LINEAR-ALGEBRA · Unit 18 · Video 1 · Interactive Practice

How Three Simple Rules Force a Formula with n! Terms

IKey Formulas

Formula	Name	Description
$\det(I) = 1$	Property 1 (Normalization)	Determinant of the identity matrix is 1
Row swap $\Rightarrow$ sign flip	Property 2 (Antisymmetry)	Exchanging any two rows negates the determinant
Linear in each row	Property 3 (Multilinearity)	Scalars factor out; row sums split the determinant
$\det(A) = \displaystyle\sum_{\sigma \in S_n} (-1)^{\mathrm{inv}(\sigma)} \prod_{i=1}^{n} a_{i,\sigma(i)}$	The Big Formula	Sum over all $n!$ permutations; sign from inversion count

IIVisualization 1 — The 2×2 Determinant

A $2\times2$ determinant is exactly two signed permutation terms: $a_{11}a_{22}$ (even, $+$ ) minus $a_{12}a_{21}$ (odd, $-$ ).

a₁₁ 3 a₁₂ 1 a₂₁ 2 a₂₂ 4

💡 Challenge: Make $\det = 0$ by giving the matrix two equal rows ( $a_{11}=a_{21}$ and $a_{12}=a_{22}$ ) — the two terms cancel and the rows are linearly dependent.

IIIVisualization 2 — All Six Permutation Terms

A $3\times3$ determinant sums $3! = 6$ signed products — one permutation per term, its sign fixed by inversion parity.

Matrix Terms shown 6

IVVisualization 3 — How Zeros Kill Terms

A permutation term dies the instant its path meets a zero, so zeros thin the $4! = 24$ terms fast.

💡 Challenge: Press upper triangular — only the identity permutation survives, which is why a triangular matrix's determinant is just the product of its diagonal.

VQuiz Questions

Question 1 · Counting Determinant Terms

The determinant of an $n \times n$ matrix is a sum of $n!$ terms in the big formula.

How many terms are in the determinant formula for a 5×5 matrix?

✅ Correct! The determinant has exactly 5! = 120 terms, one per permutation.

❌ Not quite. The key insight: each term corresponds to a permutation, and there are n! permutations of n objects.

Show solution

Solution:

The big formula sums over all permutations of $\{1, 2, 3, 4, 5\}$ . The number of permutations is:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Why the other options are wrong:

$5^2 = 25$ is the number of matrix entries, not terms
$2^5 = 32$ is unrelated to the determinant formula
$5^5 = 3125$ is the number of column-choice candidates before filtering to permutations (each of 5 rows picks from 5 columns freely)

Question 2 · Permutation Sign

The permutation $\sigma = (2, 3, 1)$ assigns row 1 $\to$ column 2, row 2 $\to$ column 3, row 3 $\to$ column 1.

What is the sign $(-1)^{\mathrm{inv}(\sigma)}$ of this permutation?

Recall: An inversion is a pair of positions where the larger column number appears before the smaller one.

✅ Correct! Two inversions (even parity), so the sign is +1.

❌ Not quite. Count inversions carefully: the pairs (2,1) and (3,1) are the only ones — that's 2 inversions, which is even.

Show solution

Solution:

For $\sigma = (2, 3, 1)$ , check all pairs of positions:

Pair of positions	Values	Inversion?
(pos 1, pos 2)	2 vs 3	No ( $2 < 3$ )
(pos 1, pos 3)	2 vs 1	Yes ( $2 > 1$ )
(pos 2, pos 3)	3 vs 1	Yes ( $3 > 1$ )

Total inversions: 2 (even)

$(-1)^2 = +1$

Cross-check by sorting: $(2,3,1) \xrightarrow{\text{swap 3 and 1}} (2,1,3) \xrightarrow{\text{swap 2 and 1}} (1,2,3)$ . Two swaps $\Rightarrow$ even $\Rightarrow$ sign $= +1$ . ✓

Question 3 · The Anti-Identity Trap

The 3×3 anti-identity matrix has $\det = -1$ . The 4×4 anti-identity has $\det = +1$ .

What is the determinant of the 5×5 anti-identity matrix?

(The anti-identity has 1s on the anti-diagonal and 0s elsewhere. The general formula is $\det = (-1)^{\lfloor n/2 \rfloor}$ .)

✅ Correct! ⌊5/2⌋ = 2 swaps (even parity), so det = +1.

❌ Not quite. Careful — don't extrapolate from 3×3 (which gives −1). Apply the formula: det = (−1)^⌊n/2⌋.

Show solution

Solution:

For the $n \times n$ anti-identity, the only surviving permutation is $(n, n{-}1, \ldots, 2, 1)$ .

For $n = 5$ , sort $(5, 4, 3, 2, 1)$ back to $(1, 2, 3, 4, 5)$ :

Swap $5 \leftrightarrow 1$ : $(1, 4, 3, 2, 5)$
Swap $4 \leftrightarrow 2$ : $(1, 2, 3, 4, 5)$ ✓

That's $\lfloor 5/2 \rfloor = 2$ swaps (even), so $\det = (-1)^2 = +1$ .

$n$	$\lfloor n/2 \rfloor$	$\det$
2	1	$-1$
3	1	$-1$
4	2	$+1$
5	2	$+1$
6	3	$-1$

⚠️ Trap: Don't assume the sign pattern from $n = 2, 3$ (both $-1$ ) continues! Always compute $\lfloor n/2 \rfloor$ .

Question 4 · True or False — Zero Column

Statement: If an $n \times n$ matrix has an entire column of zeros, then every permutation term in the determinant formula equals zero, and therefore $\det(A) = 0$ .

✅ Correct! Every permutation must pick one entry from the zero column, so every term contains a factor of zero.

❌ Not quite. Think carefully: every permutation assigns exactly one row to each column. If a column is all zeros, what happens to the product?

Show solution

Solution: True.

Every permutation $\sigma$ assigns exactly one row to each column. If column $j$ is all zeros, then for any permutation $\sigma$ , there exists some row $i$ with $\sigma(i) = j$ , so $a_{i,j} = 0$ .

Since the product $\prod_{k=1}^n a_{k,\sigma(k)}$ includes the factor $a_{i,j} = 0$ , the entire product is zero.

This holds for all $n!$ permutations, so every term is zero and $\det(A) = 0$ .

This is exactly the mechanism we saw in the 2×2 derivation: candidates that picked the same column twice left the other column empty (all zeros), and those terms died.

Solved: 0 / 4