If $\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a+b+c)(x+a+b+c)^2$, $x \neq 0$ and $a+b+c \neq 0$, then $x$ is equal to: Options: (A) $2(a+b+c)$ (B) $-(a+b+c)$ (C) $abc$ (D) $-2(a+b+c)$

### Analyzing the Setup When you first look at the determinant provided, it might seem intimidating: $$ \begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} $$ In the world of JEE Advanced, intimidation is often just a mask for elegance. Let us begin by observing the symmetry of the variables $a$, $b$, and $c$. ### The Master Transformation When you see a determinant where the rows or columns have a cyclic or symmetric nature, your first instinct should be to check the sum of the rows. Let us perform the operation $R_1 \to R_1 + R_2 + R_3$. When we add the second and third rows to the first, the first element becomes $(a-b-c) + 2b + 2c$, which simplifies to $(a+b+c)$. The second and third elements follow the same pattern, resulting in a first row consisting entirely of $(a+b+c)$. We can factor this out, leaving us with a row of ones: $$ (a+b+c) \times \begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} $$ ### Simplifying the Matrix The next step is to create zeros to simplify the evaluation. By applying the column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$, we transform the matrix into a lower triangular form: $$ (a+b+c) \times \begin{vmatrix} 1 & 0 & 0 \\ 2b & -(a+b+c) & 0 \\ 2c & 0 & -(a+b+c) \end{vmatrix} $$ Evaluating this is a breeze. The determinant is simply the product of the diagonal elements: $$ (a+b+c) \times (-(a+b+c)) \times (-(a+b+c)) = (a+b+c)^3 $$ ### Final Calculation Finally, we equate this to the right-hand side of the original equation: $$ (a+b+c)^3 = (a+b+c)(x+a+b+c)^2 $$ Assuming $a+b+c \neq 0$, we divide both sides to obtain $(a+b+c)^2 = (x+a+b+c)^2$. This leads to the linear relation $x+a+b+c = \pm(a+b+c)$. We reject the positive case because it implies $x = 0$, which is typically excluded in such problems. Thus, we are left with the final result: **$x = -2(a+b+c)$**

JEE Main 11 Jan 2019 (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: 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:

Q: Could I have just expanded the determinant directly without row operations?

Technically, yes, you could. However, expanding a $3 \times 3$ determinant with variables directly leads to a massive, messy algebraic expression that is highly prone to calculation errors. The 'JEE way' is to simplify the determinant using properties first, then expand. It is faster, cleaner, and much less likely to lead to a mistake.

Select Answer:

Visualized Solution

Analyze the Determinant Structure

Given Equation: $a - b - c 2 b 2 c 2 a b - c - a 2 c 2 a 2 b c - a - b = (a + b + c) (x + a + b + c)^{2}$
Our objective is to find the value of $x$ where $x \neq = 0$ and $a + b + c \neq = 0$ .
We will use Elementary Row Operations to simplify the determinant.

Apply Row Operation $R_{1} \to R_{1} + R_{2} + R_{3}$

Applying $R_{1} \to R_{1} + R_{2} + R_{3}$ :
New $R_{11} = (a - b - c) + 2 b + 2 c = a + b + c$
New $R_{12} = 2 a + (b - c - a) + 2 c = a + b + c$
New $R_{13} = 2 a + 2 b + (c - a - b) = a + b + c$

Factor out $(a + b + c)$ from $R_{1}$

The determinant becomes: $(a + b + c) 1 2 b 2 c 1 b - c - a 2 c 1 2 b c - a - b$
This is equal to the RHS: $(a + b + c) (x + a + b + c)^{2}$

Create Zeros using Column Operations

Applying $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ :
For $C_{2}$ : $1 - 1 = 0$ , $(b - c - a) - 2 b = - (a + b + c)$ , and $2 c - 2 c = 0$ .
For $C_{3}$ : $1 - 1 = 0$ , $2 b - 2 b = 0$ , and $(c - a - b) - 2 c = - (a + b + c)$ .

The Simplified Determinant

The determinant is now: $(a + b + c) 1 2 b 2 c 0 - (a + b + c) 0 00 - (a + b + c)$

Evaluate the Determinant

Expanding along $R_{1}$ :
Value $= (a + b + c) \times [1 \times (- (a + b + c) \times - (a + b + c) - 0)]$
Value $= (a + b + c) \times (a + b + c)^{2} = (a + b + c)^{3}$

Equate to the Given RHS

Equating LHS and RHS:
$(a + b + c)^{3} = (a + b + c) (x + a + b + c)^{2}$
Dividing by $(a + b + c)$ (since $a + b + c \neq = 0$ ):
$(a + b + c)^{2} = (x + a + b + c)^{2}$

Solve for $x$

Taking square root on both sides:
$x + a + b + c = \pm (a + b + c)$
Case 1: $x + a + b + c = + (a + b + c)$
Case 2: $x + a + b + c = - (a + b + c)$

Analyze Case 1 and Case 2

From Case 1: $x = (a + b + c) - (a + b + c) = 0$ (Rejected as $x \neq = 0$ )
From Case 2: $x = - (a + b + c) - (a + b + c)$

Final Calculation and Conclusion

$x = - 2 (a + b + c)$
Comparing with options, the correct choice is Option 4.
Key Takeaway: Use row/column sums to find common factors in symmetric determinants.

