JEE Main 2018MathematicsDeterminantsEasyMCQ

JEE Main 2018 — Determinants Question with Solution

JEE Main 2018 (08 Apr)

Question

If the system of linear equations

x+ky+3z=0
3x+ky-2z=0
2x+4y-3z=0

has a non-zero solution x, y, z, then xzy2 is equal to:

Choose an option

Show full solutionCorrect option: C
Correct answer
C10

Step-by-step explanation

1k33k-224-3=0

⇒-3k+8-3-3k-12+2-5k=0

⇒-4k+44=0 ⇒ k=11

⇒x+11y+3z=0   ...1

⇒3x+11y-2z=0   ...2

⇒2x+4y-3z=0   ...3

On solving equations1 & 2, we get

2x-5z=0

⇒x=5z2

Put it in equation 3, we get

2z+4y=0

⇒z=-2y

∴xzy2=-2y×-5yy2=10

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About this question

This is a previous-year question from JEE Main 2018, covering the Determinants chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.