3.4.1 Determinants as volume scaling

How a matrix changes area

A $2 \times 2$ matrix $A$ maps the plane to itself (lesson 3.2.2). Its first column is where $\mathbf{e}_1 = (1, 0)$ lands and its second is where $\mathbf{e}_2 = (0, 1)$ lands. So the unit square, with corners $\mathbf{0}$ , $\mathbf{e}_1$ , $\mathbf{e}_1 + \mathbf{e}_2$ and $\mathbf{e}_2$ , lands on the parallelogram whose sides are the two columns.

Take

A = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}

The square lands on the parallelogram with sides $(3, 1)$ and $(1, 2)$ . Its fourth corner is $(3, 1) + (1, 2) = (4, 3)$ .

To find its area, draw the box around it, 4 wide and 3 tall, and cut away the six pieces outside it:

piece	area
the box, 4 by 3	12
two triangles along the long sides	$1.5 + 1.5$
two triangles along the short sides	$1 + 1$
two corner squares	$1 + 1$
the parallelogram	$12 - 7 = 5$

The matrix $A$ sends the unit square to a parallelogram with sides its columns. The box around it has area 12, the six pieces outside it 7, so the parallelogram has area 5.

Every grid square lands on a copy of this parallelogram, so $A$ multiplies every area by 5. A triangle of area 2 lands on one of area 10.

The same counting with letters gives the area for any $2 \times 2$ matrix. It is $ad - bc$ , the number lesson 3.2.3 divides by to find the inverse. For $A$ , $3 \cdot 2 - 1 \cdot 1 = 5$ .

Rule: Determinant of a 2 × 2 matrix

\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc

The determinant is the signed area of the parallelogram the unit square lands on. The map multiplies every area by $\lvert \det A \rvert$ , so the determinant is a ratio: area out per unit of area in.

When the entries are plain numbers, so is the determinant. If every entry is in dollars, $ad - bc$ is in dollars squared, so changing the units changes it; case study 3.4.3 shows why that matters.

Area can stay the same while the shape changes. The stretch $\begin{pmatrix} 2 & 0 \\ 0 & 0.5 \end{pmatrix}$ turns the square into a 2 by 0.5 rectangle: $\det = 2 \cdot 0.5 - 0 \cdot 0 = 1$ . The shear $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ leans it over with the same base and height: $\det = 1$ .

(0, 0)

import numpy as np def det2(a, b, c, d): return a * d - b * c print(det2(3, 1, 1, 2)) print(4 * 3 - (1.5 + 1.5 + 1 + 1 + 1 + 1)) A = np.array([[3, 1], [1, 2]]) print(round(np.linalg.det(A), 10)) square = np.array([[0, 1, 1, 0], [0, 0, 1, 1]]) print(A @ square)

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The sign: flipped or not

Swap the columns of $A$ and the parallelogram is the same shape, yet $\det \begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix} = 1 \cdot 1 - 3 \cdot 2 = -5$ . The sign records the orientation: the direction you turn to go from the first column to the second.

For the unit square, turning from $\mathbf{e}_1$ to $\mathbf{e}_2$ is anticlockwise. Under $A$ , turning from $(3, 1)$ to $(1, 2)$ is anticlockwise too, and $\det A = 5 > 0$ . With the columns swapped, the turn from $(1, 2)$ to $(3, 1)$ is clockwise: the plane has been flipped over, as in a mirror.

matrix	$\det$	what happens to the unit square
$A$	5	area times 5, not flipped
$A$ with its columns swapped	$-5$	area times 5, flipped
stretch, $\begin{pmatrix} 2 & 0 \\ 0 & 0.5 \end{pmatrix}$	1	a 2 by 0.5 rectangle
quarter turn, $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$	1	turned, not flipped
reflection, $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$	$-1$	flipped, same size
$Z = \begin{pmatrix} 1 & 2 \\ 0.5 & 1 \end{pmatrix}$	0	flattened onto a line

The last row is lesson 3.2.3's singular matrix $Z$ . Its second column, $(2, 1)$ , is twice its first, $(1, 0.5)$ , so the square is squashed onto a line and has no area left.

Rule: What the sign says

\begin{gathered} \det A > 0: \text{orientation kept} \\ \det A < 0: \text{flipped over} \\ \det A = 0: \text{flattened} \end{gathered}

The change in area is $\lvert \det A \rvert$ in every case. A negative determinant is not a negative area: it is a positive area, flipped.

The explorer starts on $Z$ , the "Flatten" matrix, with its square squashed flat. Raise $d$ from 1 to 1.5 and the square opens to area 0.5; lower it to 0.5 and the determinant is $-0.5$ , a flip. Then try the reflection and the quarter turn.

