19. Determinant Formulas and Cofactors

Overview

The lecture derives the determinant from its defining properties, first expanding 2×2 and 3×3 cases into signed products and then generalizing to a sum of n! terms indexed by permutations. It defines cofactors as signed determinants of (n−1)×(n−1) submatrices, uses them to expand determinants along rows or columns, and derives the recurrence A_n = A_(n−1) − A_(n−2) for tridiagonal matrices of ones, whose determinants repeat with period six.

Key takeaways

The determinant of an n×n matrix is a signed sum of n! products, each selecting one entry from every row and every column.
A cofactor Cᵢⱼ equals (−1)^(i+j) times the determinant obtained by deleting row i and column j.
Cofactor expansion along any row or column reduces an n×n determinant to determinants of size n−1, and sparse rows or columns can make the expansion especially efficient.
The 4×4 sparse example has only two nonzero permutation terms, +1 and −1, so its determinant is 0 and its rows satisfy row 1 − row 2 + row 3 − row 4 = 0.
For tridiagonal matrices with 1s on the main and adjacent diagonals, determinants obey det(A_n) = det(A_(n−1)) − det(A_(n−2)) and repeat every six terms.

Chapters

0:00 Deriving the 2×2 Determinant from the Three Properties

The determinant of the identity matrix is 1, and swapping its rows changes the determinant to −1.
Linearity in each row splits a matrix with entries a, b, c, d into four simpler determinants.
Terms with a zero column vanish; the surviving diagonal and swapped-row terms give ad − bc.

6:00 Finding the Six Signed Products in a 3×3 Determinant

Splitting each of three rows into three single-entry vectors produces 3³ = 27 terms, but only terms using one entry from every row and column survive.
Each surviving term corresponds to a permutation of columns 1, 2, and 3, giving 3! = 6 products.
The sign of each product is determined by the parity of row exchanges needed to restore the columns to order; the resulting formula has three positive and three negative terms.
Diagonal direction shortcuts that work for the 3×3 case do not generalize reliably, since the counter-diagonal term in a 4×4 case is positive.

16:40 The General Determinant Formula: A Signed Sum over Permutations

For an n×n matrix, choose one entry from each row while using every column exactly once; there are n! such choices.
Each term is a product of entries a_i,σ(i), multiplied by the sign of the column permutation σ.
For the identity matrix, only the identity permutation contributes a nonzero term, so the formula gives determinant 1.
The formula is explicit but can be expensive to calculate directly, since the number of terms grows factorially.

22:30 A Sparse 4×4 Example: Two Nonzero Terms Cancel

The 4×4 determinant formula contains 24 terms, but zeros in the example matrix eliminate 22 of them.
The two surviving products are each 1: one has positive sign and the other negative sign, so the determinant is 0.
The zero determinant signals singularity; the rows satisfy row 1 − row 2 + row 3 − row 4 = 0.
Elimination or finding a row or column dependence can verify singularity without evaluating all 24 permutation terms.

28:45 Grouping the 3×3 Formula by Entries in the First Row

Collecting the six 3×3 permutation terms according to their first-row entry groups the determinant into three contributions.
The terms containing a₁₁ combine into a₁₁(a₂₂a₃₃ − a₂₃a₃₂), the determinant left after deleting row 1 and column 1.
Grouping terms containing a₁₂ or a₁₃ produces the corresponding remaining 2×2 determinants, with their signs preserved.
Each first-row entry is multiplied by a signed determinant of the submatrix left after removing its row and column.

30:00 Cofactors: Signed Minors of Size n−1

For entry aᵢⱼ, remove row i and column j; the determinant of the remaining (n−1)×(n−1) matrix is its minor.
The cofactor Cᵢⱼ is the minor multiplied by (−1)^(i+j), incorporating the sign required by the permutation formula.
Cofactor signs form a checkerboard pattern: row 1 begins positive, row 2 begins negative, and signs alternate across each row.
For a 3×3 determinant, the cofactor of a₁₂ is −(a₂₁a₃₃ − a₂₃a₃₁), while the cofactors of a₁₁ and a₁₃ have positive signs.

35:30 Cofactor Expansion Along Any Row or Column

Expanding along row 1 gives det(A) = a₁₁C₁₁ + a₁₂C₁₂ + … + a₁ₙC₁ₙ.
The same expansion works along any row or column, reducing an n×n determinant to smaller determinants.
For a 2×2 matrix, expansion along the first row gives ad + b(−c), recovering ad − bc.
Cofactor expansion can recursively reduce a determinant to 1×1 cases, while the pivot formula computes it as the product of pivots after elimination.

44:30 Using Cofactors on Tridiagonal Matrices of Ones

Define A_n as the n×n tridiagonal matrix with 1s on the main diagonal and adjacent diagonals, and 0s elsewhere.
The first cases are det(A₁) = 1, det(A₂) = 0, and det(A₃) = −1.
For A₃, subtracting row 3 from row 2 creates zeros that make a cofactor expansion straightforward; the resulting determinant is −1.
Expanding A₄ along its first row leaves determinants of A₃ and A₂, with signs that give det(A₄) = det(A₃) − det(A₂).

50:30 The Recurrence Produces a Period-Six Determinant Sequence

For every n, the tridiagonal determinants satisfy det(A_n) = det(A_(n−1)) − det(A_(n−2)).
Starting from det(A₁) = 1 and det(A₂) = 0, the sequence begins 1, 0, −1, −1, 0, 1, 1.
The values repeat with period six, so det(A₆₁) = det(A₁) = 1.
The recurrence comes from cofactor expansion and the zeros in the tridiagonal structure, avoiding a full n! permutation calculation.

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