A) \[\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\]
B) \[\frac{1}{2}\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right]\]
Correct Answer: B
Solution :
\[M=\left[ \begin{matrix} x & x \\ x & x \\ \end{matrix} \right],\,\ \forall \,x\in \,\,R\] and \[x\ne 0\] Let P be the identity element in the group \[i.e.\,,\] \[P=\left[ \begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{matrix} \right]=\frac{1}{2}\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\] P is obtained by putting \[x=\frac{1}{2}\] \[\therefore \] \[MP=\left[ \begin{matrix} x & x \\ x & x \\ \end{matrix} \right]\left[ \begin{matrix} \frac{1}{1} & \frac{1}{1} \\ \frac{1}{1} & \frac{1}{1} \\ \end{matrix} \right]\] \[=M\] and \[PM=\left[ \begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{matrix} \right]\,\,\left[ \begin{matrix} x & x \\ x & x \\ \end{matrix} \right]\] \[=\,M\] \[\therefore \] \[MP=M=PM\]
You need to login to perform this action.
You will be redirected in
3 sec
