Cramer's Rule
Theorem 1. (Cramer's Rule) Let Ax=bAx=b be a system of nn linear equations in nn unknowns, where x=(x1,...,xn)tx=(x1,...,xn)t.
If det(A)≠0det(A)≠0, then this system has a unique solution, and xk=det(Mk)det(A),∀k∈{1,...,n},xk=det(Mk)det(A),∀k∈{1,...,n}, where Mk∈Mn×n(F)Mk∈Mn×n(F) obtained from AA by replacing column kk of AA by bb.
Proof. Let y∈Fny∈Fn, and let denote yk=det(Mk)det(A),∀k∈{1,...,n}yk=det(Mk)det(A),∀k∈{1,...,n}.
Note that yk=det(Mk)det(A)=det(Mk)⋅det(A−1)=det(A−1Mk)=n∑j=1(−1)k+j(A−1Mk)kjdet(~(A−1Mk)kj)=(A−1Mk)kkdet(~(A−1Mk)kk)=(A−1b)kdet(In−1)=(A−1b)k,yk=det(Mk)det(A)=det(Mk)⋅det(A−1)=det(A−1Mk)=n∑j=1(−1)k+j(A−1Mk)kjdet(˜(A−1Mk)kj)=(A−1Mk)kkdet(˜(A−1Mk)kk)=(A−1b)kdet(In−1)=(A−1b)k, because (A−1Mk)ij={(In)ijif j≠k(A−1b)iif j=k.
Since x=A−1b is a unique solution of Ax=b, xk=(A−1b)k=yk,∀k∈{1,...,n}⟹x=y. Thus xk=det(Mk)det(A). ◼
크라메르 공식은 주어진 선형연립방정식의 해를 수학적으로 깔끔하게 제시한다는 점에서 의미가 있지만, 실제로 방정식의 해를 수치적으로 계산하고자 할 때는 별 볼일 없는 정리이다. coefficient matrix가 n×n 행렬일 때, 그 해를 구하려면 행렬식을 총 n+1번 계산해야 하기 때문에 연산량이 꽤 많은 편이다.