Determinants as Area and Volume: Examples

Before this lesson: Properties of Determinants, Linear Transformations and Matrix Transformations

What the determinant measures

Put two vectors u = (a, b) and v = (c, d) in the columns of a 2 × 2 matrix. The parallelogram with sides u and v, starting at the origin, has area |ad − bc|, which is exactly |det[u v]|. The absolute value is needed because a determinant can be negative, but an area cannot.

Here is a quick check when all four numbers are positive and v lies above u. Draw the rectangle that encloses the parallelogram: it is (a + c) by (b + d). Cutting off two triangles of area ab/2, two of area cd/2 and two rectangles of area bc leaves (a + c)(b + d) − ab − cd − 2bc = ad − bc.

The deeper reason comes from the properties of determinants. Scaling one column by k scales the area by |k|. Adding a multiple of one column to another is a shear, which slides one side of the parallelogram without changing its base or height. And det I = 1 matches the unit square. Area obeys the same three rules, so the two must agree.

The columns of A are where e₁ and e₂ land, so A sends the unit square (area 1) to a parallelogram of area |det A| = |4(3) − 2(1)| = 10.

The area and volume formulas

Area in ℝ². The parallelogram with sides u = (u₁, u₂) and v = (v₁, v₂) has

\text{Area} = \left|\det\begin{bmatrix} u_1 & v_1 \\ u_2 & v_2 \end{bmatrix}\right| = |u_1 v_2 - v_1 u_2|

Volume in ℝ³. The parallelepiped (a slanted box whose six faces are parallelograms) with edges u, v and w has

\text{Volume} = \left|\det\begin{bmatrix} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{bmatrix}\right|

Scaling by a linear transformation. If T(x) = Ax and S is any region with an area (in ℝ²) or a volume (in ℝ³), then

\text{Area of } T(S) = |\det A| \cdot \text{Area of } S \qquad \text{Volume of } T(S) = |\det A| \cdot \text{Volume of } S

This works for any region, not only parallelograms: fill S with tiny squares (or cubes), and each one becomes a tiny parallelogram (or parallelepiped) whose size is multiplied by |det A|.

How to find an area or volume with a determinant

Find the edge vectors. If the shape does not start at the origin, subtract a shared corner from its two (or three) neighboring corners.
Put the edge vectors in the columns of a matrix. (Rows work too, since det Aᵀ = det A.)
Find the determinant.
Take the absolute value. For a triangle, also take half.
For a transformed region, multiply the original area or volume by |det A|. The sign of det A tells you whether the picture was flipped.

Determinants as area and volume examples

Example 1: The area of a parallelogram

Find the area of the parallelogram with sides u = (4, 1) and v = (2, 3).

\det\begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} = 4(3) - 2(1)

Put u and v in the columns and use ad − bc (see determinant of a matrix).

= 12 - 2 = 10

Multiply, then subtract.

\text{Area} = |10| = 10

Take the absolute value.

Answer: the area is 10 square units. This is the parallelogram in the figure above.

Example 2: A parallelogram that is not at the origin

A parallelogram has vertices P(1, 2), Q(6, 4), R(3, 7) and S(8, 9), where Q and R are the two vertices next to P. Find its area.

u = Q - P = (6 - 1,\ 4 - 2) = (5, 2)

Subtract the shared corner P from its neighbor Q.

v = R - P = (3 - 1,\ 7 - 2) = (2, 5)

Subtract P from its other neighbor R.

P + u + v = (1 + 5 + 2,\ 2 + 2 + 5) = (8, 9) = S

Check that the fourth corner is S, so u and v really are the sides.

\det\begin{bmatrix} 5 & 2 \\ 2 & 5 \end{bmatrix} = 5(5) - 2(2) = 21

Put u and v in the columns and find the determinant.

\text{Area} = |21| = 21

Take the absolute value.

Answer: the area is 21 square units.

Example 3: The volume of a parallelepiped

Find the volume of the parallelepiped with edges u = (2, 0, 1), v = (1, 3, 0) and w = (0, 1, 4).

A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 3 & 1 \\ 1 & 0 & 4 \end{bmatrix}

Put u, v and w in the columns.

\det A = 2\begin{vmatrix} 3 & 1 \\ 0 & 4 \end{vmatrix} - 1\begin{vmatrix} 0 & 1 \\ 1 & 4 \end{vmatrix} + 0\begin{vmatrix} 0 & 3 \\ 1 & 0 \end{vmatrix}

Expand along row 1 with the signs + − + (see cofactor expansion).

= 2(12 - 0) - 1(0 - 1) + 0

Find each 2 × 2 determinant.

= 24 + 1 = 25

Multiply, then add.

\text{Volume} = |25| = 25

Take the absolute value.

Answer: the volume is 25 cubic units.

Example 4: How a transformation changes area

Let T(x) = Ax with A = [1 3; 2 1]. Find the area of the image of the triangle with vertices (0, 0), (2, 0) and (0, 3), and say whether T flips orientation.

