Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0$. Then for $y \neq 0$ in $\mathbb{R}$, $\begin{vmatrix} y+1 & \alpha & \beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha \end{vmatrix}$ is equal to Options: (A) $y^3$ (B) $y^3 - 1$ (C) $y(y^2 - 1)$ (D) $y(y^2 - 3)$

### Analyzing the Setup The problem presents a determinant involving complex numbers and variables. While it may appear chaotic, we must look for the hidden structure. We begin with the equation $x^2 + x + 1 = 0$. This is the defining equation for the complex cube roots of unity, where the roots are $\omega$ and $\omega^2$. By assigning $\alpha = \omega$ and $\beta = \omega^2$, we transform the intimidating determinant into a structured matrix: $$ \Delta = \begin{vmatrix} y+1 & \omega & \omega^2 \\ \omega & y+\omega^2 & 1 \\ \omega^2 & 1 & y+\omega \end{vmatrix} $$ ### The Power of Row Operations Observe the cyclic symmetry within the rows and columns. Whenever such symmetry exists, row or column operations are the most efficient path forward. Let us apply the operation $R_1 \to R_1 + R_2 + R_3$. We are essentially summing the entire matrix into the first row. When we add the columns, the elements of the first row become: $(y+1) + \omega + \omega^2$, $\omega + (y+\omega^2) + 1$, and $\omega^2 + 1 + (y+\omega)$. ### The Magic of Unity Recall the fundamental property of cube roots of unity: $1 + \omega + \omega^2 = 0$. Each element in our new first row simplifies to $y + (1 + \omega + \omega^2) = y$. Our determinant now takes the form: $$ \Delta = \begin{vmatrix} y & y & y \\ \omega & y+\omega^2 & 1 \\ \omega^2 & 1 & y+\omega \end{vmatrix} $$ We can now factor out $y$ from the first row, leaving us with a row of ones: $$ \Delta = y \begin{vmatrix} 1 & 1 & 1 \\ \omega & y+\omega^2 & 1 \\ \omega^2 & 1 & y+\omega \end{vmatrix} $$ ### Final Calculation Expanding along the first row becomes trivial with a row of ones. We calculate the minors, keeping in mind that $\omega^3 = 1$ and $\omega^4 = \omega$. As you expand, terms such as $y\omega$ and $-y\omega$ cancel out, as do $y\omega^2$ and $-y\omega^2$. The entire expression inside the bracket simplifies down to $y^2$. Multiplying by the $y$ we factored out earlier, we arrive at the final result: **$\Delta = y^3$**

JEE Main 2019 (09 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $α$ and $β$ be the roots of the equation $x^{2} + x + 1 = 0$ . Then for $y \neq = 0$ in $R$ , $y + 1 α β α y + β 1 β 1 y + α$ is equal to

Select Answer:

Progressive Hints

Hint 1

Hint 1: Look closely at the quadratic equation

x^{2} + x + 1 = 0

. In the world of JEE Advanced, this is a classic signature. The roots are not just any numbers; they are the complex cube roots of unity,

ω

and

ω^{2}

. Recognizing this is your first step to simplifying the entire problem.

Hint 2

Hint 2: When you see a determinant with cyclic symmetry—where the elements shift positions across rows and columns—do not rush to expand it. Instead, look for a row operation that makes the rows identical. Try adding all rows together:

R_{1} \to R_{1} + R_{2} + R_{3}

Hint 3

Hint 3: Once you perform the row operation, you will find that every element in the first row becomes

y + 1 + ω + ω^{2}

. Recall the fundamental identity

1 + ω + ω^{2} = 0

. Use this to simplify the first row to

[y, y, y]

and then factor out the

y

Visualized Solution

Roots of $x^{2} + x + 1 = 0$

Given equation: $x^{2} + x + 1 = 0$
The roots of this standard equation are the complex cube roots of unity.
Let the roots be $ω$ and $ω^{2}$ .
So, we can set $α = ω$ and $β = ω^{2}$ .

Substitute $α$ and $β$ into the Determinant

We need to evaluate the determinant $Δ$ .
Substitute $α = ω$ and $β = ω^{2}$ into $Δ$ .
$Δ = y + 1 ω ω^{2} ω y + ω^{2} 1 ω^{2} 1 y + ω$

Symmetry and Row Operation

Notice the cyclic symmetry in the columns and rows.
To simplify, apply the row operation: $R_{1} \to R_{1} + R_{2} + R_{3}$ .
This will add corresponding elements of the second and third rows to the first row.

