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JEE Main 20207 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2020 · 7 January · Morning Shift · Mathematics · Determinants

If the system of linear equations 2x + 2ay + az = 0 2x + 3by + bz =...

If the system of linear equations
2x+2ay+az=02x + 2ay + az = 0
2x+3by+bz=02x + 3by + bz = 0
2x+4cy+cz=0,2x + 4cy + cz = 0,
where a,b,c∈Ra, b, c \in R are non-zero and distinct; has a non-zero solution, then

Options

  1. A1a,1b,1c\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c} are in A.P.A . P .
  2. Ba,b,ca, b, c are in G.P.G . P .
  3. Ca+b+c=0a + b + c = 0
  4. Da,b,ca, b, c are in A.P.A . P .

Correct answer

A. 1a,1b,1c\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c} are in A.P.A . P .

Step-by-step solution

\left|\begin{cases}2 & 2a & a \\ 2 & 3b & b \\ 2 & 4c & c\end{cases}\right| = 0
\left|{12aa13bb14cc\begin{cases}1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c\end{cases}\right| = 0<br/>(3bc−4bc)−(2ac−4ac)+(2ab−3ab)=0<br />\left(3bc - 4bc\right) - \left(2ac - 4ac\right) + \left(2ab - 3ab\right) = 0
−bc+2ac−ab=0- bc + 2ac - ab = 0
ab+bc=2acab + bc = 2ac
a,b,cinH.P.<br/>a, b, c in H . P .<br />⇒ 1a\dfrac{1}{a}, 1b\dfrac{1}{b}, 1c\dfrac{1}{c} in A . P .
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