If the set \[G=\left\{ \left( \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right):\theta \in R \right\}\] is a group under matrix multiplication. Then, which one of the following statements in respect of G is true.
\[\left( a \right)\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\text{ is the inverse of itself}\]
(b) G is a finite group
\[\left( c \right)\left( \begin{matrix}
   \dfrac{1}{2} & \dfrac{-\sqrt{3}}{2} \\
   \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\
\end{matrix} \right)\text{ is not an element of G}\]
\[\left( d \right)\left( \begin{matrix}
   1 & 1 \\
   -1 & 1 \\
\end{matrix} \right)\text{ is an element of G}\]

Hint:To solve this question, we will separately consider all options one by one. If G is a group, then ‘a’ is an element of G is inverse of some element by b if ab = e. Also, an element is called the inverse of itself if aa = e, where e is the identity of the group G. Also, an element belongs to G if its properties match with properties of the group.

Complete step by step answer:
Given that,
\[G=\left\{ \left( \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right):\theta \in R \right\}......\left( i \right)\]
This G is a group under matrix multiplication.
Let us consider (a) first.
\[\left( a \right)\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\] is the inverse of itself. In a group, an element ‘a’ belongs to G is called the inverse of itself if aa = e where e is the identity of G. Identity of G in equation (i) is \[\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right)\] as matrix multiplication has an identity as \[\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right)\] So, here, we have \[e=\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right).\]
Now, for \[\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\] to be inverse of itself \[\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\] should be equal to the identity \[\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right)\] as stated above.
Computing it we get, \[\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right)\]
Clearly, RHS has the identity of group G as \[\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right).\]
So, we have \[\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\] as the inverse of itself.
Hence, option (a) is the right answer.
Let us consider option (b) now. G is a finite group.
Because \[\sin \theta \] and \[\cos \theta \] for \[\theta \in R\] have infinite values in between [– 1, 1]. So, as \[\sin \theta \] and \[\cos \theta \] itself have infinite values possible, \[\left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right]\] has infinite matrices possible.
Therefore, we have got that G has infinite matrices possible.
Hence, option (b) is wrong.
Let us consider option (c) now.
\[\left( c \right)\left( \begin{matrix}
   \dfrac{1}{2} & \dfrac{-\sqrt{3}}{2} \\
   \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\
\end{matrix} \right)\text{ is not an element of G}\]
Compare \[\left( \begin{matrix}
   \dfrac{1}{2} & \dfrac{-\sqrt{3}}{2} \\
   \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\
\end{matrix} \right)\] with \[\left( \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right)........\left( ii \right)\]
Comparing, we get,
\[\dfrac{1}{2}=\cos \theta \]
\[\Rightarrow \cos \theta =\dfrac{1}{2}\]
As we know that \[\cos {{60}^{\circ }}=\dfrac{1}{2}\]
\[\Rightarrow \theta ={{60}^{\circ }}\]
Now, we also know that \[\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\] and \[\dfrac{-\sqrt{3}}{2}\] is also possible at subsequent angles.
So, all the elements of equation (ii) are matched. Hence, we have \[\left( \begin{matrix}
   \dfrac{1}{2} & \dfrac{-\sqrt{3}}{2} \\
   \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\
\end{matrix} \right)\] is an element of G.
Hence, option (c) is the wrong answer.
Finally, let us consider option (d).
\[\left( d \right)\left( \begin{matrix}
   1 & 1 \\
   -1 & 1 \\
\end{matrix} \right)\text{ is an element of G}\]
When comparing this, we have,
\[\left( \begin{matrix}
   1 & 1 \\
   -1 & 1 \\
\end{matrix} \right)=\left( \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right)\]
\[\Rightarrow 1=\cos \theta \]
\[\Rightarrow \cos \theta =1\]
\[\Rightarrow \theta ={{\cos }^{-1}}\left( 1 \right)\]
\[\Rightarrow \theta ={{0}^{\circ }}\]
Now, \[\sin \theta =\sin 0=0.\]
But the entry corresponding to \[\sin \theta =1\] in the above comparison, so \[0\ne 1.\] \[\left( \begin{matrix}
   1 & 1 \\
   -1 & 1 \\
\end{matrix} \right)\] cannot be an element of G.
Hence, option (d) is wrong.
Therefore, the correct option is (a).

Note:
Students can have confusion at the step where option (c) is considered. See, if G is a group and \[a\in G\] then \[-a\in G\] also any scalar multiplication \[Ra\in G\] means if \[\left( \begin{matrix}
   \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\
   \dfrac{-\sqrt{3}}{2} & \dfrac{1}{2} \\
\end{matrix} \right)\in G\Rightarrow \left( \begin{matrix}
   \dfrac{1}{2} & \dfrac{-\sqrt{3}}{2} \\
   \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\
\end{matrix} \right)\] also belongs to G.