proof of Cramer’s rule


Since det(A)0, by properties of the determinantMathworldPlanetmath we know that A is invertiblePlanetmathPlanetmath.

We claim that this implies that the equation Ax=b has a unique solution. Note that A-1b is a solution since A(A-1b)=(AA-1)b=b, so we know that a solution exists.

Let s be an arbitrary solution to the equation, so As=b. But then s=(A-1A)s=A-1(As)=A-1b, so we see that A-1b is the only solution.

For each integer i, 1in, let ai denote the ith column of A, let ei denote the ith column of the identity matrixMathworldPlanetmath In, and let Xi denote the matrix obtained from In by replacing column i with the column vectorMathworldPlanetmath x.

We know that for any matrices A,B that the kth column of the product AB is simply the product of A and the kth column of B. Also observe that Aek=ak for k=1,,n. Thus, by multiplication, we have:

AXi=A(e1,,ei-1,x,ei+1,,en)=(Ae1,,Aei-1,Ax,Aei+1,,Aen)=(a1,,ai-1,b,ai+1,,an)=Mi

Since Xi is In with column i replaced with x, computing the determinant of Xi with cofactor expansion gives:

det(Xi)=(-1)(i+i)xidet(In-1)=1xi1=xi

Thus by the multiplicative property of the determinant,

det(Mi)=det(AXi)=det(A)det(Xi)=det(A)xi

and so xi=det(Mi)det(A) as required.

Title proof of Cramer’s rule
Canonical name ProofOfCramersRule
Date of creation 2013-03-22 13:03:24
Last modified on 2013-03-22 13:03:24
Owner rmilson (146)
Last modified by rmilson (146)
Numerical id 11
Author rmilson (146)
Entry type Proof
Classification msc 15A15