I want to prove that:
If $F$ is a free abelian group of rank $n$, then $\text{Aut}(F)$ is isomorphic to the multiplicative group of all $n\times n$ matrices over $\mathbb Z$ with $\text{det}=\pm1$.
What I have tried:
Since $F$ is a free abelian group of rank $n$ so I can write $F=\langle x_1,x_2,...,x_n\rangle$. If $\phi\in\text{Aut}(F)$ then it should map any $x_i$ to another element $x_j$ of the basis $X=\{x_1,x_2,...,x_n\}$. My idea is to map any element of $\text{Aut}(F)$ to a suitable matrix. For example when $n=2$: $\phi_1:=\{x_1\to x_1, x_2\to x_2\}\Longrightarrow \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)\\ \phi_2:=\{x_1\to x_2, x_2\to x_1\}\Longrightarrow \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$
Am I on a right approach? Or there is another better way? Thanks.