Question 9

First, calculate the product of matrices $S$ and $T$: $ST = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} (0)(1) + (-1)(0) & (0)(1) + (-1)(1) \\ (1)(1) + (0)(0) & (1)(1) + (0)(1) \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$ Comparing with $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, we get $a=0, b=-1, c=1, d=1$.

Analyze each option:

(A) $\frac{b+ia}{d+ic} = \frac{-1 + i(0)}{1 + i(1)} = \frac{-1}{1+i}$. Multiplying numerator and denominator by $(1-i)$, we get $\frac{-1(1-i)}{(1+i)(1-i)} = \frac{-1+i}{2}$. Since $\frac{-1+i}{2} \neq i$, option A is false.

(B) For $\omega = \frac{-1+i\sqrt{3}}{2}$, it is a cube root of unity, so $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. LHS: $\frac{a\omega+b}{c\omega+d} = \frac{0(\omega) - 1}{1(\omega) + 1} = \frac{-1}{\omega+1}$. Since $\omega+1 = -\omega^2$, we have $\frac{-1}{-\omega^2} = \frac{1}{\omega^2} = \frac{\omega^3}{\omega^2} = \omega$. Thus, option B is true.

(C) Let $M = ST = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$. The characteristic equation is $\det(M - \lambda I) = 0$, which gives $\lambda^2 - \text{tr}(M)\lambda + \det(M) = 0 \Rightarrow \lambda^2 - \lambda + 1 = 0$. The roots are $\lambda = e^{\pm i\pi/3}$. Since $\lambda^6 = (e^{\pm i\pi/3})^6 = e^{\pm i2\pi} = 1$, the matrix order is 6, i.e., $M^6 = I$. The equation $(ST)^2 = (ST)^m$ implies $M^m = M^2$, so $M^{m-2} = I$. This means $m-2$ must be a multiple of 6: $m-2 = 6k \Rightarrow m = 6k + 2$ for $k \in \mathbb{Z}$. For $k=1, m=8$ (multiple of 8). However, for $k=2, m=14$, which is not a multiple of 8. Thus, option C is false.

(D) Let $z = x+iy \in H$, so $y > 0$. Consider $w = \frac{az+b}{cz+d} = \frac{0z - 1}{1z + 1} = \frac{-1}{z+1}$. $w = \frac{-1}{(x+1) + iy} = \frac{-(x+1) + iy}{(x+1)^2 + y^2}$. The imaginary part of $w$ is $\text{Im}(w) = \frac{y}{(x+1)^2 + y^2}$. Since $y > 0$ and $(x+1)^2 + y^2 > 0$, we have $\text{Im}(w) > 0$. Thus $w \in H$. Option D is true.