\text{Area of triangle} = \tfrac{1}{2}(2)(3) = 3

The triangle is a right triangle with legs 2 and 3 (see area of triangles).

\det A = 1(1) - 3(2) = -5

Use ad − bc.

\text{Area of image} = |-5| \cdot 3 = 15

Multiply the area by |det A|.

T(2, 0) = (1(2) + 3(0),\ 2(2) + 1(0)) = (2, 4)

Check: find the image of the corner (2, 0).

T(0, 3) = (1(0) + 3(3),\ 2(0) + 1(3)) = (9, 3)

Find the image of the corner (0, 3). The corner (0, 0) stays at (0, 0).

\tfrac{1}{2}\left|\det\begin{bmatrix} 2 & 9 \\ 4 & 3 \end{bmatrix}\right| = \tfrac{1}{2}|6 - 36| = 15

The image is a triangle with edge vectors (2, 4) and (9, 3); half the parallelogram area matches.

Answer: the image has area 15 square units. Since det A = −5 is negative, T flips orientation, like a mirror image.

Example 5: Volume of a crystal cell

The atoms in a crystal repeat in a slanted box called a unit cell. One unit cell has edge vectors a = (4, 0, 0), b = (1, 4, 0) and c = (1, 1, 5), in ångströms (Å). Find its volume. The crystal is then squeezed so that every point (x, y, z) moves to (x, y, 0.95z). Find the new volume.

M = \begin{bmatrix} 4 & 1 & 1 \\ 0 & 4 & 1 \\ 0 & 0 & 5 \end{bmatrix}

Put a, b and c in the columns.

\det M = 4 \cdot 4 \cdot 5 = 80

M is upper triangular, so multiply the diagonal entries.

\text{Volume} = |80| = 80

Take the absolute value. The units are cubic ångströms.

B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0.95 \end{bmatrix}

Write the squeeze as a matrix: it keeps x and y and multiplies z by 0.95.

\det B = 1 \cdot 1 \cdot 0.95 = 0.95

B is diagonal, so multiply the diagonal entries.

\text{New volume} = 0.95 \cdot 80 = 76

Multiply the volume by |det B|. This equals |det(BM)| = |det B||det M|.

Answer: the unit cell has volume 80 Å³ before the squeeze and 76 Å³ after it.

Common mistakes

Mistake	Example of the error	Fix
Leaving the sign on	Calling the area −22 when det[u v] = −22	Area and volume are \|det\|, so the area is 22. The sign only tells orientation.
Using corners instead of edge vectors	Using (6, 4) and (3, 7) as the sides of the parallelogram in Example 2	Subtract the shared corner first: u = (5, 2) and v = (2, 5).
Using a diagonal as a side	Using S − P = (7, 7) as an edge in Example 2	Use the two corners next to the shared corner, not the opposite one.
Forgetting the 1/2 for a triangle	Saying the triangle with edge vectors (4, 1) and (1, 5) has area 19	A triangle is half the parallelogram: 19/2 = 9.5.
Adding scale factors for a composition	Saying B after A scales area by 2 + 1 = 3 when det A = 2 and det B = −1	Multiply: det(BA) = det B · det A = −2, so areas double and the picture is flipped.

Related ideas

Orientation. If det A is positive, T keeps the order of the sides (counterclockwise stays counterclockwise in ℝ²). If det A is negative, T reverses it, as a reflection does. If det A = 0, T squashes the plane onto a line or a point (or space onto a plane or less), so every image has area or volume 0. This is the geometric side of the invertible matrix theorem: A is invertible exactly when it does not flatten space.

Compositions. Since det(BA) = det B · det A, doing T and then another transformation multiplies the scale factors, which fits composition of linear transformations.

The triple product and Jacobians. The scalar triple product w · (u × v) in the cross product lesson equals det[u v w], so it gives the same volume. In calculus, a nonlinear change of variables stretches each tiny piece of a region by the absolute value of a determinant, the Jacobian; see change of variables with the Jacobian.

Practice problems

Find the area of the parallelogram with sides u = (3, 1) and v = (1, 4).
Find the area of the parallelogram with sides u = (−2, 5) and v = (4, 1).
A parallelogram has vertices (0, 1), (4, 2), (1, 5) and (5, 6), where (4, 2) and (1, 5) are next to (0, 1). Find its area.
Find the area of the triangle with vertices (1, 1), (5, 2) and (2, 6).
Find the volume of the parallelepiped with edges u = (1, 2, 0), v = (0, 1, 3) and w = (2, 0, 1).
Do the vectors (1, 2, 3), (4, 5, 6) and (7, 8, 9) lie in one plane? Use a determinant.
T(x) = Ax with A = [2 5; 1 4]. Find the area of the image of a disk of radius 3.
T(x) = Ax with A = [0 2; 3 1]. A square has area 4. Find the area of its image, and say whether T flips orientation.
T(x) = Ax with A = [2 0 0; 1 3 0; 4 1 1]. Find the volume of the image of the unit ball (radius 1).
A logo has area 12 cm². An app applies A = [2 1; 0 1] and then the reflection B = [1 0; 0 −1]. Find the area of the final logo and say whether it is mirrored.

