1) We consider the cofactor matrix C of A: The (i,j)-th entry c_ij of C is equal to (-1)^(i+j)*m_ij, where m_ij is equal to the determinant of the matrix A(i,j) obtained by deleting the i-th row and j-th column of A. Since the entries of A are all integers, then the entries of A(i,j) are also integers. As a result, the determinant of A(i,j) is also an integer, and so m_ij is also an integer. Therefore, the entries of C are also all integers. Consequently, the adjoint matrix of A, Adj(A)=C^T is also an nxn matrix whose entries are all integers. Furthermore, since the entries of A are integers, then det(A), the determinant of A, is also an integer. Since A is invertible, then det(A) is a nonzero integer. Therefore, the inverse of A is equal to A^-1 = (1/det(A))* Adj(A) and it is a matrix whose entries are rational numbers, that is, A^-1 is an element in Q_nn.
2) Assume that A^-1 is in Znn. Then det(A^-1) is an integer. As A*A^-1 = I_n, where I_n is the nxn identity matrix, then det(A*A^-1)=det(I_n) gives det(A)det(A^-1)=1. Since both det(A) and det(A^-1) are integers, then we see that det(A) must be 1 or -1.
On the other hand, we assume that det(A) is either 1 or -1. Since A^-1=(1/det(A))*Adj(A), then we see that A^-1 =Adj(A) or -Adj(A) depending on det(A) is 1 or -1. By the arguments in part (1), we know that the entries of Adj(A) are all integers. So we A^-1 must be an element in Z_nn.
So A^-1 is in Z_nn if and only if det(A) is either 1 or -1.