Execute $R_{1} \to R_{1} + R_{2} + R_{3}$

Adding the rows, the first row becomes:
$R_{1} = [y + 1 + ω + ω^{2}, y + ω^{2} + 1 + ω, y + 1 + ω + ω^{2}]$
$Δ = y + 1 + ω + ω^{2} ω ω^{2} y + 1 + ω + ω^{2} y + ω^{2} 1 y + 1 + ω + ω^{2} 1 y + ω$

Apply $1 + ω + ω^{2} = 0$

Recall the fundamental property of cube roots of unity: $1 + ω + ω^{2} = 0$ .
Substitute this into the first row.
The first row simplifies beautifully to $[y, y, y]$ .
$Δ = y ω ω^{2} y y + ω^{2} 1 y 1 y + ω$

Factor Out $y$ from $R_{1}$

Since $y$ is common in all elements of the first row, we can factor it out.
$Δ = y 1 ω ω^{2} 1 y + ω^{2} 1 11 y + ω$

Expand the Determinant

Expand the determinant along the simplified first row $R_{1}$ .
$Δ = y [1 ((y + ω^{2}) (y + ω) - 1) - 1 (ω (y + ω) - ω^{2}) + 1 (ω - ω^{2} (y + ω^{2}))]$

Simplify the Expansion

Expand the brackets carefully:
Term 1: $1 ((y + ω^{2}) (y + ω) - 1) = y^{2} + y ω + y ω^{2} + ω^{3} - 1$
Term 2: $- 1 (ω (y + ω) - ω^{2}) = - (y ω + ω^{2} - ω^{2})$
Term 3: $+ 1 (ω - ω^{2} (y + ω^{2})) = ω - y ω^{2} - ω^{4}$
We will use $ω^{3} = 1$ and $ω^{4} = ω$ to simplify these terms.

Final Cancellation

Substitute $ω^{3} = 1$ and $ω^{4} = ω$ :
Term 1 becomes $y^{2} + y ω + y ω^{2} + 1 - 1 = y^{2} + y ω + y ω^{2}$
Term 2 becomes $- (y ω + 0) = - y ω$
Term 3 becomes $ω - y ω^{2} - ω = - y ω^{2}$
Combining them: $Δ = y [y^{2} + y ω + y ω^{2} - y ω - y ω^{2}]$

Final Result

Notice that $+ y ω$ and $- y ω$ cancel out, as do $+ y ω^{2}$ and $- y ω^{2}$ .
The expression inside the bracket collapses to just $y^{2}$ .
$Δ = y [y^{2}] = y^{3}$
Final Answer: $y^{3}$

00:00 / 00:00

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The problem presents a determinant involving complex numbers and variables. While it may appear chaotic, we must look for the hidden structure.

We begin with the equation

x^{2} + x + 1 = 0

. This is the defining equation for the complex cube roots of unity, where the roots are

ω

and

ω^{2}

By assigning

α = ω

and

β = ω^{2}

, we transform the intimidating determinant into a structured matrix:

Δ = y + 1 ω ω^{2} ω y + ω^{2} 1 ω^{2} 1 y + ω

The Power of Row Operations

Observe the cyclic symmetry within the rows and columns. Whenever such symmetry exists, row or column operations are the most efficient path forward.

Let us apply the operation

R_{1} \to R_{1} + R_{2} + R_{3}

. We are essentially summing the entire matrix into the first row.

When we add the columns, the elements of the first row become:

(y + 1) + ω + ω^{2}

ω + (y + ω^{2}) + 1

, and

ω^{2} + 1 + (y + ω)

The Magic of Unity

Recall the fundamental property of cube roots of unity:

1 + ω + ω^{2} = 0

. Each element in our new first row simplifies to

y + (1 + ω + ω^{2}) = y

Our determinant now takes the form:

Δ = y ω ω^{2} y y + ω^{2} 1 y 1 y + ω

We can now factor out

y

from the first row, leaving us with a row of ones:

Δ = y 1 ω ω^{2} 1 y + ω^{2} 1 11 y + ω

Final Calculation

Expanding along the first row becomes trivial with a row of ones. We calculate the minors, keeping in mind that

ω^{3} = 1

and

ω^{4} = ω

As you expand, terms such as

y ω

and

- y ω

cancel out, as do

y ω^{2}

and

- y ω^{2}

. The entire expression inside the bracket simplifies down to

y^{2}

Multiplying by the

y

we factored out earlier, we arrive at the final result:

$Δ = y^{3}$

Common Traps & FAQs

Why can't I just expand the determinant directly? It seems like a lot of work, but won't I get the same answer?

You absolutely could, but in a high-stakes exam like JEE Advanced, time is your most precious currency. Expanding a

3 \times 3

determinant with complex variables directly is a recipe for algebraic errors. The 'trap' here is the sheer volume of terms. By using properties to simplify the determinant first, you reduce the complexity, minimize the chance of a sign error, and reach the answer with elegance and speed.

How do I know for sure that

α = ω

and

β = ω^{2}

? Could it be the other way around?

That is a fantastic question! In this specific determinant, the structure is symmetric with respect to

α

and

β

. Even if you swapped them, the determinant would remain unchanged. The cyclic nature of the matrix ensures that the final result is invariant to the assignment of the roots. So, rest easy—your choice won't affect the final outcome.

