Let $\mathbb{R}$ denote the set of all real numbers and let $i = \sqrt{-1}$. Consider the matrices $S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ and $T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. Let $a, b, c, d$ be real numbers such that $ST = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Let $H = \{x + iy : x, y \in \mathbb{R}$ and $y > 0\}$. Then which of the following statements is (are) TRUE ?

$S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},\quad T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ $ST = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $\text{Find } a, b, c, d \text{ (from } ST\text{) then check each option}$ $H = \{x+iy : y>0\} \text{ is the upper half-plane}$ $1.\ ST \Rightarrow a, b, c, d$ $2.\ \text{check options (1), (2), (3), (4)}$ $ST = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$ $a=0,\quad b=-1,\quad c=1,\quad d=1$ $(1):\ \frac{b+ia}{d+ic} = \frac{-1}{1+i} = \frac{-1+i}{2} \neq i$ $\Rightarrow \text{FALSE}$ $\omega^2+\omega+1=0 \Rightarrow \omega+1=-\omega^2$ $(2):\ \frac{a\omega+b}{c\omega+d} = \frac{-1}{\omega+1} = \frac{-1}{-\omega^2} = \frac{1}{\omega^2} = \omega$ $\Rightarrow \text{TRUE}$ $\det(ST)=1,\qquad \operatorname{tr}(ST)=1$ $\lambda^2-\lambda+1=0 \Rightarrow \lambda = e^{\pm i\pi/3}$ $(ST)^6 = I,\qquad \text{order of } ST = 6$ $(ST)^2=(ST)^m \Leftrightarrow (ST)^{m-2}=I \Leftrightarrow 6 \mid (m-2)$ $m = 8,\ 14,\ 20,\ldots \Rightarrow \text{not all multiples of } 8$ $\Rightarrow \text{FALSE}$ $z=x+iy,\ y>0:\quad \frac{az+b}{cz+d} = \frac{-1}{z+1}$ $= \frac{-(x+1)+iy}{(x+1)^2+y^2},\qquad \operatorname{Im} = \frac{y}{(x+1)^2+y^2} > 0$ $\Rightarrow \in H\ \Rightarrow \text{TRUE}$ $\text{True: (2) and (4)}$ $\boxed{\text{Answer: (2), (4)}}$