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

Options

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

Answer

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 Cube Root of Unity**:
The cube roots of unity are the solutions to the equation \( x^3 = 1 \). They are:
\[ 1, \quad \omega, \quad \text{and} \quad \omega^2 \]
where \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{-2\pi i / 3} \). Notably, \( \omega^3 = 1 \).

2. **Matrix Representation**:
The matrix \( A \) can be expressed as:
\[ A = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} \]

3. **Finding \( A^{100} \)**:
To find \( A^{100} \), we can use the property of diagonal matrices. For a diagonal matrix \( D = \begin{pmatrix} d_1 & 0 \\ 0 & d_2 \end{pmatrix} \), the power \( D^n \) is given by:
\[ D^n = \begin{pmatrix} d_1^n & 0 \\ 0 & d_2^n \end{pmatrix} \]
Applying this to our matrix \( A \):
\[ A^{100} = \begin{pmatrix} \omega^{100} & 0 \\ 0 & \omega^{100} \end{pmatrix} \]

4. **Calculating \( \omega^{100} \)**:
Since \( \omega^3 = 1 \), we can reduce the exponent modulo 3:
\[ 100 \mod 3 = 1 \]
Therefore, \( \omega^{100} = \omega^{1} = \omega \).

5. **Final Matrix Calculation**:
Substituting back, we get:
\[ A^{100} = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} = A \]

### Conclusion:
Thus, we conclude that:
\[ A^{100} = A \]