Invertible elements of monoid $(M_ 2(Z),\cdot)$
$A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$
I calculated $A^{-1}=\frac{1}{ad-bc}\cdot\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}$
$ad-bc\not=0$
I'm not really sure how to go from here to the final answer which is:
$\{A|$ determinant of $A\in \{\pm 1\}\}$
Could anyone point out what I'm missing? Thank you.
The invertible elements of the monoid $(M_ 2(\mathbb{Z}),\cdot)$ form a group, called $$ GL_2(\mathbb{Z}). $$ For a commutative ring $R$ with $1$, the determinant of a matrix in $GL_n(R)$ is a unit in $R$. For $R=\mathbb{Z}$ this means that $\det(A)=\pm 1$.
Related questions:
$A^{-1}$ must be coefficients in $\mathbb{Z}$. So must be,
\begin{equation} \frac{a}{ad-bc},\frac{b}{ad-bc},\frac{d}{ad-bc},\frac{d}{ad-bc} \in \mathbb{Z}. \end{equation}
So if $A$ is invertible (i.e. the above statement is true), then $\frac1{|A|}A$ is invertible too. And if $\frac1{|A|}A$ is invertible then $\frac{1}{|A|^2}A$ is invertible and so on. So, the elements of $A$ must be such that
$$\frac{a}{(ad-bc)^n},\frac{b}{(ad-bc)^n},\frac{d}{(ad-bc)^n},\frac{d}{(ad-bc)^n} \in \mathbb{Z} \quad \forall n \in \mathbb{N}.$$ The only solution is $ad-bc=1$ or $ad-bc=-1$.
Hint: If $A \in M_2(\mathbb Z)$ is invertible with inverse $B \in M_2(\mathbb Z)$, then $AB=I$ implies $(\det A)(\det B)=\det(AB)=\det(I)=1$. What does that say about $\det A$ ?
