det(AB) = det(A) det(B), almost for free
The multiplicativity of det is the deepest determinant identity, yet uniqueness makes it almost a one-liner. Fix B and define f(A) = det(AB). Check that f is, as a function of the columns of A, multilinear and alternating (because multiplying by B on the right is linear and preserves equal-column relations). So by uniqueness f(A) = c * det(A) for the constant c = f(I) = det(B). Therefore det(AB) = det(A) det(B).
Two corollaries fall out immediately. If A is invertible, det(A) det(A^-1) = det(I) = 1, so det(A^-1) = 1/det(A). And det(A) = 0 exactly when A is singular — the volume collapses iff the map is non-invertible. The volume picture from Guide 3 makes both feel inevitable.
det(A^T) = det(A)
From the Leibniz formula, det(A) = sum_sigma sgn(sigma) prod_j a_{sigma(j), j}. In the transpose, entry (i,j) becomes a_{ji}, so det(A^T) = sum_sigma sgn(sigma) prod_j a_{j, sigma(j)}. Reindex each product by tau = sigma^-1: the product is unchanged as a set of factors, and sgn(sigma) = sgn(sigma^-1). Summing over sigma is the same as summing over tau, so the two sums are equal. Hence det(A^T) = det(A).
This is the theorem promised in Guide 1: everything we proved about columns now holds verbatim for rows. Row operations affect det exactly as the corresponding column operations do.
Laplace expansion and the cofactor matrix
Group the Leibniz sum by which entry sits in row i. The terms with a_ij factor out a_ij times a sum over permutations fixing that choice — which is, up to a sign (-1)^{i+j}, the determinant of the (n-1)x(n-1) minor M_ij obtained by deleting row i and column j. Define the cofactor C_ij = (-1)^{i+j} det(M_ij). Then det(A) = sum_j a_ij C_ij: this is Laplace expansion along row i, now derived, not decreed.
Assemble the cofactors into the cofactor (adjugate) matrix: adj(A) = (C_ij)^T, the transpose of the cofactor array. The key identity is A * adj(A) = det(A) * I. The diagonal entries are Laplace expansions; the off-diagonal ones are expansions of a matrix with a repeated row, hence 0 by the alternating rule. This single identity sets up the inverse formula and Cramer's rule in Guide 5.
3x3 cofactor expansion along row 1: det(A) = a11*C11 + a12*C12 + a13*C13 C11 = +det[a22 a23; a32 a33] C12 = -det[a21 a23; a31 a33] C13 = +det[a21 a22; a31 a32] Adjugate identity (the off-diagonal zeros are the magic): A * adj(A) = det(A) * I => if det(A) != 0: A^-1 = adj(A) / det(A)