AB = Identity matrix; matrices; determinants; proof

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Let $M(n\times n, \mathbb Z)$ be the set of all $n\times n$- matrices with integer coefficients, and a matrix $A \in M$.

Proof, that: There is exactly one matrix $B \in M(n\times n, \mathbb Z)$ with $AB = I_n$, if $\det(A)=\pm1$.

I don't know how to start with this proof...Can you help please?

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If $\det(A)=\pm 1$ then $A$ is invertible and

$$A^{-1}=\frac1{\det(A)}\operatorname{adj}(A)$$ where $\operatorname{adj}(A)$ is the adjugate matrix which is the transpose of the cofactor matrix so its entries are integers. The unicity of $B$ comes from the unicity of the inverse of an invertible matrix.