Cramer's Rule - Proof

Proof

The proof for Cramer's rule uses just two properties of determinants: linearity with respect to any given column (taking for that column a linear combination of column vectors produces as determinant the corresponding linear combination of their determinants), and the fact that the determinant is zero whenever two columns are equal (which is implied by the basic property that the determinant is alternating in the columns).

Fix the index j of a column. Linearity means that if we consider only column j as variable (fixing the others arbitrarily), the resulting function Rn → R (assuming matrix entries are in R) can be given by a matrix, with one row and n columns. In fact this is precisely what Laplace expansion does, writing det(A) = C₁a_1,j + … + C_na_n,j for certain coefficients C₁,…,C_n that depend on the columns of A other than column j (the precise expression for these cofactors is not important here). The value det(A) is then the result of applying the one-line matrix L_(j) = (C₁ C₂ … C_n) to column j of A. If L_(j) is applied to any other column k of A, then the result is the determinant of the matrix obtained from A by replacing column j by a copy of column k, which is 0 (the case of two equal columns).

Now consider a system of n linear equations in n unknowns, whose coefficient matrix is A, with det(A) assumed to be nonzero:

If one combines these equations by taking C₁ times the first equation, plus C₂ times the second, and so forth until C_n times the last, then the coefficient of x_j will become C₁a_1,j + … + C_na_n,j = det(A), while the coefficients of all other unknowns become 0; the left hand side becomes simply det(A)x_j. The right hand side is C₁b₁ + … + C_nb_n, which is L_(j) applied to the column vector b of the right hand sides b_i. In fact what has been done here is multiply the matrix equation A ⋅ x = b on the left by L_(j). Dividing by the nonzero number det(A) one finds the following equation, necessary to satisfy the system:

But by construction the numerator is determinant of the matrix obtained from A by replacing column j by b, so we get the expression of Cramers rule as necessary condition for a solution. The same procedure can be repeated for other values of j to find values for the other unknowns.

The only point that remains to prove is that these values for the unknowns, the only possible ones, do indeed together form a solution. But if the matrix A is invertible with inverse A−1, then x = A−1 ⋅ b will be a solution, thus showing its existence. To see that A is invertible when det(A) is nonzero, consider the n by n matrix M obtained by stacking the one-line matrices L_(j) on top of each other for j = 1, 2, …, n (this gives the adjugate matrix for A). It was shown that L_(j) ⋅ A = (0 … 0 det(A) 0 … 0) where det(A) appears at the position j; from this it follows that M ⋅ A = det(A)I_n. Therefore

completing the proof.