Show that if A is an n x n matrix of integers, then A^-1 is a matrix of integers if and only if det(A) = +/- 1. This is actually part b of a problem. Part a) asks to prove that (x_i)det(A) = det(A_ib) where A_ib is the matrix formed by replacing the ith column of A by the column vector b. x_i is a scalar and is the component of the vector x. So since that was the previous part i tried using that for part b). So if det(A) = +/- 1, then x_i = +/- det(A_ib). but i also know that x = (A^-1)b. so the ith entry of x is also the ith row of A^-1 which i denote as a_i dot product with the vector b. So i have +/- det(A_ib) = a_i * b. i don't know what to do from here. b may or may not be a vector of integers so the det(A_ib) may or may not be an integer. and even if it were i don't know how that would show that A^-1 must be a matrix of integers. am i going about this problem in the correct way? please help.