Key Concepts and Formulas

• Complex Cube Roots of Unity: If $\omega$ is a non-real cube root of unity, then $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
• Determinant Properties: The value of a determinant remains unchanged if we apply the operation $R_i \to R_i + kR_j$.
• Determinant Expansion: A determinant can be expanded along any row or column.

Step-by-Step Solution

Step 1: Determine the Value of $\omega$

We are given that $2\omega + 1 = z$, where $z = \sqrt{-3} = i\sqrt{3}$. We want to isolate $\omega$.

$2\omega + 1 = i\sqrt{3}$ $2\omega = -1 + i\sqrt{3}$ $\omega = \frac{-1 + i\sqrt{3}}{2}$

Explanation: This step establishes the value of $\omega$ which allows us to use the properties of complex cube roots of unity in subsequent steps.

Step 2: Simplify the Determinant's Elements

We are given the determinant: \left| {\matrix{ 1 & 1 & 1 \cr 1 & { - {\omega ^2} - 1} & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^7}} \cr } } \right| = 3k We simplify the elements involving $\omega$.

• $-{\omega ^2} - 1 = -(\omega^2 + 1)$. Since $1 + \omega + \omega^2 = 0$, we have $\omega^2 + 1 = -\omega$. Thus, $-(\omega^2 + 1) = -(-\omega) = \omega$.

• ${\omega ^7} = {\omega ^{6+1}} = {\omega ^6} \cdot \omega = (\omega^3)^2 \cdot \omega = 1^2 \cdot \omega = \omega$.

Explanation: Simplifying the determinant elements using the properties of $\omega$ makes the determinant easier to work with.

Step 3: Rewrite the Simplified Determinant

Substituting the simplified elements, the determinant becomes: \left| {\matrix{ 1 & 1 & 1 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & \omega \cr } } \right| = 3k

Explanation: This step shows the simplified determinant which is now ready for row operations.

Step 4: Apply Row Operations

Apply the row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$: \left| {\matrix{ 1 & 1 & 1 \cr 0 & \omega - 1 & {{\omega ^2} - 1} \cr 0 & {{\omega ^2} - 1} & \omega - 1 \cr } } \right| = 3k

Explanation: This operation creates zeros in the first column, simplifying the determinant calculation.

Step 5: Evaluate the Determinant

Expanding the determinant along the first column: 1 \cdot \left| {\matrix{ \omega - 1 & {{\omega ^2} - 1} \cr {{\omega ^2} - 1} & \omega - 1 \cr } } \right| = 3k $(\omega - 1)^2 - (\omega^2 - 1)^2 = 3k$ $(\omega^2 - 2\omega + 1) - (\omega^4 - 2\omega^2 + 1) = 3k$ Since $\omega^3 = 1$, $\omega^4 = \omega$. $(\omega^2 - 2\omega + 1) - (\omega - 2\omega^2 + 1) = 3k$ $\omega^2 - 2\omega + 1 - \omega + 2\omega^2 - 1 = 3k$ $3\omega^2 - 3\omega = 3k$ $3(\omega^2 - \omega) = 3k$ $k = \omega^2 - \omega$

Explanation: Expanding the determinant and simplifying the expression allows us to solve for $k$ in terms of $\omega$.

Step 6: Express k in Terms of z

We have $k = \omega^2 - \omega$. We know $\omega = \frac{-1 + i\sqrt{3}}{2}$. Also, since $1+\omega+\omega^2 = 0$, $\omega^2 = -1-\omega = -1 - \frac{-1+i\sqrt{3}}{2} = \frac{-2+1-i\sqrt{3}}{2} = \frac{-1-i\sqrt{3}}{2}$. Therefore, $k = \frac{-1 - i\sqrt{3}}{2} - \frac{-1 + i\sqrt{3}}{2} = \frac{-1 - i\sqrt{3} + 1 - i\sqrt{3}}{2} = \frac{-2i\sqrt{3}}{2} = -i\sqrt{3}$ Since $z = i\sqrt{3}$, we have $k = -z$.

Explanation: This step expresses $k$ in terms of $z$ as required by the problem.

Common Mistakes & Tips

• Remember the properties of cube roots of unity: $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
• Simplify expressions involving $\omega$ before evaluating the determinant.
• Be careful with signs during algebraic manipulations.

Summary

We found the value of $\omega$ from the given equation, simplified the determinant using the properties of $\omega$, applied row operations to simplify the determinant, and finally expressed $k$ in terms of $z$. We arrived at $k = -z$.

The final answer is \boxed{-z}, which corresponds to option (D).