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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Determinants

BITSAT 2014 · Mathematics · Determinants

If | array ccc p & q-y & r-z p-x & q & r-z p-x & q-y & r array |=0,...

If \left|pq−yr−zp−xqr−zp−xq−yr\begin{array}{ccc}\mathrm{p} & \mathrm{q}-\mathrm{y} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0, then the value of px+qy+rz\frac{p}{x}+\frac{q}{y}+\frac{r}{z} is

Options

  1. A0
  2. B1
  3. C2
  4. D4pqr4 \mathrm{pqr}

Correct answer

C. 2

Step-by-step solution

\left|pq−yr−zp−xqr−zp−xq−yr\begin{array}{ccc}\mathrm{p} & \mathrm{q}-\mathrm{y} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0 Apply R1→R1−R3\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3} and R2→R2−R3,\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}, we get \left|x0−z0y−zp−xq−yr\begin{array}{ccc}\mathrm{x} & 0 & -\mathrm{z} \\ 0 & \mathrm{y} & -\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0\Rightarrow \mathrm{x}[\mathrm{yr}+\mathrm{z}(\mathrm{q}-\mathrm{y})]-\mathrm{z}[0-\mathrm{y}(\mathrm{p}-\mathrm{x})]=0 [Expansion along first row ]⇒xyr+zxq+yzp=2xyz⇒px+qy+rz=2]\Rightarrow x y r+z x q+y z p=2 x y z \Rightarrow \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2
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