Question

The identity element in the group $M=\left\{\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]|x\in R,x\ne 0\right\}$ with respect to matrix multiplication is

• A. $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$
• B. $\frac{1}{2}\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

Answer 1

Answer: (B)

$M=\left[\begin{array}{cc}x& x\\ x& x\end{array}\right],\forall x\in R$ and $x\ne 0$
Let $P$ be the identity element in the group
i.e. $P=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$
$P$ is obtained by putting $x=\frac{1}{2}$
$\therefore MP=\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]=M$
and $PM=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]=M$
$\therefore MP=M=PM$