Units of ${\rm Mat}_{n\times n}(\mathbb{Z})$ I am trying to calculate the units of the ring ${\rm Mat}_{n\times n}(\mathbb{Z})$, but I am a little bit stuck.
Here is my try:
Let's take $A\in U({\rm Mat}_{n\times n}(\mathbb{Z}))$; then, there exists $B\in {\rm Mat}_{n\times n}(\mathbb{Z})$ such that $AB=BA=Id$; taking determinants, we conclude that this implies that $\det(A)\det(B)=\det(Id)=1$; so, taking on account that the determinant of a matrix with integer coeficients is also an integer, we have two options from the last conclusion: or $\det(A)=\det(B)=1$ or $\det(A)=\det(B)=-1$; so by all of this, I would be able to conclude that $$U({\rm Mat}_{n\times n}(\mathbb{Z}))\subseteq\{A\in{\rm Mat}_{n\times n}(\mathbb{Z}): \det(A)=\pm 1\}$$
How should I proceed for finishing and giving the concrete units of this ring? Thank you in advance for your help!
 A: If $R$ is a commutative ring, then the group of invertible matrices over $R$ is
$$\mathrm{GL}_n(R) := M_n(R)^{\times} = \{A \in M_n(R) : \det(A) \in R^{\times}\}.$$
In your case, $\mathbb{Z}^{\times}=\{\pm 1\}$. Here, $\subseteq$ follows immediately from the fact that $\det(-)$ is multiplicative (as in your proof), and the converse follows from (a version of) Cramer's rule which states that for $A \in M_n(R)$ we have
$$A \cdot \mathrm{adj}(A) = \mathrm{adj}(A) \cdot A = \det(A) \cdot 1.$$
It implies that if $\det(A)$ is invertible, then $A$ is invertible with $A^{-1} = \det(A)^{-1} \cdot \mathrm{adj}(A)$.
In some texts (also on Wikipedia) you only find a proof of Cramer's rule for matrices over fields. But this implies it for arbitrary commutative rings by a standard argument: Consider the universal matrix $(X_{i,j})_{1 \leq i,j \leq n}$ over the polynomial ring $ P=\mathbb{Z}[\{X_{i,j} : 1 \leq i,j \leq n\}]$ with field of fractions $Q(P)$. Since Cramer's rule (as an equation of matrices as above) is true for the universal matrix considered as a matrix over $Q(P)$, it is also true as an equation of matrices over $P$. Now if $R$ is any commutative ring and $A=(a_{i,j}) \in M_n(R)$, there is a (unique) homomorphism $P \to R$ mapping $X_{i,j} \mapsto a_{i,j}$. Applying the induced homomorphism $M_n(P) \to M_n(R)$ on both sides of the matrix equation yields the result.
