If `A=[{:(omega,0),(0,omega):}]`, where `omega` is cube root of unity, then what is `A^(100)` equal to ?

Options

A. A
B. `-A`
C. Null matrix
D. Identity matrix

Explanation

To solve the problem, we need to find \( A^{100} \) where \( A = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} \) and \( \omega \) is a cube root of unity.

### Step-by-step Solution:

1. **Understanding the Matrix A**:
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} \]
Here, \( \omega \) is a cube root of unity, which means \( \omega^3 = 1 \).

**Hint**: Recall the properties of cube roots of unity: \( 1, \omega, \omega^2 \) where \( \omega^3 = 1 \).

2. **Finding \( A^2 \)**:
We calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} \cdot \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} = \begin{pmatrix} \omega^2 & 0 \\ 0 & \omega^2 \end{pmatrix} \]

**Hint**: When multiplying diagonal matrices, the result is another diagonal matrix with the squares of the original diagonal elements.

3. **Finding \( A^3 \)**:
Next, we calculate \( A^3 \):
\[ A^3 = A^2 \cdot A = \begin{pmatrix} \omega^2 & 0 \\ 0 & \omega^2 \end{pmatrix} \cdot \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} = \begin{pmatrix} \omega^3 & 0 \\ 0 & \omega^3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

**Hint**: Remember that \( \omega^3 = 1 \) simplifies the calculation.

4. **Finding Higher Powers of A**:
Since \( A^3 = I \) (the identity matrix), we can find \( A^{100} \) by reducing the exponent modulo 3:
\[ 100 \mod 3 = 1 \]
Therefore, \( A^{100} = A^{1} = A \).

**Hint**: Use modular arithmetic to simplify the exponent when dealing with periodic powers.

5. **Final Result**:
Thus, we conclude that:
\[ A^{100} = A = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} \]

### Summary:
The final answer is:
\[ A^{100} = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} \]