Originally Posted by

**Deveno** sure, i understand, but how do you go from that fact that det(A) = ±1, to conclude that A^-1 has integral entries?

By his question he already must know how to calculate the inverse of a matrix: $\displaystyle A^{-1}=\frac{1}{\det A}\,Adj(A)$, so it is now

obvious that the inverse is going to be an integer matrix (in fact, having determinant $\displaystyle \pm 1$ is an iff condition for an integer matrix

to have an integer inverse)...

i exhibited an orthogonal matrix for which this is NOT true.

Oh dear! The matrix we've been working with is an INTEGER one: all I've done has taken this into account all through! I never talked

about general matrices or whatever but focused on the OP where we're given an integer matrix.

in other words, i am wondering what YOU were thinking as the answer

to the question: "Why (does A^-1 have integer entries)?. What's the determinant of an orthogonal matrix?".

because if your answer is "because det(A) = ±1", that's not correct. having a determinant of ±1, does NOT imply integral entries.