Explore

Pick a matrix or set its entries. Its first column is where e₁ lands and its second where e₂ lands; the whole grid follows. Then drag the probe x and look for directions that A only stretches.

a (row 1, column 1) = 1

b (row 1, column 2) = 2

c (row 2, column 1) = 0.5

d (row 2, column 2) = 1

Probe x = (1, 2) (drag its tip)Show eigenvector lines

Matrix: $A = \begin{pmatrix} 1 & 2 \\ 0.5 & 1 \end{pmatrix}$
Columns: Ae₁ = (1, 0.5), Ae₂ = (2, 1)
det A: 1 × 1 − 2 × 0.5 = 0: the plane is flattened onto a line (or a point), so A has no inverse.
Probe: Ax = (5, 2.5): a different direction.
Eigenvalues: λ = 2 and 0

Try it Each line prints a matrix of the table and its determinant. Adding 0.0 turns a -0.0 into 0.0, as in lesson 3.2.3.

import numpy as np examples = { "A": [[3, 1], [1, 2]], "columns swapped": [[1, 3], [2, 1]], "stretch": [[2, 0], [0, 0.5]], "shear": [[1, 1], [0, 1]], "quarter turn": [[0, -1], [1, 0]], "reflection": [[1, 0], [0, -1]], "Z": [[1, 2], [0.5, 1]], } for name, M in examples.items(): print(name, round(np.linalg.det(np.array(M)), 10) + 0.0)

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Three by three: the volume of a box

A $3 \times 3$ matrix maps space to itself. The unit cube, with edges $\mathbf{e}_1$ , $\mathbf{e}_2$ and $\mathbf{e}_3$ , lands on a slanted box whose edges are the three columns. The determinant is the signed volume of that box.

Take

B = \begin{pmatrix} 2 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}

Its first two columns, $(2, 0, 0)$ and $(1, 2, 0)$ , lie flat on the floor, where the third component is 0. They span a parallelogram of area $2 \cdot 2 - 1 \cdot 0 = 4$ . The third column, $(1, 1, 2)$ , rises to height 2, so the box has volume $4 \times 2 = 8$ .

$B$ sends the unit cube to a slanted box with edges its columns. Its floor has area 4 and its height is 2, so its volume is 8.

Most boxes do not sit so neatly on the floor. For any $3 \times 3$ matrix, the determinant is built from $2 \times 2$ ones.

Write $a_{ij}$ for the entry in row $i$ and column $j$ . Its minor $M_{ij}$ is the determinant of what is left after deleting row $i$ and column $j$ .

Rule: Cofactor expansion along row 1

\begin{gathered} \det A = a_{11}M_{11} - a_{12}M_{12} \\ + a_{13}M_{13} \end{gathered}

Each entry of the first row times its minor, with signs $+$ , $-$ , $+$ . A signed minor is called a cofactor. The result is the signed volume of the box the unit cube lands on.

Take

C = \begin{pmatrix} 2 & 1 & 3 \\ 0 & 4 & 1 \\ 1 & 0 & 2 \end{pmatrix}

entry $a_{1j}$	sign	minor $M_{1j}$	term
2	$+$	$4 \cdot 2 - 1 \cdot 0 = 8$	16
1	$-$	$0 \cdot 2 - 1 \cdot 1 = -1$	1
3	$+$	$0 \cdot 0 - 4 \cdot 1 = -4$	$-12$

So $\det C = 16 + 1 - 12 = 5$ . The middle term is subtracted, and its minor is negative, so it adds 1.

Any row or column works, with the signs of a checkerboard:

\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}

Row 2 of $C$ is $(0, 4, 1)$ , with signs $-$ , $+$ , $-$ . Its minors are $M_{22} = 2 \cdot 2 - 3 \cdot 1 = 1$ and $M_{23} = 2 \cdot 0 - 1 \cdot 1 = -1$ , and the 0 saves working out $M_{21}$ . So $\det C = -0 + 4 \cdot 1 - 1 \cdot (-1) = 5$ again.

Expand along the row or column with the most zeros.

For $B$ , the first column has two zeros: $\det B = 2 \times (2 \cdot 2 - 1 \cdot 0) = 8$ . $B$ is upper triangular, with zeros below the diagonal, and its determinant is the product of the diagonal, $2 \cdot 2 \cdot 2$ .

As in the plane, a negative $3 \times 3$ determinant means the map turns space into its mirror image. The volume is still $\lvert \det A \rvert$ .

