Does an integer matrix $B$ always exist such as $A B = \det(A) * I$

Question

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.

Answer 1

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.

Answer 2

$$A B = det(A) * I \implies$$

$$ B=det(A) A^{-1} * I = $$

$$det(A) A^{-1} $$

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.

Answer 3

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.