Show answers

\det\begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix} = 3(4) - 1(1) = 11

Put u and v in the columns and use ad − bc.

\text{Area} = |11| = 11

Take the absolute value.

Answer: 11 square units.

\det\begin{bmatrix} -2 & 4 \\ 5 & 1 \end{bmatrix} = -2(1) - 4(5)

Put u and v in the columns and use ad − bc.

= -2 - 20 = -22

Multiply, then subtract.

\text{Area} = |-22| = 22

Take the absolute value.

Answer: 22 square units.

u = (4 - 0,\ 2 - 1) = (4, 1)

Subtract the shared corner (0, 1) from (4, 2).

v = (1 - 0,\ 5 - 1) = (1, 4)

Subtract (0, 1) from (1, 5).

\det\begin{bmatrix} 4 & 1 \\ 1 & 4 \end{bmatrix} = 16 - 1 = 15

Find the determinant.

Answer: 15 square units. Check: (0, 1) + u + v = (5, 6), the fourth corner.

u = (5 - 1,\ 2 - 1) = (4, 1)

Subtract the corner (1, 1) from (5, 2).

v = (2 - 1,\ 6 - 1) = (1, 5)

Subtract (1, 1) from (2, 6).

\det\begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix} = 20 - 1 = 19

Find the determinant. This is the parallelogram’s area.

\text{Area} = \tfrac{1}{2}|19| = 9.5

A triangle is half the parallelogram.

Answer: 9.5 square units.

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

Put u, v and w in the columns.

\det A = 1\begin{vmatrix} 1 & 0 \\ 3 & 1 \end{vmatrix} - 0\begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} + 2\begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix}

Expand along row 1 with the signs + − +.

= 1(1 - 0) - 0 + 2(6 - 0)

Find each 2 × 2 determinant.

= 1 + 12 = 13

Multiply, then add.

Answer: 13 cubic units.

\det\begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} = 1(45 - 48) - 4(18 - 24) + 7(12 - 15)

Expand along row 1.

= -3 + 24 - 21 = 0

Multiply, then add.

Answer: yes. The determinant is 0, so the parallelepiped is flat (volume 0), and the three vectors lie in one plane through the origin.

\det A = 2(4) - 5(1) = 3

Use ad − bc.

\text{Area of disk} = \pi(3)^2 = 9\pi

Use A = πr² (see area of a circle).

\text{Area of image} = 3 \cdot 9\pi = 27\pi

Multiply by |det A|.

Answer: 27π ≈ 84.8 square units. The image is an ellipse.

\det A = 0(1) - 2(3) = -6

Use ad − bc.

\text{Area of image} = |-6| \cdot 4 = 24

Multiply the area by |det A|.

Answer: 24 square units. The determinant is negative, so T flips orientation.

\det A = 2 \cdot 3 \cdot 1 = 6

A is lower triangular, so multiply the diagonal entries.

\text{Volume of ball} = \tfrac{4}{3}\pi(1)^3 = \tfrac{4}{3}\pi

Use V = (4/3)πr³ (see volume of a sphere).

\text{Volume of image} = 6 \cdot \tfrac{4}{3}\pi = 8\pi

Multiply by |det A|.

Answer: 8π ≈ 25.1 cubic units.

10.

\det A = 2(1) - 1(0) = 2

Use ad − bc.

\det B = 1(-1) - 0(0) = -1

Use ad − bc.

\det(BA) = (-1)(2) = -2

The final matrix is BA, and det(BA) = det B · det A.

\text{Final area} = |-2| \cdot 12 = 24

Multiply the area by |det(BA)|.

Answer: 24 cm². The determinant is negative, so the final logo is mirrored.

Determinants as area and volume FAQ

Why is the determinant the area of a parallelogram?

Both obey the same rules: scaling one side by k scales each by |k|, a shear changes neither, and the unit square gives 1. Those rules pin down a single value, so |det[u v]| must equal the area.

What does a negative determinant mean geometrically?

It means the transformation reverses orientation, like a mirror. The size of the change is still |det A|; the sign only records the flip.

How do you find the volume of a parallelepiped with a determinant?

Put its three edge vectors in the columns of a 3 × 3 matrix, find the determinant and take the absolute value. If the shape does not start at the origin, subtract a shared corner first.

What does a determinant of zero mean geometrically?

The columns lie on a line (in ℝ²) or in a plane (in ℝ³), so the parallelogram or box is flat and has no area or volume. The transformation squashes space and cannot be undone.

Summary

The parallelogram with sides u and v has area |det[u v]|; the parallelepiped with edges u, v and w has volume |det[u v w]|.
If a shape does not start at the origin, subtract a shared corner to get edge vectors. A triangle is half a parallelogram.
A linear transformation T(x) = Ax multiplies every area or volume by |det A|.
The sign of det A gives orientation: positive keeps it, negative flips it, and 0 flattens space.
For a composition, det(BA) = det B · det A, so the scale factors multiply.

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