Given that `[(1,omega,omega^(2)),(omega,omega^(2),1),(omega^(2),1,omega)][(k,1,1),(1,1,1),(1,1,1)]=[(0,0,0),(0,0,0),(0,0,0)]` then k=

Options

A. 6
B. 1
C. 8
D. 9

Answer

To solve the problem, we need to multiply the two given matrices and set the resulting matrix equal to the null matrix. Let's denote the matrices as follows:

Let \( A = \begin{pmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{pmatrix} \) and \( B = \begin{pmatrix} k & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \).

We need to find \( k \) such that:

\[ A \cdot B = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

### Step 1: Calculate the first element of the product \( A \cdot B \)

The first element of the resulting matrix is calculated as follows:

\[ (1 \cdot k) + (\omega \cdot 1) + (\omega^2 \cdot 1) = k + \omega + \omega^2 \]

Setting this equal to 0 gives us:

\[ k + \omega + \omega^2 = 0 \quad \text{(Equation 1)} \]

### Step 2: Calculate the second element of the product \( A \cdot B \)

The second element of the resulting matrix is:

\[ (\omega \cdot k) + (\omega^2 \cdot 1) + (1 \cdot 1) = \omega k + \omega^2 + 1 \]

Setting this equal to 0 gives us:

\[ \omega k + \omega^2 + 1 = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

From Equation 1, we have:

\[ k = -(\omega + \omega^2) \]

We know from the properties of the cube roots of unity that:

\[ 1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1 \]

Thus, substituting this into the expression for \( k \):

\[ k = -(-1) = 1 \]

### Step 4: Verify with Equation 2

Substituting \( k = 1 \) into Equation 2:

\[ \omega(1) + \omega^2 + 1 = \omega + \omega^2 + 1 = 0 \]

This confirms that both equations are satisfied.

### Conclusion

Thus, the value of \( k \) is:

\[ \boxed{1} \]