Unit 5, lesson 24

Determinants as Oriented Area and Volume

How the determinant measures oriented area and volume, why a 2 by 2 determinant is ad minus bc, how its sign records orientation, and why a zero determinant means dependent columns.

  • Video 24
  • Area, Volume and Orientation

Apply the matrix with columns (2,0)(2, 0) and (1,3)(1, 3) to the unit square, where a tuple such as (2,0)(2, 0) stands for a column vector. The square becomes a parallelogram of area 66. Now swap the two columns. The new matrix sends the square onto exactly the same parallelogram, yet it is a different map, because it lists the two edges in the opposite order. This lesson builds one number, the determinant, that records the area together with that order, derives the formula ad−bcad - bc, and explains what a negative or a zero determinant says about the map.

Key terms

Orientation

Definition An ordered pair (u,v)(\mathbf{u}, \mathbf{v}) of linearly independent vectors in R2\mathbb{R}^2 is positively oriented if the counterclockwise angle from u\mathbf{u} to v\mathbf{v} lies strictly between 00 and π\pi, and negatively oriented if it lies strictly between π\pi and 2π2\pi. The standard pair (e1,e2)(\mathbf{e}_1, \mathbf{e}_2), first right and then up, is positively oriented and is the reference orientation of the plane. In R3\mathbb{R}^3 an ordered triple (u,v,w)(\mathbf{u}, \mathbf{v}, \mathbf{w}) of linearly independent vectors is positively oriented, or right-handed, if w\mathbf{w} lies on the side of the plane span⁡{u,v}\operatorname{span}\{\mathbf{u}, \mathbf{v}\} where the right thumb points when the fingers point along u\mathbf{u} and curl toward v\mathbf{v} through the smaller angle. Otherwise the triple is negatively oriented, or left-handed. The standard triple (e1,e2,e3)(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3) is right-handed. A dependent pair or triple has no orientation.

In plain words Orientation belongs to an ordered list of directions, not to the region they span. Listing the same two independent vectors in the opposite order always gives the opposite orientation.

Example The vector (2,0)(2, 0) points right, and the counterclockwise turn from it to (1,3)(1, 3) is about 71.6∘71.6^\circ, so the pair ((2,0),(1,3))((2, 0), (1, 3)) is positively oriented. The counterclockwise turn from (1,3)(1, 3) back to (2,0)(2, 0) is about 288.4∘288.4^\circ, so the reversed pair is negatively oriented. Both pairs span the same parallelogram.

Determinant

Definition The oriented area of an ordered pair (u,v)(\mathbf{u}, \mathbf{v}) of vectors in R2\mathbb{R}^2 is the area of the parallelogram

P(u,v)={su+tv:s,t∈[0,1]}P(\mathbf{u}, \mathbf{v}) = \{ s\mathbf{u} + t\mathbf{v} : s, t \in [0, 1] \}

with a plus sign when the pair is positively oriented and a minus sign when it is negatively oriented. When u\mathbf{u} and v\mathbf{v} are dependent, the parallelogram is a segment or a point and the oriented area is 00. The determinant of a real 2×22 \times 2 matrix A=[a1a2]A = \begin{bmatrix} \mathbf{a}_1 & \mathbf{a}_2 \end{bmatrix} is the oriented area of its ordered columns (a1,a2)(\mathbf{a}_1, \mathbf{a}_2). Since Ae1=a1A\mathbf{e}_1 = \mathbf{a}_1 and Ae2=a2A\mathbf{e}_2 = \mathbf{a}_2, the image of the unit square is the parallelogram P(a1,a2)P(\mathbf{a}_1, \mathbf{a}_2), so ∣det⁡A∣\lvert \det A \rvert is its area and the sign records the order of its two edges. In the same way, the determinant of a real 3×33 \times 3 matrix is the oriented volume of its three ordered columns, the volume of the parallelepiped they span with a plus sign for a right-handed triple, a minus sign for a left-handed one, and 00 for a dependent one. Only square matrices have determinants. The next lesson defines the determinant of an n×nn \times n matrix for every nn and shows how to compute it.

In plain words The absolute value of the determinant is the area of the parallelogram that the unit square becomes. The sign says whether the two image edges keep the order of e1\mathbf{e}_1 and e2\mathbf{e}_2 or reverse it. A negative determinant does not mean a negative area.

Example The matrix

A=[2103]A = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}

has columns (2,0)(2, 0) and (1,3)(1, 3). They span a parallelogram with base 22 on the horizontal axis and height 33, so its area is 66. The pair is positively oriented, so det⁡A=6\det A = 6. Swapping the columns gives a matrix that covers the same parallelogram with the reversed pair, so its determinant is −6-6.

