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

### Analyzing the Setup We are given the determinant equation: $$ \begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \end{vmatrix} = 0 $$ We are tasked with finding the value of the expression: $$ \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} $$ The constraint $a \neq p$, $b \neq q$, and $c \neq r$ ensures that the denominators are non-zero, allowing us to perform algebraic manipulations without the risk of division by zero. ### Sculpting the Determinant To bridge the gap between the determinant and our target expression, we must generate the terms $(p-a)$, $(q-b)$, and $(r-c)$ within the matrix. We apply the following row operations: First, perform $R_1 \to R_1 - R_2$: $$ \begin{vmatrix} p-a & b-q & 0 \\ a & q & c \\ a & b & r \end{vmatrix} = 0 $$ Next, perform $R_2 \to R_2 - R_3$: $$ \begin{vmatrix} p-a & b-q & 0 \\ 0 & q-b & c-r \\ a & b & r \end{vmatrix} = 0 $$ ### Simplifying the Structure We can factor out $(p-a)$ from the first column, $(q-b)$ from the second column, and $(r-c)$ from the third column. Note that $(b-q) = -(q-b)$ and $(c-r) = -(r-c)$. Factoring these terms out, we obtain: $$ (p-a)(q-b)(r-c) \begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \frac{a}{p-a} & \frac{b}{q-b} & \frac{r}{r-c} \end{vmatrix} = 0 $$ Since the product $(p-a)(q-b)(r-c) \neq 0$, we can divide the entire equation by this product. ### Final Calculation Expanding the remaining determinant along the first row: $$ 1 \left( \frac{r}{r-c} + \frac{b}{q-b} \right) - (-1) \left( 0 - \left( -\frac{a}{p-a} \right) \right) = 0 $$ This simplifies to: $$ \frac{r}{r-c} + \frac{b}{q-b} + \frac{a}{p-a} = 0 $$ To reach our target, we rewrite $a$ as $p - (p-a)$ and $b$ as $q - (q-b)$: $$ \left( \frac{p}{p-a} - 1 \right) + \left( \frac{q}{q-b} - 1 \right) + \frac{r}{r-c} = 0 $$ Grouping the terms, we find: $$ \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} - 2 = 0 $$ Thus, the final value is: **2**

JEE Advanced 1991

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: 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 $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ .

Q: Wait, why do we divide by $(p-a)(q-b)(r-c)$? Is that allowed?

Great question! The problem explicitly states $a \neq p, b \neq q, c \neq r$. This is the problem setter's way of giving you a 'green light' to divide. Because these terms are non-zero, dividing by them is perfectly valid and is the key to unlocking the target expression.

Q: How do I know which row operation to pick?

Look at the target! The target denominators are $(p-a)$, $(q-b)$, and $(r-c)$. Row operations are designed to create these differences. Always let the target expression dictate your strategy. If you see a difference in the denominator, create that difference in the matrix!

Enter Numerical Value:

Visualized Solution

Given Determinant and Target

Given Determinant: $p a a b q b c c r = 0$
Target Expression: $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$
Constraints: $a \neq = p, b \neq = q, c \neq = r$ (Ensures denominators are non-zero)

Strategy: Row Operations

Strategy: Use row operations to generate the terms $(p - a)$ , $(q - b)$ , and $(r - c)$ within the determinant.
First Operation: $R_{1} \to R_{1} - R_{2}$

Applying $R_{1} \to R_{1} - R_{2}$

Applying $R_{1} \to R_{1} - R_{2}$ :
$p - a a a b - q q b 0 c r = 0$

Applying $R_{2} \to R_{2} - R_{3}$

Applying $R_{2} \to R_{2} - R_{3}$ :
$p - a 0 a b - q q - b b 0 c - r r = 0$

Factoring Out from Columns

Factor out $(p - a)$ from $C_{1}$ , $(q - b)$ from $C_{2}$ , and $(r - c)$ from $C_{3}$ :
$(p - a) (q - b) (r - c) 10 \frac{a}{p - a} \frac{b - q}{q - b} 1 \frac{b}{q - b} 0 \frac{c - r}{r - c} \frac{r}{r - c} = 0$

Simplifying the Determinant

Notice that $\frac{b - q}{q - b} = - 1$ and $\frac{c - r}{r - c} = - 1$ .
Since $(p - a) (q - b) (r - c) \neq = 0$ , we can divide it out:
$10 \frac{a}{p - a} - 1 1 \frac{b}{q - b} 0 - 1 \frac{r}{r - c} = 0$

Expanding the Determinant

Expanding along $R_{1}$ :
$1 \cdot (1 \cdot \frac{r}{r - c} - (- 1) \cdot \frac{b}{q - b}) - (- 1) \cdot (0 \cdot \frac{r}{r - c} - (- 1) \cdot \frac{a}{p - a}) = 0$
$(\frac{r}{r - c} + \frac{b}{q - b}) + (\frac{a}{p - a}) = 0$
$\frac{a}{p - a} + \frac{b}{q - b} + \frac{r}{r - c} = 0$

The Numerator Substitution Trick

We need $\frac{p}{p - a}$ and $\frac{q}{q - b}$ , but we have $\frac{a}{p - a}$ and $\frac{b}{q - b}$ .
Rewrite the numerators: $a = p - (p - a)$ and $b = q - (q - b)$ .
Substitute these into the equation:
$\frac{p - ( p - a )}{p - a} + \frac{q - ( q - b )}{q - b} + \frac{r}{r - c} = 0$

