If `omega` is cube roots of unity, prove that `{[(1,omega,omega^2),(omega,omega^2,1),(omega^2,1,omega)]+[(omega,omega^2,1),(omega^2,1,omega),(omega,omega^2,1)]} [(1),(omega),(omega^2)]=[(0),(0),(0)]`

Options

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

Answer

To prove the statement given in the question, we will follow these steps:

### Step 1: Define the Matrices
Let:
\[ A = \begin{pmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{pmatrix} \]
and
\[ B = \begin{pmatrix} \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \\ \omega & \omega^2 & 1 \end{pmatrix} \]

We need to prove that:
\[ (A + B) \begin{pmatrix} 1 \\ \omega \\ \omega^2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]

### Step 2: Add the Matrices A and B
Now, we will add matrices \( A \) and \( B \):
\[ A + B = \begin{pmatrix} 1 + \omega & \omega + \omega^2 & \omega^2 + 1 \\ \omega + \omega^2 & \omega^2 + 1 & 1 + \omega \\ \omega^2 + 1 & 1 + \omega & \omega + \omega^2 \end{pmatrix} \]

### Step 3: Simplify Each Entry
Using the property of cube roots of unity, we know:
\[ 1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1 \]
Thus, we can simplify each entry:
- \( 1 + \omega = -\omega^2 \)
- \( \omega + \omega^2 = -1 \)
- \( \omega^2 + 1 = -\omega \)

So, we can rewrite the matrix \( A + B \):
\[ A + B = \begin{pmatrix} -\omega^2 & -1 & -\omega \\ -1 & -\omega & -\omega^2 \\ -\omega & -\omega^2 & -1 \end{pmatrix} \]

### Step 4: Multiply the Resulting Matrix by the Vector
Now we will multiply \( (A + B) \) by the vector \( \begin{pmatrix} 1 \\ \omega \\ \omega^2 \end{pmatrix} \):
\[ (A + B) \begin{pmatrix} 1 \\ \omega \\ \omega^2 \end{pmatrix} = \begin{pmatrix} -\omega^2 \cdot 1 - 1 \cdot \omega - \omega \cdot \omega^2 \\ -1 \cdot 1 - \omega \cdot \omega - \omega^2 \cdot \omega^2 \\ -\omega \cdot 1 - \omega^2 \cdot \omega - 1 \cdot \omega^2 \end{pmatrix} \]

Calculating each entry:
1. First entry:
\[ -\omega^2 - \omega - \omega^3 = -\omega^2 - \omega - 1 \quad (\text{since } \omega^3 = 1) \]
Using \( 1 + \omega + \omega^2 = 0 \):
\[ = 0 \]

2. Second entry:
\[ -1 - \omega^2 - \omega^4 = -1 - \omega^2 - \omega \quad (\text{since } \omega^4 = \omega) \]
Using \( 1 + \omega + \omega^2 = 0 \):
\[ = 0 \]

3. Third entry:
\[ -\omega - \omega^3 - \omega^2 = -\omega - 1 - \omega^2 \]
Using \( 1 + \omega + \omega^2 = 0 \):
\[ = 0 \]

### Step 5: Conclusion
Thus, we have shown that:
\[ (A + B) \begin{pmatrix} 1 \\ \omega \\ \omega^2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
This completes the proof.