{u}, \mathbf{v}). Same parallelogram, and \mathbf{v} is to the left of \mathbf{u} exactly when \mathbf{u} is to the right of \mathbf{v}.
- Scale: S(\mathbf{u}, k\mathbf{v}) = k\,S(\mathbf{u}, \mathbf{v}). The tip of k\mathbf{v} is \lvert k \rvert times as far from the line through \mathbf{u}, on the same side when k > 0 and the other side when k < 0. By the swap rule the same holds in the first slot.
- Shear: S(\mathbf{u}, \mathbf{v} + k\mathbf{u}) = S(\mathbf{u}, \mathbf{v}). Adding k\mathbf{u} slides the tip of \mathbf{v} parallel to \mathbf{u}: same base, same height, same side. By the swap rule, S(\mathbf{u} + k\mathbf{v}, \mathbf{v}) = S(\mathbf{u}, \mathbf{v}).
Lemma. Any rule F on ordered pairs obeying swap, scale and shear has F(\mathbf{u}, \mathbf{v}) = (ad - bc)\,F(\mathbf{e}_1, \mathbf{e}_2).
If a \neq 0, shear \mathbf{v} by \displaystyle -\frac{b}{a}\mathbf{u}, which leaves (0, h) with \displaystyle h = d - \frac{bc}{a} = \frac{ad - bc}{a}. Pull out h by scaling, then shear \mathbf{u} by -c\,\mathbf{e}_2 and pull out a:
and ha = ad - bc. If a = c = 0, then \mathbf{u} = \mathbf{0} = 0\,\mathbf{u}, so the scale rule gives F(\mathbf{u}, \mathbf{v}) = 0 = ad - bc. If a = 0 and c \neq 0, then \mathbf{u} = c\,\mathbf{e}_2, and shearing \mathbf{v} by \displaystyle -\frac{d}{c}\mathbf{u} = -d\,\mathbf{e}_2 leaves b\,\mathbf{e}_1:
which is (ad - bc)F(\mathbf{e}_1, \mathbf{e}_2) because a = 0.
Theorem 1 is the lemma for F = S, since S(\mathbf{e}_1, \mathbf{e}_2) = 1. This is the general proof that Computing a determinant promised.
Theorem 2. Put F(\mathbf{u}, \mathbf{v}) = S(A\mathbf{u}, A\mathbf{v}). Because A is linear, A(\mathbf{v} + k\mathbf{u}) = A\mathbf{v} + kA\mathbf{u} and A(k\mathbf{v}) = kA\mathbf{v}, so F obeys all three rules. Also F(\mathbf{e}_1, \mathbf{e}_2) = S(\mathbf{a}_1, \mathbf{a}_2) = \det A. The lemma gives F(\mathbf{u}, \mathbf{v}) = \det A \cdot S(\mathbf{u}, \mathbf{v}).
Theorem 3 is a limit argument, given here as a sketch. If \det A = 0, Theorem 4 puts A(R) on a line, of area 0. Otherwise cover the plane with a grid of squares of side s. A square \mathbf{p} + sQ, with Q the unit square, goes to A\mathbf{p} + s\,A(Q), a parallelogram of area s^2 \lvert \det A \rvert by Theorem 2. The squares inside R have images inside A(R) that overlap only along edges (a map with \det A \neq 0 sends different points to different points), and the squares touching R have images covering A(R). So \operatorname{area}(A(R)) lies between \lvert \det A \rvert times the inner total and \lvert \det A \rvert times the outer total. As s \to 0 both totals approach \operatorname{area}(R); that is what having an area means, taken here without proof.
Theorem 4. By Computing a determinant, \det A = 0 exactly when the columns are parallel. Then every A\mathbf{x} = x_1\mathbf{a}_1 + x_2\mathbf{a}_2 lies on their common line (or is \mathbf{0}). If \mathbf{a}_1 = \mathbf{0}, then A\mathbf{e}_1 = \mathbf{0}; otherwise \mathbf{a}_2 = k\mathbf{a}_1 and A(k, -1) = k\mathbf{a}_1 - \mathbf{a}_2 = \mathbf{0}. Two different inputs, that vector and \mathbf{0}, share one output, so no map can send it back. When \det A \neq 0 the map can be undone, by Inverse of a 2 by 2 matrix, and solving with it.
