Skip to main content
Mathematics
LINEAR ALGEBRA · MEASURING WHAT VECTORS ENCLOSE

Determinants: area & volume

Stretch a photograph to twice its width and its area doubles. That is easy to picture. What if we slant it instead? An edge can become longer without the area increasing. Judging by what “looks bigger” is no longer reliable enough.

We need a comparison we can calculate. Recall matrices and linear transformations: a matrix is a table of numbers specifying how points move. Its two columns tell us where the arrows for the two unit directions go. Let those arrows bound a square with sides of length 1. After the transformation, it usually becomes a parallelogram. The original area was 1, so the new area tells us directly how much areas have been enlarged or reduced.

Try changing one column and notice which edge it moves. Then bring the two edges onto the same line. The edges are still there, but the enclosed area has disappeared. We will work out how ad − bc records these changes, keeping area size separate from a reversal of orientation. The picture will help us see why the subtraction is needed.

01Why ad − bc? Watch the area stay put.

Start both vectors at the origin. Complete their opposite sides to enclose a parallelogram. Read the matrix down its columns: u=(a,c)u=(a,c) and v=(b,d)v=(b,d). Dashed component lines show exactly which horizontal and vertical displacement each letter means.

0%
xy0bc = 1d = 2u (3, 1)v (1, 2)a = 3c = 1b = 1d = 2
Area throughout5
ad − bc6 − 1 = 5

Original matrix and columns

A=(abcd)=(3112)A=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}3&1\\1&2\end{pmatrix}
u=(ac)\mathbf u=\begin{pmatrix}a\\c\end{pmatrix}v=(bd)\mathbf v=\begin{pmatrix}b\\d\end{pmatrix}

Here a, b, c and d are positive and ad > bc. We start with this simple case so each area can be seen directly.

h=d−bcah=d-\frac{bc}{a}
S=ah=a(d−bca)=ad−bcS=ah=a\left(d-\frac{bc}{a}\right)=ad-bc

Both rearrangements are shears: they slide parallel strips without stretching those strips. That is why we can measure the final rectangle instead of the original parallelogram. The small strokes merely make the sliding visible; the map applies to the whole continuous region.

What if we use the rows (a,b) and (c,d) instead? Making these the new columns exchanges b and c. This is called transposing the matrix, written Aᵀ. Apply the same two-by-two formula: the result is ad−cb. Since cb=bc, it equals ad−bc. Thus this row construction encloses the same area, even though its shape may differ. We have checked this fact directly for the two-by-two case used here.

AT=(acbd)det⁡AT=ad−cb=ad−bc=det⁡A\begin{gathered}A^T=\begin{pmatrix}a&c\\b&d\end{pmatrix}\\\begin{aligned}\det A^T&=ad-cb\\&=ad-bc\\&=\det A\end{aligned}\end{gathered}

02Zero area means one direction was lost.

The determinant is a signed area. Its absolute value is the geometric area; its sign tells the orientation of the ordered pair u, v. Now bring v onto the same line as u. The parallelogram becomes thinner until it has no area.

0%
xy0u (3, 1)v (1, 2)
det A5
Geometric area |det A|5
A=(3112),det⁡A=5A=\begin{pmatrix}3&1\\1&2\end{pmatrix},\quad\det A=5
Independent vectors open up a 2D area. Approaching the same line reduces the perpendicular height, while the first vector stays fixed.
v=12u⟹det⁡A=0\mathbf v=\tfrac12\mathbf u\quad\Longrightarrow\quad\det A=0

This is linear dependence: one vector can be made by scaling the other. A zero vector also creates dependence. For a nonzero determinant, turning from u to v through the smaller angle is counterclockwise for a positive sign and clockwise for a negative sign. A negative determinant does not mean negative geometric area.

Area=∣ad−bc∣\text{Area}=|ad-bc|

03In three dimensions, area becomes volume.

Use three vectors from the same origin. Their sums give the eight corners of a parallelepiped: a slanted box. These three vectors are the columns of a 3×33 \times 3 matrix. The absolute value of its determinant is the volume of that box. Drag the model to look around it.

0%
Drag to orbit · x, y, z are fixed reference axes
det A6.12
Geometric volume |det A|6.12
At the endpointPlane
A=(20.50.401.70.4001.8)A=\begin{pmatrix}2&0.5&0.4\\0&1.7&0.4\\0&0&1.8\end{pmatrix}

u · column 1　v · column 2　w · column 3

Keep u and v independent. Bring w into their plane until w = 0.6u + 0.5v. No pair is parallel, yet the set of three is dependent: w contributes no third direction.

Initially the base lies in the xy-plane: base area = 2×1.7=3.42 \times 1.7 = 3.4, perpendicular height = 1.8, and volume = 3.4×1.8=6.123.4 \times 1.8 = 6.12. A sloping edge is not the perpendicular height.

V=Sbase h=∣det⁡A∣V=S_{\rm base}\,h=|\det A|

Dependence means a missing dimension.

Vectors are linearly dependent if a combination of them equals zero with coefficients that are not all zero. Equivalently, at least one vector can be expressed using the others. In 2D, two dependent vectors lie on a line (or both are zero). In 3D, three dependent vectors lie in a plane, a line, or reduce to the origin.

Let us explain the missing-direction statement before extending it. For two vectors, keep a nonzero first vector as the base. Subtract from the second vector its part parallel to the base. This slides sideways without changing the perpendicular height or area. The remaining part is exactly that height, with a sign indicating which side it lies on. It is zero precisely when the second vector was a multiple of the first. If the first vector was zero, the area was already zero.

For three vectors, first do the same within the base made by the first two. If the base has collapsed, the box has zero volume. Otherwise subtract from the third vector its part in the base plane. Sliding each horizontal slice does not change its area, so the box keeps its volume. The remaining perpendicular length is its height. A nonzero base times this height is zero exactly when the third vector lies in the base plane, meaning it is a combination of the first two.

In n dimensions we define volume recursively by the same rule: the previous-dimensional base volume times the next perpendicular height. The determinant is its signed version, with the usual unit-coordinate box assigned +1. At each stage, subtract the new vector’s projections onto the earlier perpendicular unit directions. A projection here means the component along a direction; along unit u, it is (v⋅u) u(v\cdot u)\,u. The remainder is perpendicular because its dot product with each earlier u is v⋅u−v⋅u=0v\cdot u-v\cdot u=0.

∣det⁡A∣=h1h2⋯hn|\det A|=h_1h_2\cdots h_n

Each h is the nonnegative length of one remainder. The product vanishes exactly when at least one height is zero. At that stage, the new column is made entirely from earlier columns. Conversely, if the columns are dependent, take the first column that can be made from earlier ones; its remainder is zero. With no such column, every remainder has positive length and their product is positive. This proves both directions of the claim. A zero first column counts as an immediate zero height.

det⁡A=0⟺columns of A are linearly dependent\det A=0\quad\Longleftrightarrow\quad\text{columns of A are linearly dependent}

For an n×nn \times n square matrix, the same idea measures n-dimensional signed volume. If the columns fail to supply all nn independent directions, that n-dimensional volume is zero. A nonzero determinant means all nn directions are present.

Revisit matrices as transformations →
Keep going

What would you like to do next?

Determinants
Determinants
Next in MathematicsFour fundamental subspacesStill unsure about a step?Ask about it in the forumAll 28 modelsBack to Mathematics

Sign in to see more

Sign in to continue the rest of this lesson and keep your progress in sync.

Sign in / Sign up