Question

Show that the set $G$ of all matrices of the form $\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]$ where $x\in R-\left\{0\right\}$, is a group under matrix multiplication.

Answer 1

Let $A,B,C\in G$

Let $A=\left[\begin{array}{cc}a& a\\ a& a\end{array}\right],B=\left[\begin{array}{cc}b& b\\ b& b\end{array}\right],C=\left[\begin{array}{cc}c& c\\ c& c\end{array}\right]$

$AB=\left[\begin{array}{cc}a& a\\ a& a\end{array}\right]\left[\begin{array}{cc}b& b\\ b& b\end{array}\right]=\left[\begin{array}{cc}2ab& 2ab\\ 2ab& 2ab\end{array}\right]\in G$

So for $A,B\in G$,$AB\in G$ hence closure

Now $AB=\left[\begin{array}{cc}2ab& 2ab\\ 2ab& 2ab\end{array}\right]$

$BC=\left[\begin{array}{cc}2bc& 2bc\\ 2bc& 2bc\end{array}\right]$

$AB\cdot C=\left[\begin{array}{cc}2ab& 2ab\\ 2ab& 2ab\end{array}\right]\left[\begin{array}{cc}c& c\\ c& c\end{array}\right]=\left[\begin{array}{cc}4abc& 4abc\\ 4abc& 4abc\end{array}\right]$

$A\cdot BC=\left[\begin{array}{cc}a& a\\ a& a\end{array}\right]\left[\begin{array}{cc}2bc& 2bc\\ 2bc& 2bc\end{array}\right]=\left[\begin{array}{cc}4abc& 4abc\\ 4abc& 4abc\end{array}\right]$

$\left(AB\right)\cdot C=A\cdot \left(BC\right)$ hence associative

$\left[\begin{array}{cc}0.5& 0.5\\ 0.5& 0.5\end{array}\right]\in G$ is multiplicative identity

$\left[\begin{array}{cc}0.5a& 0.5a\\ 0.5a& 0.5a\end{array}\right]\in G$ is multiplicative inverse of $A\in G$

So $G$ is a group