Question

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.

Answer 1

Hey james once you are done with this please help me....

Answer 2

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
\[ \det(A) = \det(A^{-1}) = -1 \]

Answer 3

Hey turn towards Unanswered question james....
Adherent points

Answer 4

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.

Answer 5

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?