Let α be a root of the equation (a - c)x² + (b - a)x + (c - b) = 0 where a, b, c are distinct real n

Q: Wait, why is $x=1$ a root of the quadratic equation?

This is a classic property of polynomials! If the sum of all coefficients in a polynomial equation is zero, then $x=1$ is always a root. In our case, $(a-c) + (b-a) + (c-b) = 0$, so $x=1$ must be a root.

Q: Why is the matrix singular if $\alpha=1$?

A matrix is singular if its determinant is zero. When we substitute $\alpha=1$ into the matrix, the first row becomes $[1, 1, 1]$ and the second row is already $[1, 1, 1]$. Since two rows are identical, the determinant is automatically zero, confirming the matrix is singular.

Q: Why can we assume $\alpha=1$?

The problem states that $\alpha$ is a root of the quadratic equation and that the matrix is singular. We found that $x=1$ is a root of the quadratic, and substituting $\alpha=1$ makes the matrix singular. This consistency confirms that $\alpha=1$ is indeed the root we are looking for.

Q: How does $X+Y+Z=0$ lead to $X^3+Y^3+Z^3=3XYZ$?

This is a fundamental algebraic identity. If $X+Y+Z=0$, then $X+Y = -Z$. Cubing both sides gives $(X+Y)^3 = (-Z)^3$, which expands to $X^3 + Y^3 + 3XY(X+Y) = -Z^3$. Substituting $X+Y = -Z$ back in, we get $X^3 + Y^3 - 3XYZ = -Z^3$, which rearranges to $X^3 + Y^3 + Z^3 = 3XYZ$.

Question

Let $\alpha$ be a root of the equation $(a-c) x^{2}+(b-a) x+(c-b)=0$ where $a, b, c$ are distinct real numbers such that the matrix $\begin{bmatrix} \alpha^{2} & \alpha & 1 \ 1 & 1 & 1 \ a & b & c \end{bmatrix}$ is singular. Then the value of $\frac{(a-c)^{2}}{(b-a)(c-b)}+\frac{(b-a)^{2}}{(a-c)(c-b)}+\frac{(c-b)^{2}}{(a-c)(b-a)}$ is

Options:

(A) 6

(B) 3

(C) 9

(D) 12

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the precipice of a complex problem, staring at a quadratic equation that seems to defy simplicity:

$$(a-c)x^2 + (b-a)x + (c-b) = 0$$

At first glance, the variables $a$, $b$, and $c$ might seem like a chaotic jumble. But in the world of JEE Advanced, chaos is often just order in disguise. Let us embark on a journey to unravel this mystery.

### The Quadratic Mystery

The first step in our journey is to observe the coefficients of our quadratic equation. We have $A = (a-c)$, $B = (b-a)$, and $C = (c-b)$.

Notice the cyclic nature of these terms? It is almost like a dance where each variable takes a turn being subtracted from the next. Whenever you encounter such a structure, your mathematical intuition should scream, "Add them up!"

Let us calculate the sum of these coefficients:

$$(a-c) + (b-a) + (c-b)$$

As we expand this, we see $a - c + b - a + c - b$. The $a$'s cancel, the $b$'s cancel, and the $c$'s cancel. The sum is exactly zero.

A fundamental theorem of polynomials states that if the sum of the coefficients of a polynomial is zero, then $x=1$ is guaranteed to be a root. Thus, we have identified that $\alpha = 1$.

### The Matrix Dance

Now, the problem introduces a matrix $M$ and tells us it is singular:

$$M = \begin{bmatrix} \alpha^2 & \alpha & 1 \ 1 & 1 & 1 \ a & b & c \end{bmatrix}$$

What does it mean for a matrix to be singular? It means its determinant is zero. Let us test our finding, $\alpha = 1$, by substituting it into the matrix.

The matrix becomes:

$$\begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ a & b & c \end{bmatrix}$$

Look at the first two rows. They are identical! In the realm of linear algebra, if any two rows of a matrix are identical, the determinant is zero. This confirms our finding: $\alpha = 1$ is indeed the correct root.

### The Algebraic Symphony

We are now left with the final challenge: evaluating the expression:

$$E = \frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(a-c)(c-b)} + \frac{(c-b)^2}{(a-c)(b-a)}$$

This looks intimidating, but let us simplify it using the substitution method. Let $X = a-c$, $Y = b-a$, and $Z = c-b$. We already know that $X+Y+Z = 0$.

Our expression now transforms into a much cleaner form:

$$E = \frac{X^2}{YZ} + \frac{Y^2}{XZ} + \frac{Z^2}{XY}$$

To add these fractions, we find the least common multiple of the denominators, which is $XYZ$. The expression becomes:

$$E = \frac{X \cdot X^2 + Y \cdot Y^2 + Z \cdot Z^2}{XYZ} = \frac{X^3 + Y^3 + Z^3}{XYZ}$$

### The Grand Finale

We have arrived at the climax of our journey. We need to evaluate $X^3 + Y^3 + Z^3$.

We recall the powerful algebraic identity: if $X+Y+Z = 0$, then $X^3 + Y^3 + Z^3 = 3XYZ$. This identity is a cornerstone of competitive mathematics.

Substituting this into our expression, we get:

$$E = \frac{3XYZ}{XYZ}$$

Since $a$, $b$, and $c$ are distinct, $X$, $Y$, and $Z$ are non-zero, allowing us to cancel $XYZ$ from the numerator and denominator.

The result is simply **3**.

Select Answer:

Visualized Solution

Analyze the Quadratic Equation

Sum of Coefficients Property

Identify the Root $α$

Singular Matrix Condition

Verify Determinant

Define New Variables

Check the Sum of $X, Y, Z$

Rewrite the Target Expression

Simplify the Fractions

Apply Algebraic Identity

Final Calculation

Summary and Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Quadratic Mystery

The Matrix Dance

The Algebraic Symphony

The Grand Finale

Similar Questions

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:

Let $β$ be a real number. Consider the matrix $A = β 23 011 1 - 2 - 2$ . If $A^{7} - (β - 1) A^{6} - β A^{5}$ is a singular matrix, then the value of $9 β$ is ______.

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 $a \neq = p, b \neq = q, c \neq = r$ and $p a a b q b c c r = 0$ . Then find the value of $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ .

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

If $a - b - c 2 b 2 c 2 a b - c - a 2 c 2 a 2 b c - a - b = (a + b + c) (x + a + b + c)^{2}$ , $x \neq = 0$ and $a + b + c \neq = 0$ , then $x$ is equal to:

Which of the following values of $α$ satisfy the equation $(1 + α)^{2} (2 + α)^{2} (3 + α)^{2} (1 + 2 α)^{2} (2 + 2 α)^{2} (3 + 2 α)^{2} (1 + 3 α)^{2} (2 + 3 α)^{2} (3 + 3 α)^{2} = - 648 α$ ?

The value of the determinant $111 a b c a^{2} - b c b^{2} - c a c^{2} - ab$ is .........

If matrix $A = a b c b c a c a b$ where $a, b, c$ are real positive numbers, $ab c = 1$ and $A^{T} A = I$ , then find the value of $a^{3} + b^{3} + c^{3}$ .

The sum of the real roots of the equation $x 2 - 3 - 6 - 3 x 2 x - 1 x - 3 x + 2 = 0$ , is equal to :