Splitting the Fractions

Split the fractions with the new numerators:
$(\frac{p}{p - a} - \frac{p - a}{p - a}) + (\frac{q}{q - b} - \frac{q - b}{q - b}) + \frac{r}{r - c} = 0$
$(\frac{p}{p - a} - 1) + (\frac{q}{q - b} - 1) + \frac{r}{r - c} = 0$

Final Calculation and Result

Group the target terms together:
$\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} - 1 - 1 = 0$
Move the constants to the right side:
$\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2$
Final Answer: $2$

00:00 / 00:00

The Sigma Insight: Properties of Determinants

Analyzing the Setup

We are given the determinant equation:

p a a b q b c c r = 0

We are tasked with finding the value of the expression:

\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}

The constraint $a eq p$, $b eq q$, and $c eq r$ ensures that the denominators are non-zero, allowing us to perform algebraic manipulations without the risk of division by zero.

Sculpting the Determinant

To bridge the gap between the determinant and our target expression, we must generate the terms

(p - a)

(q - b)

, and

(r - c)

within the matrix. We apply the following row operations:

First, perform

R_{1} \to R_{1} - R_{2}

p - a a a b - q q b 0 c r = 0

Next, perform

R_{2} \to R_{2} - R_{3}

p - a 0 a b - q q - b b 0 c - r r = 0

Simplifying the Structure

We can factor out

(p - a)

from the first column,

(q - b)

from the second column, and

(r - c)

from the third column. Note that

(b - q) = - (q - b)

and

(c - r) = - (r - c)

Factoring these terms out, we obtain:

(p - a) (q - b) (r - c) 10 \frac{a}{p - a} - 1 1 \frac{b}{q - b} 0 - 1 \frac{r}{r - c} = 0

Since the product $(p-a)(q-b)(r-c) eq 0$, we can divide the entire equation by this product.

Final Calculation

Expanding the remaining determinant along the first row:

1 (\frac{r}{r - c} + \frac{b}{q - b}) - (- 1) (0 - (- \frac{a}{p - a})) = 0

This simplifies to:

\frac{r}{r - c} + \frac{b}{q - b} + \frac{a}{p - a} = 0

To reach our target, we rewrite

a

p - (p - a)

and

b

q - (q - b)

(\frac{p}{p - a} - 1) + (\frac{q}{q - b} - 1) + \frac{r}{r - c} = 0

Grouping the terms, we find:

\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} - 2 = 0

Thus, the final value is:

Common Traps & FAQs

Why can't we just expand the determinant directly? It looks like a standard 3x3.

You could, but it would be an algebraic nightmare! Expanding directly leads to a massive polynomial that is very hard to factor. Row operations are the 'surgical' tools of linear algebra—they allow us to simplify the structure before we do the heavy lifting. Always look for ways to simplify the matrix before expanding.

Wait, why do we divide by

(p - a) (q - b) (r - c)

? Is that allowed?

Great question! The problem explicitly states

a \neq = p, b \neq = q, c \neq = r

. This is the problem setter's way of giving you a 'green light' to divide. Because these terms are non-zero, dividing by them is perfectly valid and is the key to unlocking the target expression.

How do I know which row operation to pick?

Look at the target! The target denominators are

(p - a)

(q - b)

, and

(r - c)

. Row operations are designed to create these differences. Always let the target expression dictate your strategy. If you see a difference in the denominator, create that difference in the matrix!

Animated Solution for Mathematics - Matrices and Determinants: 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 $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ .

Enter Numerical Value:

Visualized Solution

Given Determinant and Target

Strategy: Row Operations

Applying $R_{1} \to R_{1} - R_{2}$

Applying $R_{2} \to R_{2} - R_{3}$

Factoring Out from Columns

Simplifying the Determinant

Expanding the Determinant

The Numerator Substitution Trick

Splitting the Fractions

Final Calculation and Result

The Sigma Insight: Properties of Determinants

Analyzing the Setup

Sculpting the Determinant

Simplifying the Structure

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

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For all values of $A, B, C$ and $P, Q, R$ show that $cos (A - P) cos (B - P) cos (C - P) cos (A - Q) cos (B - Q) cos (C - Q) cos (A - R) cos (B - R) cos (C - R) = 0$ .

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=p,b=q,c=r and ​paa​bqb​ccr​​=0. Then find the value of p−ap​+q−bq​+r−cr​.

Enter Numerical Value:

Visualized Solution

Given Determinant and Target

Strategy: Row Operations

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

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

Factoring Out from Columns

Simplifying the Determinant

Expanding the Determinant

The Numerator Substitution Trick

Splitting the Fractions

Final Calculation and Result

The Sigma Insight: Properties of Determinants

Analyzing the Setup

Sculpting the Determinant

Simplifying the Structure

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Matrices and Determinants: 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 $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ .

Applying $R_{1} \to R_{1} - R_{2}$

Applying $R_{2} \to R_{2} - R_{3}$

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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:

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 :

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

Prove that for all values of $θ$ , $sin θ sin (θ + \frac{2 π}{3}) sin (θ - \frac{2 π}{3}) cos θ cos (θ + \frac{2 π}{3}) cos (θ - \frac{2 π}{3}) sin 2 θ sin (2 θ + \frac{4 π}{3}) sin (2 θ - \frac{4 π}{3}) = 0$ .

For all values of $A, B, C$ and $P, Q, R$ show that $cos (A - P) cos (B - P) cos (C - P) cos (A - Q) cos (B - Q) cos (C - Q) cos (A - R) cos (B - R) cos (C - R) = 0$ .