Q20 / 124 · NTA Abhyas (JEE Main) 2020 · Mathematics · Determinants

Let 2x+ay+6z=8, x+2y+bz=5 and x+y+3z=4 be three equations. If these 3 equations are consistent, then

  • A
    b=3, a≠2
  • Correct answer
    B
    a=2, b≠3
  • C
    a≠2, b≠3
  • D
    a≠2, b=4

Solution

From Cramer’s rule, 
for consistent system △=△1=△2=△3=0
△= 2 a 6 1 2 b 1 1 3 =2 6−b −a 3−b +6 1−2
=12-2b-3a+ab-6
=6-2b-3a+ab
=a-2b-3
△1=8a652b413=86-b-a15-4b+65-8
=48-8b-15a+4ab-18
=30-15a-8b+4ab
=(4b-15)(a-2)
△2=28615b143=215-4b-83-b+64-5
=30-8b-24+8b-6=0
△3=2a8125114=28-5-a4-5+81-2=a-2
For consistent system a=2

Hence, the correct answer is option: 'B'.

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