2-by-2 determinant formula

Definition For all real numbers a,b,c,da, b, c, d,

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

To derive it, suppose first that a≠0a \ne 0. The shear (x,y)↦(x,y−cax)(x, y) \mapsto (x, y - \tfrac{c}{a}x) slides each vertical line along itself, so it keeps areas and does not flip the plane. It sends the first column (a,c)(a, c) to (a,0)(a, 0) and the second column (b,d)(b, d) to (b,d−bca)(b, d - \tfrac{bc}{a}). The first edge now lies on the horizontal axis with signed length aa, and the second edge has signed height d−bcad - \tfrac{bc}{a}. The area is base times height, and the pair is positively oriented exactly when the signed length and the signed height have the same sign, so the oriented area is

a(d−bca)=ad−bc.a\left(d - \frac{bc}{a}\right) = ad - bc.

If a=0a = 0 and c≠0c \ne 0, the first edge (0,c)(0, c) is vertical. Sliding the second edge parallel to it, from (b,d)(b, d) to (b,0)(b, 0), keeps the oriented area, and the pair ((0,c),(b,0))((0, c), (b, 0)) has oriented area −bc-bc, which is ad−bcad - bc when a=0a = 0. For instance, when bb and cc are positive, the first edge points up and the second points right, a clockwise quarter turn, so the oriented area −bc-bc is negative. If a=c=0a = c = 0, the first column is zero and both sides equal 00.

In plain words Multiply down the main diagonal, then subtract the product of the other diagonal. The order of the subtraction matters, and the result is the oriented area of the columns, not a product of their lengths.

Example For the matrix with columns (1,2)(1, 2) and (2,1)(2, 1),

det⁡[1221]=1⋅1−2⋅2=−3.\begin{aligned} \det \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} &= 1 \cdot 1 - 2 \cdot 2 \\[0.4em] &= -3. \end{aligned}

Every entry is positive, but the counterclockwise turn from (1,2)(1, 2) to (2,1)(2, 1) is more than a half turn, so the pair is negatively oriented and spans area 33. For the shear with columns (1,0)(1, 0) and (s,1)(s, 1), the formula gives 1⋅1−s⋅0=11 \cdot 1 - s \cdot 0 = 1 for every real ss. The image of the unit square leans further as ∣s∣\lvert s \rvert grows, but its area stays 11.

Volume scale factor

Definition Let AA be a real n×nn \times n matrix with n=2n = 2 or n=3n = 3. For every bounded region RR in Rn\mathbb{R}^n that has an area (when n=2n = 2) or a volume (when n=3n = 3), the image A(R)={Ax:x∈R}A(R) = \{A\mathbf{x} : \mathbf{x} \in R\} satisfies

vol⁡(A(R))=∣det⁡A∣vol⁡(R),\operatorname{vol}(A(R)) = \lvert \det A \rvert \operatorname{vol}(R),

where vol⁡\operatorname{vol} means area in the plane. The number ∣det⁡A∣\lvert \det A \rvert is the volume scale factor of AA, called the area scale factor in the plane. For n≥4n \ge 4 the same number is the scale factor of nn-dimensional volume.

In plain words A linear map multiplies every area or volume by the same factor, the absolute value of the determinant. Cover a region with tiny squares or cubes. Each one goes to a tiny parallelogram or parallelepiped whose size is ∣det⁡A∣\lvert \det A \rvert times as large, so the whole region is scaled by that factor too. The factor alone does not determine individual lengths.

Example The diagonal matrix D=diag⁡(2,3,−1)D = \operatorname{diag}(2, 3, -1) sends the unit cube to a box with edge lengths 22, 33 and 11, so its volume scale factor is 66. The third edge now points down instead of up, which makes the column triple left-handed, so det⁡D=−6\det D = -6. A solid ball of radius 11, with volume 43π\tfrac{4}{3}\pi, goes to a solid ellipsoid of volume 8π8\pi. In the plane, the matrix diag⁡(2,12)\operatorname{diag}(2, \tfrac{1}{2}) has determinant 11 and keeps every area, yet it doubles the length of e1\mathbf{e}_1 and halves the length of e2\mathbf{e}_2.

Orientation-reversing map

Definition Let A=[a1a2]A = \begin{bmatrix} \mathbf{a}_1 & \mathbf{a}_2 \end{bmatrix} be an invertible real 2×22 \times 2 matrix. The map x↦Ax\mathbf{x} \mapsto A\mathbf{x} is orientation-reversing if it sends the positively oriented pair (e1,e2)(\mathbf{e}_1, \mathbf{e}_2) to a negatively oriented pair (a1,a2)(\mathbf{a}_1, \mathbf{a}_2), that is, if det⁡A<0\det A < 0. It is orientation-preserving if det⁡A>0\det A > 0. An orientation-reversing map sends every positively oriented pair (u,v)(\mathbf{u}, \mathbf{v}) to a negatively oriented pair (Au,Av)(A\mathbf{u}, A\mathbf{v}), because

det⁡[AuAv]=det⁡A det⁡[uv],\begin{aligned} &\det \begin{bmatrix} A\mathbf{u} & A\mathbf{v} \end{bmatrix} \\[0.4em] &= \det A \, \det \begin{bmatrix} \mathbf{u} & \mathbf{v} \end{bmatrix}, \end{aligned}

a case of the product rule proved in the next lesson. The same definitions apply to an invertible 3×33 \times 3 matrix, with right-handed and left-handed triples. A singular matrix neither keeps nor reverses orientation, since its image vectors are dependent.

