Why is $\gcd(a,c)=1$?

Question

Consider the $2\times2$ matrix, $A\in M_2(\mathbb Z)$: $$A = \left(\begin{array}{cc} a & b\\ c & d \end{array}\right)$$

Assume that $\det A = \pm 1$. Why is it true that $\gcd (a,c) = 1$?

Answer 1

Let $a=a'm$ and $c=c'm$, where $m=\gcd(a,c)>0$; then $$ \det\begin{pmatrix}a&b\\c&d\end{pmatrix} = \det\begin{pmatrix}a'm&b\\c'm&d\end{pmatrix} = m\det\begin{pmatrix}a'&b\\c'&d\end{pmatrix} $$ by the properties of the determinant.

If the determinant is $\pm1$, then $m$ has to be $1$.

Similarly, $\gcd(a,b)=1$, $\gcd(c,d)=1$ and $\gcd(b,d)=1$. This easily generalizes to matrices of any size.

Answer 2

you have that: $$ad-cb=\pm1$$ It follows that if n divide a and c say $a=np$ and $c=nq$ then n automatically divide 1 since we get $$ad-bc= n(pd-qb)=\pm1$$ that is $n=\pm 1$. Thus $gcd(a,c)=1$