C

import numpy as np def det3(M): (a, b, c), (d, e, f), (g, h, i) = M return a * (e * i - f * h) - b * (d * i - f * g) + c * (d * h - e * g) B = [[2, 1, 1], [0, 2, 1], [0, 0, 2]] C = [[2, 1, 3], [0, 4, 1], [1, 0, 2]] print(det3(B), det3(C)) for j in range(3): minor = np.delete(np.delete(np.array(C), 0, axis=0), j, axis=1) print(j + 1, C[0][j], round(np.linalg.det(minor), 10) + 0.0) print(np.linalg.det(np.array(C)), round(np.linalg.det(np.array(C)), 10))

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Determinant 0: no inverse

A determinant is 0 exactly when the box is flat: the columns lie in one plane, or on one line. Then some column is a combination of the others, and the columns are called dependent (lesson 3.5.1 studies this). Elimination finds fewer than $n$ pivots, so the rank is below $n$ (lesson 3.3.2).

A flattened space cannot be unflattened, so $A$ has no inverse: it is singular, as lesson 3.2.3 called $Z$ . The inverse formula of lesson 3.2.3 divides by $ad - bc$ , which is 0 here.

Rule: The invertibility test

\det A \neq 0 \iff A^{-1} \text{ exists}

For a square $n \times n$ matrix, four statements go together: $\det A \neq 0$ , independent columns, rank $n$ , and exactly one solution of $A\mathbf{x} = \mathbf{b}$ for every $\mathbf{b}$ . When $\det A = 0$ , $A\mathbf{x} = \mathbf{b}$ has no solution or infinitely many.

A $3 \times 3$ example:

S = \begin{pmatrix} 1 & 3 & 4 \\ 3 & 1 & 4 \\ 1 & 6 & 7 \end{pmatrix}

Its third column is the sum of the first two, so the box is flat. Along row 1, the minors are $-17$ , 17 and 17, and $1 \cdot (-17) - 3 \cdot 17 + 4 \cdot 17 = 0$ . Its rank is 2.

On a computer, np.linalg.det(S) prints a tiny number, here $7.5 \times 10^{-15}$ , not 0; the exact digits depend on the computer's maths library. As lesson 3.3.2 found for rank, rounding hides zeros: elimination in decimals leaves a tiny number where a 0 pivot should be. Ask np.linalg.matrix_rank instead, which allows for rounding.

Z

import numpy as np Z = np.array([[1, 2], [0.5, 1]]) S = np.array([[1, 3, 4], [3, 1, 4], [1, 6, 7]]) print(np.linalg.det(Z), np.linalg.matrix_rank(Z)) print(S[:, 0] + S[:, 1] - S[:, 2]) print(f"{np.linalg.det(S):.1e}", np.linalg.matrix_rank(S))

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Rules for products, transposes and rows

Doing one map after another multiplies their area factors: the product $AB$ is " $B$ , then $A$ " (lesson 3.2.2). So $\det(AB) = \det A \cdot \det B$ .

With $A$ as above, $\det A = 5$ , and $B = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ , $\det B = 3$ . Both $AB$ and $BA$ have determinant 15, although $AB \neq BA$ (lesson 3.2.1). There is no rule for sums: $A + B = \begin{pmatrix} 4 & 3 \\ 1 & 5 \end{pmatrix}$ has determinant 17, not $5 + 3$ .

Two more rules follow. $A^{-1}$ undoes $A$ , so it divides every area by $\det A$ : $\det C^{-1} = \frac{1}{5}$ . And $\det A^\top = \det A$ : rows and columns play the same part, so each rule for rows below holds for columns too.

Row operations (lesson 3.3.1) change the determinant in set ways. Starting from $C$ , with $\det C = 5$ :

change to $C$	new $\det$
swap rows 1 and 3	$-5$
multiply row 2 by 3	15
subtract 2 × row 3 from row 1	5
multiply every entry by 2	$2^3 \times 5 = 40$

Multiplying every entry by 2 multiplies each of the 3 rows by 2, so the determinant grows by $2 \times 2 \times 2$ . In the box picture, every edge doubles and the volume grows 8 times.

Rule: Determinant rules

\begin{gathered} \det(AB) = \det A \cdot \det B \\ \det A^\top = \det A \\ \det A^{-1} = 1 / \det A \\ \det(cA) = c^n \det A \end{gathered}

For an $n \times n$ matrix. A row swap flips the sign, and multiplying one row by $c$ multiplies the determinant by $c$ . Adding a multiple of one row to another changes nothing.

