Theorem: Cramer's Rule
Let an n x n matrix A represent a system of linear equations such that AX = B.
Then it follows that if Det(A) ≠ 0, then X has only one unique solution and xi = det(A(Ci ↔ B))/det(A).
Proof:
(1) Assume that Det(A) ≠ 0
(2) Then A is invertible [see Theorem 4, here]
(3) Now, A-1B is a unique solution to AX = B [See Lemma 2, here]
(4) From Corollary 4.1, here, we have:
X = A-1B = 1/det(A)(adj A)B
(5) Therefore, we have:
xi = 1/det(A)*∑(j=1,n) enti,j(adj A))bj
[See Definition 1 here for definition of matrix multiplication if needed]
= 1/det(A)*∑(j=1,n) bj*(cofj,i(A)) [See Definition 1 here for definition of adj A]
(6) Using Corollary 4.1 here, this gives us:
1/det(A)*∑(j=1,n)bj*(cofj,i(A)) = 1/det(A)*det(A(Ci ↔ B)).
(7) Putting it all together gives us:
xi = 1/det(A)*det(A(Ci ↔ B)).
QED
References
- Charles G. Cullen, Matrices and Linear Transformations, Dover Publications, Inc., 1972.
- "Cramer's Rule", PlanetMath.org
3 comments :
Thank you very much sir!
Your posts and your blog "Math Refresher" is a real help for me in my course. You are great at explaining concepts! :P
yeah "Math Refresher" is really good at explainning concepts, I was looking for something simple like this :P
in my last course of linar algebra, cramer's rule got in a real mess for me...
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