Theorem 5. The columns of AB are A\mathbf{b}_1 and A\mathbf{b}_2, so by Theorem 2
Read as maps (Composing transformations as a matrix product): B scales signed area by \det B, then A scales the result by \det A.
Worked examples
The press on a collar and a patch
Tshepo’s press acts by A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}, lengths in centimeters. A collar piece fills the rectangle 0 \le x \le 4, 0 \le y \le 2, and a round patch is a disk of radius 3. How much fabric does each cover after pressing?
Solution\det A = 3 \cdot 2 - 1 \cdot 1 = 5, so by Theorem 3 every area is multiplied by 5. The collar has area 8, so it covers 40 square centimeters. Check it directly: the rectangle is spanned by (4, 0) and (0, 2), which go to A(4, 0) = (12, 4) and A(0, 2) = (2, 4), and
The patch has area 9\pi and becomes an ellipse of area 45\pi, about 141.4 square centimeters. The factor belongs to the map.
A shear keeps area, not shape
Sopho’s press slants the fabric by H = \begin{bmatrix} 1 & \tfrac34 \\ 0 & 1 \end{bmatrix}. What happens to the area and the perimeter of the square piece 0 \le x \le 4, 0 \le y \le 4?
Solution\det H = 1 \cdot 1 - \tfrac34 \cdot 0 = 1, so the area stays 16. The sides go to H(4, 0) = (4, 0) and H(0, 4) = (3, 4), and
The slanted side has length \displaystyle \sqrt{3^2 + 4^2} = 5, so the perimeter grows from 16 to 2(4 + 5) = 18.
A mirror reverses orientation
A sleeve piece is the triangle P(1, 0), Q(3, 1), K(1, 2), and the fabric is printed through the reflection R = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}. Compare the triangle and its image.
SolutionThe sides from P are Q - P = (2, 1) and K - P = (0, 2), with
so the triangle has area \displaystyle \frac{4}{2} = 2 and P \to Q \to K runs counterclockwise. The images are P'(-1, 0), Q'(-3, 1), K'(-1, 2). Since R is linear, Q' - P' = R(Q - P) = (-2, 1) and K' - P' = (0, 2), and by Theorem 2 their signed area is \det R \cdot 4 = -4:
The image has the same area 2, but P' \to Q' \to K' runs clockwise: a left sleeve comes out as a right sleeve.
A press that flattens
A faulty press acts by M = \begin{bmatrix} 2 & -1 \\ -4 & 2 \end{bmatrix}. Where does it send the plane, and can its effect be undone?
Solution\det M = 4 - 4 = 0. The columns (2, -4) and (-1, 2) are parallel, the second \displaystyle -\frac{1}{2} times the first, so
and every image lies on the line y = -2x. The unit square goes to the segment from (-1, 2) to (2, -4), of area 0. Also M(1, 2) = (0, 0), so (3, 1) and (4, 3), which differ by (1, 2), both land on (5, -10). So M has no inverse, as When a matrix or transformation can be undone develops.
One press after another
Dineo first pre-stretches the cloth by B = \begin{bmatrix} 2 & 0 \\ 1 & -1 \end{bmatrix} and then runs it through Tshepo’s press A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}. Find the signed area factor of the combined map, and compare it with doing the press first.
SolutionStretch first, press second is the matrix AB. Since \det B = -2 - 0 = -2 and \det A = 5, Theorem 5 predicts \det(AB) = -10: areas grow by 10 and the cloth comes out flipped. Directly, AB = \begin{bmatrix} 7 & -1 \\ 4 & -2 \end{bmatrix} and \det(AB) = -14 + 4 = -10. The other order is BA = \begin{bmatrix} 6 & 2 \\ 2 & -1 \end{bmatrix}, with \det(BA) = -6 - 4 = -10.
AB \neq BA, yet both orders multiply signed area by (-2)(5).
