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JEE Main 202126 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
26 February
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2021 · 26 February · Evening Shift · Mathematics · Determinants

Consider the following system of equations: x + 2y - 3z = a 2x + 6y...

Consider the following system of equations:

x+2y−3z=ax + 2y - 3z = a

2x+6y−11z=b2x + 6y - 11z = b

x−2y+7z=cx - 2y + 7z = c

where a,ba, b and cc are real constants. Then the system of equations :

Options

  1. Ahas a unique solution when 5a=2b+c5a = 2b + c
  2. Bhas no solution for all a,ba, b and cc
  3. Chas infinite number of solutions when 5a=2b+c5a = 2b + c
  4. Dhas a unique solution for all a,ba, b and cc

Correct answer

C. has infinite number of solutions when 5a=2b+c5a = 2b + c

Step-by-step solution

D = \left|\begin{cases}1 & 2 & - 3 \\ 2 & 6 & - 11 \\ 1 & - 2 & 7\end{cases}\right|

=20−2(25)−3(−10)= 20 - 2 ( 25 ) - 3 ( - 10 )

=20−50+30=0= 20 - 50 + 30 = 0

{D}_{1} = \left|\begin{cases}a & 2 & - 3 \\ b & 6 & - 11 \\ c & - 2 & 7\end{cases}\right|

=20a−2(7b+11c)−3(−2b−6c)= 20a - 2 ( 7b + 11c ) - 3 ( - 2b - 6c )

=20a−14b−22c+6b+18c= 20a - 14b - 22c + 6b + 18c

=20a−8b−4c= 20a - 8b - 4c

=4(5a−2b−c)= 4 ( 5a - 2b - c )

{D}_{2} = \left|\begin{cases}1 & a & - 3 \\ 2 & b & - 11 \\ 1 & c & 7\end{cases}\right|

=7b+11c−a(25)−3(2c−b)= 7b + 11c - a ( 25 ) - 3 ( 2c - b )

=7b+11c−25a−6c+3b= 7b + 11c - 25a - 6c + 3b

=−25a+10b+5c= - 25a + 10b + 5c

=−5(5a−2b−c)= - 5 ( 5a - 2b - c )

{D}_{3} = \left|\begin{cases}1 & 2 & a \\ 2 & 6 & b \\ 1 & - 2 & c\end{cases}\right|

=6c+2b−2(2c−b)−10a= 6c + 2b - 2 ( 2c - b ) - 10a

=−10a+4b+2c= - 10a + 4b + 2c

=−2(5a−2b−c)= - 2 ( 5a - 2b - c )

for infinite solution

D=D1=D2=D3=0D = {D}_{1} = {D}_{2} = {D}_{3} = 0

⇒5a=2b+c⇒ 5a = 2b + c

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