Let $R$ denote the set of all real numbers and let $i=\sqrt{-1}$. Consider the matrices

$S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$.

Let $a,b,c,d$ be real numbers such that $ST=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Let $H=\{x+iy:x,y\in R\text{and}y>0\}$. Then which of the following statements is (are) TRUE?

The correct options are:
1. If $\omega =\frac{-1+i\sqrt{3}}{2}$, then $\frac{a\omega +b}{c\omega +d}=\omega$
2. If $z\in H$, then $\frac{az+b}{cz+d}\in H$

Step 1: Determine the values of $a,b,c,d$ by computing matrix multiplication $ST$.

We are given matrices:

$S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$

Multiplying $S$ and $T$:

$ST=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}0·1+(-1)·0& 0·1+(-1)·1\\ 1·1+0·0& 1·1+0·1\end{array}\right)=\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$

Comparing this with $ST=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, we get:

$a=0$, $b=-1$, $c=1$, $d=1$

Step 2: Analyze the second statement involving the complex cube root of unity $\omega$.

Given $\omega =\frac{-1+i\sqrt{3}}{2}$, which is the non-real cube root of unity satisfying ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$.

Substitute the values of $a,b,c,d$ into the expression:

$\frac{a\omega +b}{c\omega +d}=\frac{0·\omega +(-1)}{1·\omega +1}=\frac{-1}{\omega +1}$

Using the identity $1+\omega +{\omega }^2=0$, we have $\omega +1=-{\omega }^2$:

$\frac{-1}{-{\omega }^2}=\frac{1}{{\omega }^2}=\frac{{\omega }^3}{{\omega }^2}=\omega$

Hence, the statement "If $\omega =\frac{-1+i\sqrt{3}}{2}$, then $\frac{a\omega +b}{c\omega +d}=\omega$" is TRUE.

Step 3: Analyze the fourth statement involving the upper half-plane set $H$.

Set $H=\{z=x+iy:x,y\in R\text{and}y>0\}$, representing complex numbers with a positive imaginary part.

Let $z=x+iy\in H$, so $y>0$. We consider:

$W=\frac{az+b}{cz+d}=\frac{0·z-1}{1·z+1}=\frac{-1}{z+1}=\frac{-1}{(x+1)+iy}$

To express $W$ in standard form, multiply the numerator and denominator by the complex conjugate of the denominator:

$W=\frac{-1·\left(\right(x+1)-iy)}{{(x+1)}^2+y^2}=\frac{-(x+1)+iy}{{(x+1)}^2+y^2}$

The imaginary part of $W$ is given by:

$\text{Im}\left(W\right)=\frac{y}{{(x+1)}^2+y^2}$

Since $y>0$ and the denominator ${(x+1)}^2+y^2>0$, it follows that $\text{Im}\left(W\right)>0$. Thus, $W\in H$.

Hence, the statement "If $z\in H$, then $\frac{az+b}{cz+d}\in H$" is TRUE.