Matrix Multiplication and Complex Transformation on Upper Half Plane

To determine which of the given statements are TRUE, we first evaluate the product of the matrices $S$ and $T$.

Given: $S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \quad \text{and} \quad T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$

Multiplying $S$ 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 $ST = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, we get: $a = 0, \quad b = -1, \quad c = 1, \quad d = 1$

Now, we evaluate each of the given options:

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Option (A):

We test the expression $\frac{b+ia}{d+ic}$: $\frac{b+ia}{d+ic} = \frac{-1 + i(0)}{1 + i(1)} = \frac{-1}{1+i}$

Multiplying the numerator and denominator by the complex conjugate $1-i$: $\frac{-1(1-i)}{(1+i)(1-i)} = \frac{-1+i}{1 - i^2} = \frac{-1+i}{2} = -\frac{1}{2} + \frac{1}{2}i \neq i$

Thus, Option (A) is FALSE.

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Option (B):

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

We evaluate the expression $\frac{a\omega+b}{c\omega+d}$: $\frac{a\omega+b}{c\omega+d} = \frac{0 \cdot \omega + (-1)}{1 \cdot \omega + 1} = \frac{-1}{\omega + 1}$

Since $1 + \omega + \omega^2 = 0$, we have $\omega + 1 = -\omega^2$. Substituting this gives: $\frac{-1}{-\omega^2} = \frac{1}{\omega^2} = \frac{\omega^3}{\omega^2} = \omega$

Thus, Option (B) is TRUE.

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Option (C):

Let $A = ST = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$.

The characteristic equation of $A$ is given by: $\det(A - \lambda I) = 0 \implies \lambda^2 - \text{tr}(A)\lambda + \det(A) = 0 \implies \lambda^2 - \lambda + 1 = 0$

By the Cayley-Hamilton theorem: $A^2 - A + I = 0 \implies A^2 = A - I$

Multiplying by $(A + I)$: $(A + I)(A^2 - A + I) = 0 \implies A^3 + I = 0 \implies A^3 = -I \implies A^6 = I$

Thus, the order of matrix $A = ST$ is $6$.

Given $(ST)^m = (ST)^2$ for $m > 2$, since $ST$ is invertible ($\det(ST) = 1 \neq 0$), we can multiply both sides by $(ST)^{-2}$: $(ST)^{m-2} = I$

This implies that $m - 2$ must be a multiple of $6$: $m - 2 = 6k \implies m = 6k + 2 \quad \text{for } k \in \{1, 2, 3, \dots\}$

• For $k = 1$: $m = 8$ (which is a multiple of 8).
• For $k = 2$: $m = 14$ (which satisfies $(ST)^{14} = (ST)^2$, but $14$ is not an integer multiple of 8).

Since $m$ is not necessarily a multiple of 8 for all valid $m$, Option (C) is FALSE.

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Option (D):

Let $z \in H$, which means $z = x + iy$ with $x, y \in \mathbb{R}$ and $y > 0$.

We need to check whether $w = \frac{az+b}{cz+d} \in H$: $w = \frac{0 \cdot z - 1}{1 \cdot z + 1} = \frac{-1}{z+1} = \frac{-1}{(x+1) + iy}$

Multiplying the numerator and denominator by $(x+1) - iy$: $w = \frac{-((x+1) - iy)}{((x+1) + iy)((x+1) - iy)} = \frac{-(x+1) + iy}{(x+1)^2 + y^2}$

The imaginary part of $w$ is: $\text{Im}(w) = \frac{y}{(x+1)^2 + y^2}$

Since $y > 0$ and $(x+1)^2 + y^2 > 0$ for all real $x, y$, we clearly have: $\text{Im}(w) > 0 \implies w \in H$

Thus, Option (D) is TRUE.

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Conclusion:

The correct statements are (B) and (D).