Let $A = \begin{bmatrix} 1 & \frac{-1-i\sqrt{3}}{2} \\ \frac{-1+i\sqrt{3}}{2} & 1 \end{bmatrix}$. Then $A^{(0)} = 2^k \cdot A$ where $k$ is
Step-by-Step Solution
Key Concept: Recognize that $\frac{-1-i\sqrt{3}}{2} = \omega^2$ and $\frac{-1+i\sqrt{3}}{2} = \omega$ where $\omega$ is a primitive cube root of unity. Then use the recurrence relation $A^n = 2^{n-1}A$ (obtained by computing $A^2 = 2A$) to find that $A^{100} = 2^{99}A$.
Given matrix $A = \begin{bmatrix} 1 & ω^2 \\ ω & 1 \end{bmatrix}$ where $ω$ is a cube root of unity, we compute $A^2 = \begin{bmatrix} 1+ω^3 & 2ω^2 \\ 2ω & ω^3+1 \end{bmatrix} = 2\begin{bmatrix} 1 & ω^2 \\ ω & 1 \end{bmatrix} = 2A$. Then $A^3 = A^2 \cdot A = 2A \cdot A = 2A^2 = 4A$, and continuing this pattern we get $A^{100} = 2^{99}A$. Therefore $k = 99$.
Correct Answer: 99