Key Concepts and Formulas

This problem involves evaluating a $3 \times 3$ determinant whose elements are powers of $\omega$, where $1, \omega, \omega^2$ are the cube roots of unity. To solve this, we will utilize the fundamental properties of cube roots of unity and determinant operations.

1. Properties of Cube Roots of Unity:

   • $\omega^3 = 1$: This means that any power of $\omega$ can be simplified by taking its exponent modulo 3. For any integer $k$, $\omega^{3k} = (\omega^3)^k = 1^k = 1$.
   • $1 + \omega + \omega^2 = 0$: The sum of the three cube roots of unity is zero.
   • Generalization for $\omega^n$:
     • If $n$ is a multiple of 3 (i.e., $n=3k$ for some integer $k$), then $\omega^n = 1$ and $\omega^{2n} = 1$. In this case, $1 + \omega^n + \omega^{2n} = 1+1+1 = 3$.
     • If $n$ is NOT a multiple of 3 (i.e., $n \not\equiv 0 \pmod 3$), then $\omega^n \neq 1$. In this case, the sum $1 + \omega^n + \omega^{2n}$ is a geometric series sum $\frac{(\omega^n)^3 - 1}{\omega^n - 1} = \frac{\omega^{3n} - 1}{\omega^n - 1} = \frac{1 - 1}{\omega^n - 1} = 0$.
     • Therefore, $1 + \omega^n + \omega^{2n}$ is either 0 (if $n$ is not a multiple of 3) or 3 (if $n$ is a multiple of 3).

2. Determinant Properties:

   • Column Operations: Applying an operation like $C_i \to C_i + k C_j$ (adding a multiple of one column to another column) does not change the value of the determinant.
   • Common Factor: If all elements of a row or column have a common factor, that factor can be taken out of the determinant.
   • Zero Column/Row: If any column or row consists entirely of zeros, the value of the determinant is zero.
   • Identical/Proportional Columns/Rows: If two columns or two rows are identical or proportional, the value of the determinant is zero.

Step-by-Step Solution

1. State the given determinant: We are given the determinant: \Delta = \left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr {{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr {{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr } } \right|

2. Apply a strategic column operation: We observe the cyclic nature of the elements in the determinant. A common strategy for such determinants (especially those involving roots of unity) is to perform a column (or row) operation that creates a common factor. Let's apply the column operation $C_1 \to C_1 + C_2 + C_3$.

• Why this step? This operation sums the elements across each row into the first column. This is particularly useful here because we know the sum $1 + \omega^n + \omega^{2n}$ simplifies significantly based on the properties of cube roots of unity.

After applying the operation, the elements in the first column become:

• First element: $1 + \omega^n + \omega^{2n}$
• Second element: $\omega^n + \omega^{2n} + 1$
• Third element: $\omega^{2n} + 1 + \omega^n$

Notice that all elements in the new first column are identical: $(1 + \omega^n + \omega^{2n})$. So, the determinant transforms to: \Delta = \left| {\matrix{ {1 + {\omega ^n} + {\omega ^{2n}}} & {{\omega ^n}} & {{\omega ^{2n}}} \cr {1 + {\omega ^n} + {\omega ^{2n}}} & {{\omega ^{2n}}} & 1 \cr {1 + {\omega ^n} + {\omega ^{2n}}} & 1 & {{\omega ^n}} \cr } } \right|

3. Factor out the common term from the first column: We can factor out the common term $(1 + \omega^n + \omega^{2n})$ from the first column.

• Why this step? Factoring out a common term simplifies the determinant, making it easier to evaluate or recognize further properties. \Delta = (1 + {\omega ^n} + {\omega ^{2n}}) \left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr 1 & {{\omega ^{2n}}} & 1 \cr 1 & 1 & {{\omega ^n}} \cr } } \right|

4. Analyze the value of the factor $(1 + \omega^n + \omega^{2n})$: As established in the "Key Concepts" section, the sum $1 + \omega^n + \omega^{2n}$ depends on whether $n$ is a multiple of 3.

• Case A: If $n$ is a multiple of 3 (e.g., $n=0, 3, 6, \dots$) In this case, $\omega^n = 1$ and $\omega^{2n} = 1$. Therefore, the factor $1 + \omega^n + \omega^{2n} = 1 + 1 + 1 = 3$. Substituting this into the determinant expression: \Delta = 3 \left| {\matrix{ 1 & 1 & 1 \cr 1 & 1 & 1 \cr 1 & 1 & 1 \cr } } \right|

  • Why this step? We need to evaluate the determinant for specific conditions of $n$ to cover all possibilities. The determinant on the right has all three rows identical. A determinant with identical rows (or columns) is equal to zero. Thus, $\Delta = 3 \times 0 = 0$.

• Case B: If $n$ is NOT a multiple of 3 (e.g., $n=1, 2, 4, 5, \dots$) In this case, $\omega^n \neq 1$. Therefore, the factor $1 + \omega^n + \omega^{2n} = 0$. Substituting this into the determinant expression: \Delta = 0 \cdot \left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr 1 & {{\omega ^{2n}}} & 1 \cr 1 & 1 & {{\omega ^n}} \cr } } \right|

  • Why this step? This covers the other condition for $n$. Since the multiplying factor is 0, the entire determinant $\Delta$ must be 0. Thus, $\Delta = 0$.

Conclusion: In both cases, whether $n$ is a multiple of 3 or not, the value of the determinant $\Delta$ is 0. This means the determinant is always 0 for any integer $n$.

The final answer is $\boxed{\text{0}}$.

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Important Note: The provided "Correct Answer" for this problem is (A) $\omega^2$. However, based on the fundamental properties of cube roots of unity and determinant algebra, the determinant for the given matrix structure consistently evaluates to 0 for any integer $n$. This is a well-known result for this type of cyclic determinant involving powers of $\omega$. If a problem were designed to yield $\omega^2$, the structure of the determinant itself would need to be different. As an expert teacher, I must present the mathematically sound derivation.

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Tips for Success & Common Mistakes:

• Master Cube Roots of Unity Properties: The properties $1+\omega+\omega^2=0$ and $\omega^3=1$ are indispensable for problems involving $\omega$. Always simplify powers of $\omega$ using $\omega^3=1$ first.
• Recognize Cyclic Determinants: Determinants with cyclically permuted elements (like this one) often simplify greatly by adding rows/columns. Look for patterns like $a, b, c; b, c, a; c, a, b$ or $a, b, c; c, a, b; b, c, a$.
• Strategic Row/Column Operations: The choice of $C_1 \to C_1 + C_2 + C_3$ (or $R_1 \to R_1 + R_2 + R_3$) is very common when elements sum to a useful quantity (like 0 in this case).
• Avoid Calculation Errors: When expanding determinants or simplifying powers, be meticulous with signs and exponents.
• Consider All Cases: For expressions like $1 + \omega^n + \omega^{2n}$, always consider the cases where $n$ is a multiple of 3 and when it is not.

Summary/Key Takeaway:

This problem demonstrates a classic application of the properties of cube roots of unity in evaluating determinants. The key insight is that the sum $1 + \omega^n + \omega^{2n}$ is either 0 or 3, which dramatically simplifies the determinant after a suitable column operation. For the given determinant, the value is consistently 0, regardless of the integer value of $n$.