## anonymous 4 years ago 2. Suppose that A is an n×n matrix with integer entries, and that A is invertible. Since A is invertible, we can compute the matrix A^−1. The computations seem much cleaner (and friendlier) when A^−1 also has only integer entries. The purpose of this question is to figure out when that can happen. Prove: That (if A is an invertible n × n matrix, with only integer entries) A^−1 has integer entries if and only if det(A) = ±1 (where “= ±1” means equals 1 or equals −1). Reminder: Since this is an “if and only if”, don’t forget that means that you have two directions to prove.

1. anonymous

2. JamesJ

One way is easy. If A has all integral entries, then det(A) is also an (non-zero) integer. If A also has all integral entries, then det(A^-1) is also integral . Now, as $1 = \det(I) = \det(AA^{-1}) = \det(A).\det(A^{-1})$ the only way that can be the case is if $\det(A) = \det(A^{-1}) = 1$ or $\det(A) = \det(A^{-1}) = -1$

3. anonymous

4. JamesJ

Now the other way. We know that A has all integral entries and we have that $$\det A = \pm 1$$. This must mean that $$\det A^{-1} = \pm 1$$ respectively also. Now think about why this can imply the entries of A^-1 are all integers; or why if they're not all integers this leads to a contradiction.

5. anonymous

It implies that they're all integers right? Because ... if the inverse of detA is equal to pm1, then they must be integers so that they could undo themselves?