If AX = B is an n X n system of equations over the integers, how do I show that it has integer solutions if det(A) = +1 or -1
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If det A = +1 or -1, then A is invertible, since it has nonzero determinant.
Therefore, the unique solution to Ax = b is x = inv(A)*b.
The question is, why would x have only integers?
Well, we know that b is a column vector of only integers.
Can we show that the matrix A only has integers too?
Actually, inv(A) does only have integers.
One of the theorems about inverses and determinants
(after you’ve learned about cofactor expansion of determinants)
Is that inv(A) = (1/det A) * adj(A), where adj(A) is the matrix of cofactors from A.
But the matrix A is all integers, so the cofactors will all be integers.
Therefore, adj(A) is full of integers, and so inv(A) is too.
So in the end, x will be a solution with only integers.
For more on cofactors or the theorem, go to http://tutorial.math.lamar.edu/Classes/LinAlg/MethodOfCofactors.aspx