Prove that the determinant of an invertible matrix \(A\) is equal to \(\pm 1\)
when all of the entries of \(A\) and \(A^{-1}\) are integers. Getting Started:
Denote det( \(A\) ) as \(x\) and \(\operatorname{det}\left(A^{-1}\right)\) as \(y\)
Note that \(x\) and \(y\) are real numbers. To prove that \(\operatorname{det}(A)\)
is equal to \(\pm 1,\) you must show that both \(x\) and \(y\) are integers such
that their product \(x y\) is equal to 1
(i) Use the property for the determinant of a matrix product to show that \(x
y=1\)
(ii) Use the definition of a determinant and the fact that the entries of \(A\)
and \(A^{-1}\) are integers to show that both \(x=\operatorname{det}(A)\) and
\(y=\operatorname{det}\left(A^{-1}\right)\) are integers.
(iii) Conclude that \(x=\operatorname{det}(A)\) must be either 1 or \(-1\) because
these are the only integer solutions to the equation \(x y=1\)