Question

Let \( A = \begin{bmatrix} 1 & \frac{-1-i\sqrt{3}}{2} \\ \frac{-1+i\sqrt{3}}{2} & 1 \end{bmatrix} \). Then \( A^{100} \) is equal to:

• A. $2^{100} A$
• B. $2^{99} A$
• C. $2^{98} A$
• D. $A$
• E. $A^2$

Hint:

When you see the specific values $\frac{-1 \pm i\sqrt{3}}{2}$, immediately substitute $\omega$ and $\omega^2$. It transforms complex matrix multiplication into simple algebraic manipulation.

Answer 1

The Correct Option is B

Concept: The elements of the matrix involve the complex cube roots of unity. Let $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$. The matrix can be rewritten using these values to find a pattern in higher powers.
• Properties: $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
• For a matrix of the form $A = \begin{bmatrix} 1 & \omega^2
\omega & 1 \end{bmatrix}$, we calculate initial powers to observe the behavior of $A^n$.

Step 1: Rewrite the matrix and find $A^2$.
$A = \begin{bmatrix} 1 & \omega^2
\omega & 1 \end{bmatrix}$. \[ A^2 = \begin{bmatrix} 1 & \omega^2 \omega & 1 \end{bmatrix} \begin{bmatrix} 1 & \omega^2 \omega & 1 \end{bmatrix} = \begin{bmatrix} 1 + \omega^3 & \omega^2 + \omega^2 \omega + \omega & \omega^3 + 1 \end{bmatrix} = \begin{bmatrix} 2 & 2\omega^2 2\omega & 2 \end{bmatrix} = 2A \]

Step 2: Generalize to $A^n$.
Using the property $A^2 = 2A$: \[ A^3 = A^2 \cdot A = (2A) \cdot A = 2A^2 = 2(2A) = 2^2 A \] \[ A^4 = A^3 \cdot A = (2^2 A) \cdot A = 2^2 A^2 = 2^2 (2A) = 2^3 A \] By induction, $A^n = 2^{n-1} A$.

Step 3: Calculate $A^{100}$.
Substituting $n = 100$: \[ A^{100} = 2^{100-1} A = 2^{99} A \]