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JEE Main 201325 AprilOnline
Exam
JEE Main
Year
2013
Date
25 April
Mode
Online
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2013 · 25 April · Online · Mathematics · Hyperbola

Consider the system of equations : x+a y=0, y+a z=0 and z+a x=0 ....

Consider the system of equations : x+ay=0,y+az=0x+a y=0, y+a z=0 and z+ax=0z+a x=0. Then the set of all real values of ' aa ' for which the system has a unique solution is:

Options

  1. AR−{1}\mathrm{R}-\{1\}
  2. BR−{−1}\mathrm{R}-\{-1\}
  3. C{1,−1}\{1,-1\}
  4. D{1,0,−1}\{1,0,-1\}

Correct answer

B. R−{−1}\mathrm{R}-\{-1\}

Step-by-step solution

Given system of equations is homogeneous which is x+ay=0y+az=0z+ax=0\begin{aligned}&x+a y=0 \\ &y+a z=0 \\ &z+a x=0 \end{aligned}It can be writtenin matrix form as\mathrm{A}=\left(1a001aa01\begin{array}{lll}1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{array}$\right) Now,|\mathrm{A}|=\left[1-a\left(-a^2\right)\right]=1+a^3 \neq 0So,systemhasonlytrivialsolution.Now,So, system has only trivial solution. Now,|\mathrm{A}|=0onlywhenonly whena=-1So,systemofequationshasinfinitelymanysolutionswhichisnotpossiblebecauseitisgiventhatsystemhasauniquesolution.Hencesetofallrealvaluesof′So, system of equations has infinitely many solutions which is not possible because it is given that system has a unique solution. Hence set of all real values of 'a′is' is\mathrm{R}-$\{-1\}
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