Linear-Algebra-Done-Right · Unit 28 · Video 5 · Interactive Practice

The Determinant Formula: A Signed Sum over Permutations

IKey Formulas

Formula	Name	What it says
$(v_1, \dots, v_n) \mapsto \det\begin{pmatrix} v_1 & \cdots & v_n \end{pmatrix}$	Determinant of the columns	The alternating $n$ -linear form on $\mathbf{F}^n$ equal to $1$ on $e_1, \dots, e_n$
$\det A = \displaystyle\sum_{(j_1, \dots, j_n) \in \operatorname{perm} n} \operatorname{sign}(j_1, \dots, j_n)\, A_{j_1, 1} \cdots A_{j_n, n}$	The permutation formula	One entry from each column, rows named by the permutation
$\det A = A_{1,1} A_{2,2} - A_{2,1} A_{1,2}$	$n = 2$	Two permutations: $(1, 2)$ and $(2, 1)$
$\det A = \lambda_1 \cdots \lambda_n$	Upper triangular, diagonal $\lambda_1, \dots, \lambda_n$	Every other term picks an entry below the diagonal

Key Insight: the formula has $n!$ terms, already $3{,}628{,}800$ for $n = 10$ . It is a tool for proofs, not for computation.

IISix Signed Products

Each permutation of $1, 2, 3$ names one row in each column; the determinant is the signed sum of the six products.

Matrix Selected entry 2

IIIWhy Upper Triangular Leaves One Term

Every permutation other than the identity has some $j_k > k$ , which picks a zero below the diagonal.

IVQuiz Questions

Problem 1 · A 2-by-2 Determinant

Given: $A = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}$ . Find $\det A$ .

✅ Correct!

5 \cdot 4 - 3 \cdot 2 = 14

❌ Not quite. The permutation

(2, 1)

has one inversion, so its term

A_{2,1}A_{1,2}

is subtracted.

❌ Not quite. The identity term

A_{1,1}A_{2,2}

has sign

+1

❌ Not quite.

A

is not triangular, so the term of

(2, 1)

is not zero.

❌ Not quite. Use

\det A = A_{1,1}A_{2,2} - A_{2,1}A_{1,2}

Show solution

The two permutations of $1, 2$ are $(1, 2)$ with sign $+1$ and $(2, 1)$ with sign $-1$ :

\det A = A_{1,1}A_{2,2} - A_{2,1}A_{1,2} = 5 \cdot 4 - 3 \cdot 2 = 20 - 6 = 14.

Problem 2 · One Term of a 3-by-3 Determinant

Given: a $3 \times 3$ matrix $A$ . Which signed product is the term of the permutation $(2, 3, 1)$ in the formula for $\det A$ ?

✅ Correct! Column

k

contributes the entry in row

j_k

, and

(2, 3, 1)

has two inversions, so the sign is

+1

❌ Not quite.

(2, 3, 1)

has two inversions (

1

comes after

2

and after

3

), so its sign is

+1

❌ Not quite. That reads the permutation as column indices. In

A_{j_k, k}

the permutation gives the rows;

A_{1,2}A_{2,3}A_{3,1}

is the term of

(3, 1, 2)

❌ Not quite.

A_{1,1}A_{2,3}A_{3,2}

belongs to the permutation

(1, 3, 2)

, and its sign is

-1

Show solution

The term of $(j_1, j_2, j_3)$ is $\operatorname{sign}(j_1, j_2, j_3)\, A_{j_1, 1} A_{j_2, 2} A_{j_3, 3}$ . With $(j_1, j_2, j_3) = (2, 3, 1)$ the entries are $A_{2,1}$ , $A_{3,2}$ , $A_{1,3}$ .

Inversions of $(2, 3, 1)$ : the pairs $\{1, 2\}$ and $\{1, 3\}$ , so $N = 2$ and the sign is $+1$ . The term is $+A_{2,1}A_{3,2}A_{1,3}$ .

Problem 3 · An Upper Triangular Matrix

Given: $A = \begin{pmatrix} 2 & 7 & -1 & 4 \\ 0 & -3 & 5 & 1 \\ 0 & 0 & 1 & 8 \\ 0 & 0 & 0 & 2 \end{pmatrix}$ . Find $\det A$ .

✅ Correct!

2 \cdot (-3) \cdot 1 \cdot 2 = -12

; the entries above the diagonal play no part.

❌ Not quite.

2

is the sum of the diagonal entries; the determinant is their product.

❌ Not quite. Keep the sign of the diagonal entry

-3

❌ Not quite. Only the identity term survives, and its product contains no zero.

Show solution

For any permutation other than the identity, some $j_k > k$ , so the factor $A_{j_k, k}$ lies below the diagonal and is $0$ . Only the identity term survives, with sign $+1$ :

\det A = A_{1,1}A_{2,2}A_{3,3}A_{4,4} = 2 \cdot (-3) \cdot 1 \cdot 2 = -12.

Problem 4 · Operating on the Columns

Given: $v_1, v_2, v_3 \in \mathbf{F}^3$ with $\det\begin{pmatrix} v_1 & v_2 & v_3 \end{pmatrix} = 4$ . Find $\det\begin{pmatrix} v_2 & v_1 & 3v_3 + v_1 \end{pmatrix}$ .

✅ Correct!

3\det(v_2, v_1, v_3) + \det(v_2, v_1, v_1) = 3 \cdot (-4) + 0 = -12

❌ Not quite. Swapping the first two columns multiplies the determinant by

-1

❌ Not quite. Linearity in column

3

pulls out the factor

3

❌ Not quite. Only the

v_1

part of column

3

repeats a column; the

3v_3

part does not.

❌ Not quite. Use that the determinant is an alternating

3

-linear form in the columns.

Show solution

The determinant is linear in column $3$ :

\det(v_2, v_1, 3v_3 + v_1) = 3\det(v_2, v_1, v_3) + \det(v_2, v_1, v_1).

The second term has two equal columns, so it is $0$ . Swapping the first two columns changes the sign: $\det(v_2, v_1, v_3) = -\det(v_1, v_2, v_3) = -4$ . So the answer is $3 \cdot (-4) = -12$ .

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