Cramer's rule, read off the adjugate
From A * adj(A) = det(A) I we get the explicit inverse A^-1 = adj(A)/det(A) whenever det(A) != 0. Apply it to Ax = b: x = A^-1 b, and the i-th component works out to Cramer's rule: x_i = det(A_i)/det(A), where A_i is A with its i-th column replaced by b. Each unknown is a ratio of two determinants — a closed form, no elimination.
There is a slick volume proof too: replace column i of A by b = sum x_j c_j. By multilinearity, det(A_i) = sum_j x_j det(.. with column i = c_j ..). Every term with j != i has a repeated column and vanishes; only j = i survives, giving det(A_i) = x_i det(A). Divide — uniqueness and alternating do all the work.
The determinant of an operator
A linear transformation T: V -> V has many matrices, one per basis, related by change of basis B = P^-1 A P. By multiplicativity, det(B) = det(P^-1) det(A) det(P) = det(A), since det(P^-1) det(P) = 1. The number is basis-independent, so we may define the determinant of an operator det(T) as the determinant of any matrix representing it.
This makes det a genuine invariant of T, alongside the trace. The product of the eigenvalues equals det(T) and their sum equals the trace — the det-trace identities you can read straight off the characteristic polynomial as its constant term and (signed) second coefficient.
Three famous determinants
The theory gives clean formulas. The Vandermonde determinant of nodes x_1,...,x_n equals product over i<j of (x_j - x_i); it is nonzero exactly when the nodes are distinct, which is why polynomial interpolation has a unique solution. The Wronskian is the determinant of a function-and-derivatives matrix; a nonzero Wronskian certifies that solutions of a differential equation are linearly independent.
Vandermonde (3 nodes):
V = det[1 x1 x1^2; 1 x2 x2^2; 1 x3 x3^2]
= (x2 - x1)(x3 - x1)(x3 - x2)
zero <=> two nodes coincide <=> columns dependent
Wronskian of f, g at a point:
W = det[f g ; f' g'] = f*g' - f'*g
W != 0 somewhere => f, g linearly independentThe payoff: why a top-form exists at all
Step back. We assumed a normalized alternating multilinear form exists and is unique — but WHY is the space of such forms exactly one-dimensional? The structural answer is the top exterior power. The space of alternating n-forms on an n-dimensional V is the dual of the top exterior power Lambda^n(V), and dim Lambda^n(V) = C(n,n) = 1. A one-dimensional space has, up to scale, exactly one nonzero element — that is the uniqueness of Guide 2, now explained, not just proved.
From this height the whole track snaps together. An operator T induces a map on the one-dimensional Lambda^n(V); a linear map on a 1D space IS a scalar — and that scalar is exactly det(T). Multiplicativity becomes the functoriality Lambda^n(ST) = Lambda^n(S) Lambda^n(T). The determinant is not a clever formula bolted onto matrices; it is the unique top-dimensional volume form, and every rule you proved is its shadow.