I got stuck after factoring out the

y

. The expansion still looks messy. Is there a trick?

The trick is to stay calm and trust the process. Once you have a row of ones (after factoring out

y

), you are in a very strong position. Expand along that row. Yes, the algebra looks intimidating, but remember that

ω^{3} = 1

and

ω^{4} = ω

. These identities are designed to make the 'messy' terms vanish. Keep going; the cancellation is almost always cleaner than you expect.

Animated Solution for Mathematics - Matrices and Determinants: Let $α$ and $β$ be the roots of the equation $x^{2} + x + 1 = 0$ . Then for $y \neq = 0$ in $R$ , $y + 1 α β α y + β 1 β 1 y + α$ is equal to

Select Answer:

Visualized Solution

Roots of $x^{2} + x + 1 = 0$

Substitute $α$ and $β$ into the Determinant

Symmetry and Row Operation

Execute $R_{1} \to R_{1} + R_{2} + R_{3}$

Apply $1 + ω + ω^{2} = 0$

Factor Out $y$ from $R_{1}$

Expand the Determinant

Simplify the Expansion

Final Cancellation

Final Result

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Power of Row Operations

The Magic of Unity

Final Calculation

Similar Questions

If $a + x = b + y = c + z + 1$ , where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $x y z a + y b + y c + y x + a y + b z + c$ is equal to :

If $x + 1 x x x x + λ x x x x + λ = (103 x + 81)$ , then $λ, \frac{λ}{3}$ are the roots of the equation

The values of $α$ , for which $11 2 α + 3 \frac{3}{2} \frac{1}{3} 3 α + 1 α + \frac{3}{2} α + \frac{1}{3} 0 = 0$ , lie in the interval

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The determinant $a b a α + b b c b α + c a α + b b α + c 0$ is equal to zero, if

If $ω (\neq = 1)$ is a cube root of unity, then $1 1 - i - i 1 + i + ω^{2} - 1 - i + ω - 1 ω^{2} ω^{2} - 1 - 1 =$

If $Δ_{1} = x - sin θ cos θ sin θ - x 1 cos θ 1 x$ and $Δ_{2} = x - sin 2 θ cos 2 θ sin 2 θ - x 1 cos 2 θ 1 x$ , $x \neq = 0$ ; then for all $θ \in (0, \frac{π}{2})$ :

Without expanding a determinant at any stage, show that $x^{2} + x 2 x^{2} + 3 x - 1 x^{2} + 2 x + 3 x + 1 3 x 2 x - 1 x - 2 3 x - 3 2 x - 1 = x A + B$ , where $A$ and $B$ are determinants of order $3$ not involving $x$ .

If $a > 0$ and discriminant of $a x^{2} + 2 b x + c$ is $- v e$ , then $a b a x + b b c b x + c a x + b b x + c 0$ is equal to

Select Answer:

Visualized Solution

Roots of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+x+1=0

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β into the Determinant

Symmetry and Row Operation

Execute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​→R1​+R2​+R3​

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+ω+ω2=0

Factor Out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Expand the Determinant

Simplify the Expansion

Final Cancellation

Final Result

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Power of Row Operations

The Magic of Unity

Final Calculation

Similar Questions

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Roots of $x^{2} + x + 1 = 0$

Substitute $α$ and $β$ into the Determinant

Execute $R_{1} \to R_{1} + R_{2} + R_{3}$

Apply $1 + ω + ω^{2} = 0$

Factor Out $y$ from $R_{1}$

If $a + x = b + y = c + z + 1$ , where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $x y z a + y b + y c + y x + a y + b z + c$ is equal to :

If $x + 1 x x x x + λ x x x x + λ = (103 x + 81)$ , then $λ, \frac{λ}{3}$ are the roots of the equation

The values of $α$ , for which $11 2 α + 3 \frac{3}{2} \frac{1}{3} 3 α + 1 α + \frac{3}{2} α + \frac{1}{3} 0 = 0$ , lie in the interval

If $α \neq = a, β \neq = b, γ \neq = c$ and $α a a b β b c c γ = 0$ , then $\frac{a}{α - a} + \frac{b}{β - b} + \frac{γ}{γ - c}$ is equal to:

The determinant $a b a α + b b c b α + c a α + b b α + c 0$ is equal to zero, if

If $ω (\neq = 1)$ is a cube root of unity, then $1 1 - i - i 1 + i + ω^{2} - 1 - i + ω - 1 ω^{2} ω^{2} - 1 - 1 =$

If $Δ_{1} = x - sin θ cos θ sin θ - x 1 cos θ 1 x$ and $Δ_{2} = x - sin 2 θ cos 2 θ sin 2 θ - x 1 cos 2 θ 1 x$ , $x \neq = 0$ ; then for all $θ \in (0, \frac{π}{2})$ :

If $a > 0$ and discriminant of $a x^{2} + 2 b x + c$ is $- v e$ , then $a b a x + b b c b x + c a x + b b x + c 0$ is equal to