Let `omega` be a complex number such that `2omega+1=z` where `z=sqrt(-3)`. If `|{:(1,1,1),(1,-omega^(2)-1,omega^(2)),(1,omega^(2),omega^(7))|=3k`, then k is equal to

Options

A. 1
B. `-z`
C. z
D. -1

Answer

To solve the problem step by step, we will follow the instructions provided in the video transcript and break down the solution into clear steps.

### Step 1: Find the value of \( \omega \)

We are given that \( z = \sqrt{-3} \). We can rewrite this as:
\[ z = i\sqrt{3} \]
where \( i = \sqrt{-1} \).

Next, we know that:
\[ 2\omega + 1 = z \]
Substituting the value of \( z \):
\[ 2\omega + 1 = i\sqrt{3} \]
Now, we can solve for \( \omega \):
\[ 2\omega = i\sqrt{3} - 1 \]
\[ \omega = \frac{i\sqrt{3} - 1}{2} \]

### Step 2: Write the determinant

The determinant we need to evaluate is:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -\omega^2 - 1 & \omega^2 \\ 1 & \omega^2 & \omega^7 \end{vmatrix} \]
We know that \( \omega^3 = 1 \), so \( \omega^7 = \omega \).

Thus, we can rewrite the determinant as:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -\omega^2 - 1 & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix} \]

### Step 3: Simplify the determinant

We can perform column operations on the determinant. We will add the first column to the second and third columns:
\[ D = \begin{vmatrix} 1 & 1 + 1 & 1 + 1 \\ 1 & -\omega^2 & \omega^2 + 1 \\ 1 & \omega^2 + 1 & \omega + 1 \end{vmatrix} \]
This simplifies to:
\[ D = \begin{vmatrix} 1 & 2 & 2 \\ 1 & -\omega^2 & \omega^2 + 1 \\ 1 & \omega^2 + 1 & \omega + 1 \end{vmatrix} \]

### Step 4: Calculate the determinant

Now we can calculate the determinant using expansion:
\[ D = 1 \cdot \begin{vmatrix} -\omega^2 & \omega^2 + 1 \\ \omega^2 + 1 & \omega + 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & \omega^2 + 1 \\ 1 & \omega + 1 \end{vmatrix} \]

Calculating the first determinant:
\[ \begin{vmatrix} -\omega^2 & \omega^2 + 1 \\ \omega^2 + 1 & \omega + 1 \end{vmatrix} = -\omega^2(\omega + 1) - (\omega^2 + 1)(\omega^2) \]
Calculating the second determinant:
\[ \begin{vmatrix} 1 & \omega^2 + 1 \\ 1 & \omega + 1 \end{vmatrix} = 1(\omega + 1) - 1(\omega^2 + 1) = \omega + 1 - \omega^2 - 1 = \omega - \omega^2 \]

### Step 5: Substitute back into the determinant equation

Now substituting these back into the determinant:
\[ D = -\omega^2(\omega + 1) - \omega^2 - 1 - 2(\omega - \omega^2) \]
This simplifies to:
\[ D = -\omega^3 - \omega^2 - \omega^2 - 1 - 2\omega + 2\omega^2 \]
Using \( \omega^3 = 1 \):
\[ D = -1 - 2\omega^2 - 2\omega \]

### Step 6: Relate to \( 3k \)

We know from the problem statement that:
\[ D = 3k \]
Thus, we can equate:
\[ -1 - 2\omega^2 - 2\omega = 3k \]
From here, we can solve for \( k \):
\[ k = \frac{-1 - 2\omega^2 - 2\omega}{3} \]

### Step 7: Substitute the value of \( \omega \)

Substituting \( \omega = \frac{i\sqrt{3} - 1}{2} \) into the equation will give us the final value of \( k \).