\det(AB)

import numpy as np def det(M): return round(np.linalg.det(np.array(M)), 10) + 0.0 A = np.array([[3, 1], [1, 2]]) B = np.array([[1, 2], [0, 3]]) print(det(A @ B), det(B @ A), det(A) * det(B), det(A + B)) C = np.array([[2, 1, 3], [0, 4, 1], [1, 0, 2]]) print(det(C), det(C.T), det(np.linalg.inv(C))) tripled = C.copy() tripled[1] = 3 * C[1] added = C.copy() added[0] = C[0] - 2 * C[2] print(det(C[[2, 1, 0]]), det(tripled), det(added), det(2 * C))

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How a computer finds a determinant

Elimination (lesson 3.3.1) uses only swaps and "add a multiple of one row to another". The second kind leaves the determinant alone, and each swap flips its sign. Elimination ends at a triangular matrix, whose determinant is the product of its diagonal, the pivots.

For $C$ : subtract $0.5 \times$ row 1 from row 3, giving $(0, -0.5, 0.5)$ . Then add $0.125 \times$ row 2, giving $(0, 0, 0.625)$ . The pivots are 2, 4 and 0.625, and $2 \times 4 \times 0.625 = 5$ .

np.linalg.det works this way, in decimals, which is where its rounding comes from. A small determinant alone does not mean a matrix is nearly singular. Take $0.001\,I$ , where $I$ is the $3 \times 3$ identity: its determinant is $10^{-9}$ , yet its columns point along the three axes and its rank is 3.

C

import numpy as np U = np.array([[2, 1, 3], [0, 4, 1], [1, 0, 2]], dtype=float) U[2] = U[2] - 0.5 * U[0] print(U[2]) U[2] = U[2] + 0.125 * U[1] print(U[2]) print(U[0, 0] * U[1, 1] * U[2, 2]) print(f"{np.linalg.det(0.001 * np.eye(3)):.1e}", np.linalg.matrix_rank(0.001 * np.eye(3)))

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Case study 3.4.3 uses determinants to spot redundant assets, and lesson 3.6.1 finds eigenvalues by solving $\det(A - \lambda I) = 0$ .

Common mistakes

Multiply along the diagonals. $\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$ , not $ab - cd$ .
The sign is orientation, not negative area. A determinant of $-5$ means area times 5, flipped. The area is $\lvert \det A \rvert$ .
Mind the cofactor signs. Along row 1 they are $+$ , $-$ , $+$ ; along row 2, $-$ , $+$ , $-$ . Forgetting the minus on the middle term is the usual slip.
Scaling a matrix scales its determinant by $c^n$ . $\det(2C) = 8 \det C$ for a $3 \times 3$ matrix, not $2 \det C$ . Scaling one row multiplies by $c$ once.
There is no rule for sums. $\det(A + B)$ is not $\det A + \det B$ ; only products split.
Only square matrices have determinants. A $3 \times 2$ matrix has a rank but no determinant.
Do not test det == 0 on a computer. Rounding leaves tiny numbers such as $10^{-15}$ ; use matrix_rank. And a tiny determinant can come from small units, as for $0.001\,I$ .

Determinants

0 of 12 correct

1.
Find $\det \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$
2.
The unit square is mapped by $\begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix}$ . What is the area of its image?
3.
A matrix has determinant $-2$ . It maps a triangle of area 3. What is the area of the image?
4.
Find the determinant of $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ , which swaps the two axes
5.
For which $k$ is $\begin{pmatrix} 2 & k \\ 3 & 6 \end{pmatrix}$ singular?
6.
Find $\det \begin{pmatrix} 1 & 2 & 0 \\ 3 & 1 & 2 \\ 0 & 1 & 1 \end{pmatrix}$ by cofactor expansion along row 1
7.
A box has edges $(3, 0, 0)$ , $(1, 2, 0)$ and $(0, 1, 4)$ . What is its volume?
8.
$\det A = 4$ and $\det B = -2$ . Find $\det(AB)$
9.
$A$ is $3 \times 3$ with $\det A = 5$ . Find $\det(2A)$
10.
$\det A = 7$ . Swap two rows of $A^\top$ . What is the new determinant?
11.
$\det A = 4$ . Find $\det A^{-1}$
12.
A $3 \times 3$ matrix has determinant 0, and two of its columns are not multiples of each other. What is its rank?

How to type answers

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\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc

Determinants as volume scaling

How a matrix changes area

Rule: Determinant of a 2 × 2 matrix

The sign: flipped or not

Rule: What the sign says

Three by three: the volume of a box

Rule: Cofactor expansion along row 1

Determinant 0: no inverse

Rule: The invertibility test

Rules for products, transposes and rows

Rule: Determinant rules

How a computer finds a determinant

Common mistakes

Determinants

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