00:00 / 00:00

The Sigma Insight: Properties of Determinants

Analyzing the Setup

When you first look at the determinant provided, it might seem intimidating:

a - b - c 2 b 2 c 2 a b - c - a 2 c 2 a 2 b c - a - b

In the world of JEE Advanced, intimidation is often just a mask for elegance. Let us begin by observing the symmetry of the variables

a

b

, and

c

The Master Transformation

When you see a determinant where the rows or columns have a cyclic or symmetric nature, your first instinct should be to check the sum of the rows. Let us perform the operation

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

When we add the second and third rows to the first, the first element becomes

(a - b - c) + 2 b + 2 c

, which simplifies to

(a + b + c)

. The second and third elements follow the same pattern, resulting in a first row consisting entirely of

(a + b + c)

We can factor this out, leaving us with a row of ones:

(a + b + c) \times 1 2 b 2 c 1 b - c - a 2 c 1 2 b c - a - b

Simplifying the Matrix

The next step is to create zeros to simplify the evaluation. By applying the column operations

C_{2} \to C_{2} - C_{1}

and

C_{3} \to C_{3} - C_{1}

, we transform the matrix into a lower triangular form:

(a + b + c) \times 1 2 b 2 c 0 - (a + b + c) 0 00 - (a + b + c)

Evaluating this is a breeze. The determinant is simply the product of the diagonal elements:

(a + b + c) \times (- (a + b + c)) \times (- (a + b + c)) = (a + b + c)^{3}

Final Calculation

Finally, we equate this to the right-hand side of the original equation:

(a + b + c)^{3} = (a + b + c) (x + a + b + c)^{2}

Assuming $a+b+c eq 0$, we divide both sides to obtain

(a + b + c)^{2} = (x + a + b + c)^{2}

. This leads to the linear relation

x + a + b + c = \pm (a + b + c)

We reject the positive case because it implies

x = 0

, which is typically excluded in such problems. Thus, we are left with the final result:

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

Common Traps & FAQs

Why did we add all the rows together? Is there a specific rule for this?

Great question! In JEE Advanced problems involving determinants with cyclic or symmetric variables, the sum of rows or columns is often a hidden factor. By performing

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

, we are essentially 'extracting' the symmetry of the matrix. It is a powerful tool to simplify complex expressions into something manageable.

Why did we reject the case where

x = 0

Always pay attention to the constraints given in the problem statement! The problem explicitly stated

x \neq = 0

. In competitive exams, ignoring these constraints is a common trap. We solved for

x

and found two possibilities; since one violated the given condition, we had to discard it.

Could I have just expanded the determinant directly without row operations?

Technically, yes, you could. However, expanding a

3 \times 3

determinant with variables directly leads to a massive, messy algebraic expression that is highly prone to calculation errors. The 'JEE way' is to simplify the determinant using properties first, then expand. It is faster, cleaner, and much less likely to lead to a mistake.

Animated Solution for Mathematics - Matrices and Determinants: 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:

Select Answer:

Visualized Solution

Analyze the Determinant Structure

Apply Row Operation $R_{1} \to R_{1} + R_{2} + R_{3}$

Factor out $(a + b + c)$ from $R_{1}$

Create Zeros using Column Operations

The Simplified Determinant

Evaluate the Determinant

Equate to the Given RHS

Solve for $x$

Analyze Case 1 and Case 2

Final Calculation and Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Master Transformation

Simplifying the Matrix

Final Calculation

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:

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

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

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Let $a, b, c$ be such that $b (a + c) \neq = 0$ . If $a - b c a + 1 b + 1 c - 1 a - 1 b - 1 c + 1 + a + 1 a - 1 (- 1)^{n + 2} a b + 1 b - 1 (- 1)^{n + 1} b c - 1 c + 1 (- 1)^{n} c = 0$ , then the value of $n$ is:

If $x - 4 2 x 2 x 2 x x - 4 2 x 2 x 2 x x - 4 = (A + B x) (x - A)^{2}$ , then the ordered pair $(A, B)$ is equal to :

If $a b c a^{2} b^{2} c^{2} 1 + a^{3} 1 + b^{3} 1 + c^{3} = 0$ and vectors $(1, a, a^{2}), (1, b, b^{2})$ and $(1, c, c^{2})$ are non-coplanar, then the product $ab c$ equals

If $Δ = x - 2 2 x - 3 3 x - 5 2 x - 3 3 x - 4 5 x - 8 3 x - 4 4 x - 5 10 x - 17 = A x^{3} + B x^{2} + C x + D$ , then $B + C$ is equal to :

Select Answer:

Visualized Solution

Analyze the Determinant Structure

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

Factor out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a+b+c) from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Create Zeros using Column Operations

The Simplified Determinant

Evaluate the Determinant

Equate to the Given RHS

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Analyze Case 1 and Case 2

Final Calculation and Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Master Transformation

Simplifying the Matrix

Final Calculation

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Apply Row Operation $R_{1} \to R_{1} + R_{2} + R_{3}$

Factor out $(a + b + c)$ from $R_{1}$

Solve for $x$

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:

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

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

The value of $(a + 1) (a + 2) (a + 2) (a + 3) (a + 3) (a + 4) a + 2 a + 3 a + 4 111$ is

If $x - 4 2 x 2 x 2 x x - 4 2 x 2 x 2 x x - 4 = (A + B x) (x - A)^{2}$ , then the ordered pair $(A, B)$ is equal to :

If $a b c a^{2} b^{2} c^{2} 1 + a^{3} 1 + b^{3} 1 + c^{3} = 0$ and vectors $(1, a, a^{2}), (1, b, b^{2})$ and $(1, c, c^{2})$ are non-coplanar, then the product $ab c$ equals

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