If a ≠ p, b ≠ q, c ≠ r and , then the value of is

Engineering

Mathematics

Introduction to Determinants

Question

If a ≠ p, b ≠ q, c ≠ r and $∣ \begin{array}{ccc} p & b & c \\ a & q & c \\ a & b & r \end{array} ∣ = 0$ , then the value of $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ is

– 1

– 2

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$∣ \begin{array}{ccc} p & b & c \\ a & q & c \\ a & b & r \end{array} ∣ = 0$

R₁ → R₁ – R₃

$∣ \begin{array}{ccc} p - a & 0 & c - r \\ a & q & c \\ a & b & r \end{array} ∣ = 0$

R₂ → R₂ – R₃

∣ \begin{array}{ccc} p - a & 0 & c - r \\ 0 & q - b & c - r \\ a & b & r \end{array} ∣

p – a(r(q – b) – b)(c – r) + (c – r)(– a(q – b))
R₃ → R₁ + R₃

∣ \begin{array}{ccc} p - a & 0 & c - r \\ 0 & q - b & c - r \\ p & b & c \end{array} ∣ = 0

∣ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ \frac{p}{p - a} & \frac{b}{q - b} & \frac{c}{c - r} \end{array} ∣ = 0

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

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

substituting gets value = – 2

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