This problem masterfully combines concepts from complex numbers (specifically, roots of unity) and determinants. The key to solving it efficiently lies in recognizing the properties of the roots of the given quadratic equation and applying strategic determinant operations.

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1. Key Concepts: Roots of Unity and Properties

The foundation of this problem is understanding the roots of the equation $x^2 + x + 1 = 0$. Definition: The roots of $x^2 + x + 1 = 0$ are the non-real cube roots of unity. These are commonly denoted as $\omega$ and $\omega^2$. Let's assign $\alpha = \omega$ and $\beta = \omega^2$ (or vice-versa).

From the theory of roots of unity, we know the following fundamental properties:

• Sum of roots: $\alpha + \beta = \omega + \omega^2 = -1$ (from Vieta's formulas, or directly).
• Product of roots: $\alpha \beta = \omega \cdot \omega^2 = \omega^3 = 1$ (from Vieta's formulas, or directly).
• Crucial identity: $1 + \omega + \omega^2 = 0$. Therefore, $1 + \alpha + \beta = 0$.
• Difference squared: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (-1)^2 - 4(1) = 1 - 4 = -3$.

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2. Setting up the Determinant

We are asked to evaluate the determinant: D = \left| {\matrix{ {y + 1} & \alpha & \beta \cr \alpha & {y + \beta } & 1 \cr \beta & 1 & {y + \alpha } \cr } } \right| The variable $y$ is a non-zero real number.

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3. Applying Column Operations to Simplify

To simplify determinants, we often look for opportunities to create zeros or common factors using row or column operations. Strategy: We notice the identity $1 + \alpha + \beta = 0$. If we add all columns to the first column, we can leverage this identity. Apply the column operation $C_1 \to C_1 + C_2 + C_3$:

• The new first element of $C_1$ will be $(y + 1) + \alpha + \beta = y + (1 + \alpha + \beta) = y + 0 = y$.
• The new second element of $C_1$ will be $\alpha + (y + \beta) + 1 = y + (\alpha + \beta + 1) = y + 0 = y$.
• The new third element of $C_1$ will be $\beta + 1 + (y + \alpha) = y + (\beta + 1 + \alpha) = y + 0 = y$.

So the determinant becomes: D = \left| {\matrix{ y & \alpha & \beta \cr y & {y + \beta } & 1 \cr y & 1 & {y + \alpha } \cr } } \right|

Explanation: This step is taken because it creates a common factor ($y$) in the first column, which can then be factored out, simplifying the determinant significantly.

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4. Factoring out the Common Term

Now, factor out $y$ from the first column: D = y \left| {\matrix{ 1 & \alpha & \beta \cr 1 & {y + \beta } & 1 \cr 1 & 1 & {y + \alpha } \cr } } \right|

Explanation: Factoring out a common term from a row or column makes the numbers inside the determinant smaller, making subsequent calculations easier.

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5. Creating Zeros in the First Column

To further simplify, we can create zeros in the first column using row operations. Apply the row operations:

• $R_2 \to R_2 - R_1$
• $R_3 \to R_3 - R_1$

The determinant becomes: D = y \left| {\matrix{ 1 & \alpha & \beta \cr 1 - 1 & (y + \beta) - \alpha & 1 - \beta \cr 1 - 1 & 1 - \alpha & (y + \alpha) - \beta \cr } } \right| D = y \left| {\matrix{ 1 & \alpha & \beta \cr 0 & {y + \beta - \alpha } & {1 - \beta } \cr 0 & {1 - \alpha } & {y + \alpha - \beta } \cr } } \right|

Explanation: Creating zeros in a row or column is a standard technique to reduce the order of the determinant when expanding it, as terms multiplied by zero vanish.

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6. Expanding the Determinant

Now, expand the determinant along the first column (which contains two zeros): D = y \cdot \left[ 1 \cdot \left| {\matrix{ {y + \beta - \alpha } & {1 - \beta } \cr {1 - \alpha } & {y + \alpha - \beta } \cr } } \right| - 0 + 0 \right] $D = y \left[ (y + \beta - \alpha)(y + \alpha - \beta) - (1 - \beta)(1 - \alpha) \right]$

Explanation: Expanding along a column with zeros simplifies the calculation to a single $2 \times 2$ determinant. The formula for a $2 \times 2$ determinant \left| {\matrix{ a & b \cr c & d \cr } } \right| = ad - bc is applied.

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7. Evaluating the Expressions

Let's evaluate the two terms inside the square bracket separately using the properties of $\alpha$ and $\beta$:

Term 1: $(y + \beta - \alpha)(y + \alpha - \beta)$ This expression is of the form $(A+B)(A-B) = A^2 - B^2$, where $A=y$ and $B=(\alpha - \beta)$. So, $(y + (\beta - \alpha))(y - (\beta - \alpha)) = y^2 - (\beta - \alpha)^2$.

Now, substitute the value of $(\beta - \alpha)^2$: We know $(\alpha - \beta)^2 = -3$. Since $(\beta - \alpha)^2 = (-1)^2 (\alpha - \beta)^2 = (\alpha - \beta)^2$, then $(\beta - \alpha)^2 = -3$. Therefore, Term 1 $= y^2 - (-3) = y^2 + 3$.

Term 2: $(1 - \beta)(1 - \alpha)$ Expand this product: $(1 - \beta)(1 - \alpha) = 1 - \alpha - \beta + \alpha\beta$ Factor out the negative sign from $\alpha$ and $\beta$: $= 1 - (\alpha + \beta) + \alpha\beta$ Substitute the known values $\alpha + \beta = -1$ and $\alpha\beta = 1$: $= 1 - (-1) + 1 = 1 + 1 + 1 = 3$.

Explanation: These evaluations directly use the properties derived in Section 1. This is where the specific nature of $\alpha$ and $\beta$ as roots of unity becomes critical.

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8. Final Calculation

Substitute the evaluated terms back into the determinant expression: $D = y \left[ (y^2 + 3) - 3 \right]$ $D = y \left[ y^2 \right]$ $D = y^3$

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9. Tips and Common Mistakes

• Recognize Roots of Unity: Immediately identify $x^2+x+1=0$ with $\omega$ and $\omega^2$. This is a common pattern in JEE problems.
• Properties are Key: Memorize or be able to quickly derive the key properties like $1+\omega+\omega^2=0$, $\omega^3=1$, and $(\omega-\omega^2)^2 = -3$.
• Strategic Operations: Don't just expand the determinant immediately. Look for row/column operations that simplify the matrix (e.g., creating common factors or zeros).
• Algebraic Precision: Be careful with signs, especially when dealing with expressions like $(\alpha-\beta)^2$. A common mistake is to forget the negative sign.

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10. Summary / Key Takeaway

This problem demonstrates how combining knowledge of complex numbers (roots of unity) with determinant properties can lead to an elegant solution. The key steps were:

1. Identifying the roots $\alpha, \beta$ as $\omega, \omega^2$ and listing their properties.
2. Using $1+\alpha+\beta=0$ to simplify the determinant via column operations, extracting a common factor $y$.
3. Further simplification by creating zeros and expanding the determinant.
4. Carefully substituting the properties of $\alpha, \beta$ to evaluate the resulting terms.

The final answer is $\boxed{\text{y}^3}$.