Volume in three dimensions
Sopho molds a shoulder pad from a 2 \times 3 \times 4 centimeter block of foam with the map C = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 1 & 3 \\ 1 & 0 & 1 \end{bmatrix}. Find the volume of the box spanned by the columns of C, and the volume of the molded pad.
SolutionExpand along row 1, as in Computing a determinant; larger cases belong to Determinants by cofactor expansion.
The columns (2, 0, 1), (1, 1, 0), (0, 3, 1) span a box of volume 5. The block has volume 24, so by the three-dimensional statement the pad has volume 5 \cdot 24 = 120 cubic centimeters. The sign is positive, so the pad is not mirror-reversed.
Boundaries
- Only a square matrix has a determinant. A 2 \times 3 matrix maps \mathbb{R}^3 to \mathbb{R}^2, and comparing a volume with an area makes no sense.
- One factor for every region needs a linear map. A shift, A\mathbf{x} + \mathbf{b}, keeps the factor \lvert \det A \rvert; a curving map such as (x, y) \mapsto (x^2, y) stretches different places by different amounts.
- The determinant measures area or volume only. Lengths, angles and perimeters may change when \det A = 1, as in Example 2.
- Areas scale by \lvert \det A \rvert, never by a negative number; the sign is orientation.
- Theorem 3 needs a region that has an area.
- The determinant does not split over sums: \det(A + B) \neq \det A + \det B in general.
Where this goes next
When a matrix or transformation can be undone turns Theorem 4 into a test: a square matrix can be undone exactly when its determinant is not zero, one item of the The invertible matrix theorem. Determinants by cofactor expansion and Properties of determinants: row operations, products, transposes compute determinants of any size and prove the product rule algebraically. A map that flattens has a nonzero kernel, and asking when A - \lambda I flattens leads to the The characteristic equation and finding eigenvalues and Eigenvalues and eigenvectors.
Exercises
Area scale factors
- A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}. Find the area of the image of the unit square, and of a piece of area 7.
- A = \begin{bmatrix} 5 & -2 \\ 3 & 1 \end{bmatrix}. A disk of radius 2 is pressed by A. Find the area of the ellipse it becomes.
- A = \begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix}. Find the area of the image of the rectangle 0 \le x \le 3, 0 \le y \le 1 twice: from the images of its sides, and from \det A.
- A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}. Find the area of the image of the triangle with vertices (0, 0), (2, 0), (0, 4).
- Find the area factor of -2I_2, and say whether it keeps orientation.
- Find every k for which \begin{bmatrix} k & 2 \\ 1 & 3 \end{bmatrix} multiplies areas by 4.
Orientation and shape
- Which of these keep orientation? \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}, \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}
- Find S(\mathbf{u}, \mathbf{v}) for \mathbf{u} = (2, 1), \mathbf{v} = (-1, 3), then for their images under P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.
- The shear \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} acts on the square 0 \le x \le 3, 0 \le y \le 3. Find the area and perimeter of the image.
Flattening
- M = \begin{bmatrix} 6 & -9 \\ -4 & 6 \end{bmatrix}. Find \det M, a nonzero \mathbf{x} with M\mathbf{x} = \mathbf{0}, and the line every image lies on.
- Find every k for which \begin{bmatrix} k & 4 \\ k & k \end{bmatrix} flattens the plane, and the line it lands on in each case.
- Show that \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 1 & 3 & 7 \end{bmatrix} flattens space.
Products and volume
- A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} and B = \begin{bmatrix} 1 & 3 \\ 1 & 1 \end{bmatrix}. Find \det(AB), \det(BA), \det(A^2) and \det(A^3B) without multiplying out.
- A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix}. Find \det(A^{-1}), then check it from the inverse.
- Find the volume of the box spanned by (1, 0, 2), (3, 1, 0) and (0, 2, 1).
- Find the volume of the image of the unit ball under \begin{bmatrix} 1 & 0 & 2 \\ 0 & 3 & 0 \\ 1 & 0 & 4 \end{bmatrix}.