In plain words An orientation-reversing map turns the plane over, like a mirror. Its negative determinant records the reversed order of the edges, not a negative area. Reversing one axis changes the orientation, and reversing a second axis changes it back.

Example Reflection across the vertical axis has columns (−1,0)(-1, 0) and (0,1)(0, 1) and determinant −1-1. It keeps every area and reverses orientation. The map x↦−x\mathbf{x} \mapsto -\mathbf{x} of the plane, with columns (−1,0)(-1, 0) and (0,−1)(0, -1), reverses both axes. Its determinant is (−1)(−1)−0⋅0=1(-1)(-1) - 0 \cdot 0 = 1, and it is the half turn about the origin, which keeps orientation.

Zero determinant and dependent columns

Definition A real 2×22 \times 2 matrix AA has det⁡A=0\det A = 0 if and only if its columns are linearly dependent. In that case AA sends the whole plane into a line through the origin, or onto the origin alone, and the image of the unit square has area 00. This follows from the definition, since independent columns span a genuine parallelogram with positive area. Likewise, a real 3×33 \times 3 matrix has determinant 00 exactly when its columns are dependent, and then it sends space into a plane, a line or the single point 0\mathbf{0}, and the image of the unit cube has volume 00.

In plain words A zero determinant means collapse. The map loses at least one dimension, and no area or volume survives. The determinant reports only that full dimension is lost, not how much of the space remains.

Example The matrix with columns (1,2)(1, 2) and (2,4)(2, 4) has determinant 1⋅4−2⋅2=01 \cdot 4 - 2 \cdot 2 = 0, since the second column is twice the first. Every output is

x[12]+y[24]=(x+2y)[12],x\begin{bmatrix} 1 \\ 2 \end{bmatrix} + y\begin{bmatrix} 2 \\ 4 \end{bmatrix} = (x + 2y)\begin{bmatrix} 1 \\ 2 \end{bmatrix},

a multiple of (1,2)(1, 2), so the image is a line. The zero matrix also has determinant 00, but its image is the single point 0\mathbf{0}.

Check yourself

  1. Let AA be the matrix with columns (1,2)(1, 2) and (4,3)(4, 3). Find det⁡A\det A, decide whether AA keeps or reverses orientation, and find the area of the image of the triangle with vertices (0,0)(0, 0), (2,0)(2, 0) and (0,1)(0, 1).
Show the answer

The matrix is

A=[1423],A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix},

so det⁡A=1⋅3−4⋅2=−5\det A = 1 \cdot 3 - 4 \cdot 2 = -5. The sign is negative, so AA reverses orientation. The triangle has area 11, so its image has area ∣−5∣⋅1=5\lvert -5 \rvert \cdot 1 = 5. As a check, the image has vertices (0,0)(0, 0), (2,4)(2, 4) and (4,3)(4, 3), and half of ∣2⋅3−4⋅4∣\lvert 2 \cdot 3 - 4 \cdot 4 \rvert is 55.

  1. For which numbers kk does the matrix with columns (k,8)(k, 8) and (2,k)(2, k) have determinant 00? For each such kk, describe the image of the plane.
Show the answer

The determinant is k⋅k−2⋅8=k2−16k \cdot k - 2 \cdot 8 = k^2 - 16, which is zero for k=4k = 4 and k=−4k = -4. For k=4k = 4 the columns are (4,8)(4, 8) and (2,4)(2, 4), the first is twice the second, and the image is the line spanned by (1,2)(1, 2). For k=−4k = -4 the columns are (−4,8)(-4, 8) and (2,−4)(2, -4), the first is −2-2 times the second, and the image is the line spanned by (1,−2)(1, -2). In both cases the plane collapses onto a line, not onto a point, because the columns are not zero.

  1. The map x↦−x\mathbf{x} \mapsto -\mathbf{x} of the plane reverses both axes and keeps orientation. Does the map x↦−x\mathbf{x} \mapsto -\mathbf{x} of R3\mathbb{R}^3 keep orientation?
Show the answer

No. Its matrix is −I=diag⁡(−1,−1,−1)-I = \operatorname{diag}(-1, -1, -1). The unit cube goes to a cube of volume 11, and all three edges are reversed. Two reversals cancel, but the third remains, so det⁡(−I)=(−1)3=−1\det(-I) = (-1)^3 = -1 and the map reverses orientation while keeping every volume. Indeed the image triple (−e1,−e2,−e3)(-\mathbf{e}_1, -\mathbf{e}_2, -\mathbf{e}_3) is left-handed, since the right thumb for −e1-\mathbf{e}_1 and −e2-\mathbf{e}_2 points along e3\mathbf{e}_3, away from −e3-\mathbf{e}_3.

How this lesson connects

It also uses areas of rectangles, triangles and parallelograms.