Group made of $\begin{pmatrix} a &b \\ c & d \\ \end{pmatrix}$ where $ad-bc\not=0,a,b,c,d\in \mathbb Z_3$

Question

I have very straightforward easy question, but I could not figure it out.

You are given a group $\left( \left\{ \begin{pmatrix} a &b \\ c & d \\ \end{pmatrix}\;\big|\;ad-bc\not=0,a,b,c,d\in \mathbb Z_3 \right \},Matrix\; Multiplication \right)$

Question: Show that $|G|=48$

Given answer:

I do not get the red underlined sentence.

Answer 1

The underlined sentence means that we must exclude the three cases $(c,d) \neq \lambda (a,b)$ for $\lambda \in \mathbb{Z}_3 = \{0,1,2\}$.

• If we would have $(c,d) = \lambda (a,b)$ for some $\lambda \in \mathbb{Z}_3$, then we would get $ad - bc = a(\lambda b) - b (\lambda a) = 0$, meaning that the matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is not in the set.

Answer 2

If the second row is a multiple of the first row, say that $c=ka$ and $d=kb$, then $ad-bc=abk-bak=0$.

Answer 3

If $A$ is a matrix with two rows and two columns, $\det A=0$ if and only if one of the rows is a multiple of the other one.