- Show that every rotation \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} has determinant 1, and say why a picture predicts it.
Reasoning
- Explain why a map with \det A = 0 cannot be undone, using what it does to the unit square.
- Show that \displaystyle \det(A^{-1}) = \frac{1}{\det A} for an invertible A, and read it as areas.
- Show that if \det(AB) = 0 then \det A = 0 or \det B = 0.
- Explain why \det(cA) = c^2 \det A for a 2 \times 2 matrix and c^3 \det A for a 3 \times 3 one.
- Refute with a counterexample: “a map with \det A = 1 keeps every length.”
Answers
- 10 and 70. \det A = 12 - 2 = 10.
- 44\pi. \det A = 5 + 6 = 11, and the disk has area 4\pi.
- 21. The sides go to (6, -3) and (1, 3), with \begin{vmatrix} 6 & 1 \\ -3 & 3 \end{vmatrix} = 21; also \det A = 7 and 7 \cdot 3 = 21.
- 28. \det A = -1 - 6 = -7, the triangle has area 4, and 7 \cdot 4 = 28. The image vertices (0, 0), (2, 6), (8, -4) give \displaystyle \frac{1}{2}\lvert -8 - 48 \rvert = 28.
- 4, and it keeps orientation. \det(-2I_2) = (-2)(-2) = 4 > 0: the map is a turn by 180^\circ with lengths doubled.
- k = 2 or \displaystyle k = -\frac{2}{3}. The factor is \lvert 3k - 2 \rvert = 4, so 3k - 2 = 4 or -4.
- The first and third. Their determinants are 1, -1, 1, -5.
- 7, then -7. \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} = 7; the images are (1, 2), (3, -1), with -1 - 6 = -7, and \det P = -1.
- Area 9, perimeter \displaystyle 6 + 6\sqrt{5}. The sides go to (3, -6), of length \displaystyle 3\sqrt{5}, and (0, 3); the determinant is 1.
- \det M = 36 - 36 = 0; \mathbf{x} = (3, 2) (or any multiple); the line \displaystyle y = -\frac{2}{3}x, along the column (6, -4).
- k = 0 or k = 4, from k^2 - 4k = 0. At k = 0 the columns (0, 0), (4, 0) put everything on the x-axis; at k = 4 both columns are (4, 4), on the line y = x.
- Along row 1, 1(7 - 12) - 2(0 - 4) + 3(0 - 1) is -5 + 8 - 3 = 0. Row 3 is row 1 plus row 2.
- -6, -6, 9, -54. \det A = 3, \det B = 1 - 3 = -2; then 3(-2), (-2)3, 3^2, 3^3(-2).
- \displaystyle \frac{1}{10}. \det A = 10; \displaystyle A^{-1} = \frac{1}{10}\begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix} has determinant \displaystyle \frac{12 - 2}{100} = \frac{1}{10}.
- 13. With those columns, along row 1: 1(1 - 0) - 3(0 - 4) + 0 = 13.
- 8\pi. Along row 2 only the 3 survives: 3(4 - 2) = 6, and \displaystyle 6 \cdot \frac{4}{3}\pi = 8\pi.
- \cos^2\theta + \sin^2\theta = 1. A rotation keeps every area and does not flip the plane, so the factor must be +1.
- The unit square goes to a segment or a point, so two different points land together (Theorem 4). An undoing map would have to send that one image back to two places.
- AA^{-1} = I_2, so \det A \det(A^{-1}) = \det I_2 = 1. If A multiplies areas by \lvert \det A \rvert, the map that undoes it must divide them by the same number.
- \det A \det B = 0, and a product of two real numbers is 0 only when one of them is. In maps: if neither flattens, doing one after the other cannot flatten.
- cA multiplies every column by c, and the scale rule pulls one c out of each column: two in the plane, three in space. Lengths scaled by c scale areas by c^2 and volumes by c^3.
- \begin{bmatrix} 2 & 0 \\ 0 & \tfrac12 \end{bmatrix} has determinant 1, yet it sends \mathbf{e}_1, of length 1, to (2, 0), of length 2.