If `omega=1` is the complex cube root of unity and matrix `H=|{:(,omega,0),(,0,omega):}|`, then `H^(70)` is equal to:

Options

A. H
B. 0
C. `-H`
D. `H^2`

Answer

To solve the problem, we need to find \( H^{70} \) where \( H = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} \) and \( \omega \) is a complex cube root of unity.

### Step-by-step Solution:

1. **Understanding the Matrix \( H \)**:
The matrix \( H \) can be expressed as:
\[ H = \omega I \]
where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

2. **Finding \( H^{70} \)**:
Using the property of matrix exponentiation:
\[ H^{70} = (\omega I)^{70} = \omega^{70} I^{70} \]
Since \( I^{70} = I \), we have:
\[ H^{70} = \omega^{70} I \]

3. **Calculating \( \omega^{70} \)**:
We know that \( \omega \) is a cube root of unity, which means:
\[ \omega^3 = 1 \]
To simplify \( \omega^{70} \), we can reduce the exponent modulo 3:
\[ 70 \mod 3 = 1 \]
Thus:
\[ \omega^{70} = \omega^1 = \omega \]

4. **Final Result**:
Therefore, substituting back, we get:
\[ H^{70} = \omega I = \omega \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} \]

### Conclusion:
The final result is:
\[ H^{70} = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} \]