Does an integer matrix $B$ always exist such as $A B = \det(A) * I$ Given $A$ an invertible matrix with integer coefficients, does an integer matrix $B$ always exist such as:
$$
A B = \det(A) * I
$$
I tried with several matrices and I couldn't find a counter example.
I also tried to adapt a Gaussian elimination algorithm to avoid any division and return $B$ along with some divisor. It turns out the divisor always end up being a multiple of $\det(A)$, but I didn't manage to tweak the algorithm to get $\det(A)$ itself.
 A: For every matrix $A$, we have $A \operatorname{adj}(A) = (\det A) I$, where $\operatorname{adj}(A)$ is the adjugate matrix. This holds even if $A$ is not invertible.
If $A$ has integer entries, then so has $\operatorname{adj}(A)$, because its entries are determinants of submatrices of $A$, aka as minors or cofactors.
A: $$A B = det(A) * I \implies$$
$$ B=det(A) A^{-1} * I = $$
$$det(A) A^{-1}  $$
That is , $B$ is uniquely determined by $A$
Note that $$A^{-1} = (1/det(A)) M$$ where $$ M=adj (A) $$is a matrix found by cofactors of A.
Therefore $$ B= det(A) A^{-1}= det(A)(1/det(A) M =M$$
Since terms of A are integers, terms of M are also integers, thus the answer is yes.
A: Slightly more general statement is this:

Let $A$ be an invertible matrix with integer coefficients. Then $A^{-1}$ also has integer coefficients if and only if $\det A = \pm 1$.

Assume $A^{-1}$ has integer coefficients:
$$\det A \cdot \det A^{-1} = \det(AA^{-1}) = \det I = 1$$
Hence $\det A = \det A^{-1} = \pm 1$.
Conversely, if $\det A = \pm 1$, then
$$A^{-1} = \frac{1}{\det A} \operatorname{adj}(A) = \pm \operatorname{adj}(A)$$
which certainly has integer coefficients, as explaned elsewhere.
