If `omega` is a complex cube root of unit and
`A=[(omega,0,0),(0,omega^(2),0),(0,0,1)]` then `A^(-1)=` . . .

Options

A. `[(omega^(2),0,0),(0,omega,0),(0,0,1)]`
B. `[(1,0,0),(0,1,0),(0,0,1)]`
C. `[(1,0,0),(0,omega^(2),0),(0,0,omega)]`
D. `[(0,0,omega),(0,omega^(2),0),(1,0,0)]`

Explanation

To find the inverse of the matrix \( A \), we start with the given matrix:

\[ A = \begin{pmatrix} \omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

where \( \omega \) is a complex cube root of unity. This means that \( \omega^3 = 1 \) and \( \omega \neq 1 \).

### Step 1: Understand the properties of the matrix
The matrix \( A \) is a diagonal matrix. The inverse of a diagonal matrix can be found by taking the reciprocal of each of the diagonal elements.

### Step 2: Calculate the inverse
The diagonal elements of \( A \) are \( \omega \), \( \omega^2 \), and \( 1 \). Therefore, the inverse \( A^{-1} \) will have the diagonal elements \( \frac{1}{\omega} \), \( \frac{1}{\omega^2} \), and \( 1 \).

\[ A^{-1} = \begin{pmatrix} \frac{1}{\omega} & 0 & 0 \\ 0 & \frac{1}{\omega^2} & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Simplify the elements
Since \( \omega^3 = 1 \), we can express \( \frac{1}{\omega} \) and \( \frac{1}{\omega^2} \) in terms of \( \omega \):
- \( \frac{1}{\omega} = \omega^2 \)
- \( \frac{1}{\omega^2} = \omega \)

Thus, we can rewrite \( A^{-1} \) as:

\[ A^{-1} = \begin{pmatrix} \omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Final Result
The inverse of the matrix \( A \) is:

\[ A^{-1} = \begin{pmatrix} \omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1 \end{pmatrix} \]