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JEE Main 20188 April
Exam
JEE Main
Year
2018
Date
8 April
Subject
Mathematics
Chapter
Determinants

JEE Main 2018 · 8 April · Mathematics · Determinants

If the system of linear equations x + ky + 3z = 0 3x + ky - 2z = 0...

If the system of linear equations

x+ky+3z=0x + ky + 3z = 0
3x+ky−2z=03x + ky - 2z = 0
2x+4y−3z=02x + 4y - 3z = 0

has a non-zero solution (x,y,z),\left(x, y, z\right), then xzy2\dfrac{xz}{{y}^{2}} is equal to:

Options

  1. A3030
  2. B−10- 10
  3. C1010
  4. D−30- 30

Correct answer

C. 1010

Step-by-step solution

\left|\begin{cases}1 & k & 3 \\ 3 & k & - 2 \\ 2 & 4 & - 3\end{cases}\right| = 0

⇒(−3k+8)−3(−3k−12)+2(−5k)=0⇒ \left(- 3k + 8\right) - 3\left(- 3k - 12\right) + 2\left(- 5k\right) = 0

⇒−4k+44=0⇒ - 4k + 44 = 0 ⇒⇒ k=11k = 11

⇒x+11y+3z=0...(1)⇒ x + 11y + 3z = 0 . . . \left(1\right)

⇒3x+11y−2z=0...(2)⇒ 3x + 11y - 2z = 0 . . . \left(2\right)

⇒2x+4y−3z=0...(3)⇒ 2x + 4y - 3z = 0 . . . \left(3\right)

On solving equations\left(1\right) \text{ and } \left(2\right),weget<br/><br/>, we get<br /><br />2x - 5z = 0<br/><br/><br /><br />⇒ x = \dfrac{5z}{2}<br/><br/>Putitinequation(3)<br /><br />Put it in equation \left(3\right), we get

2z+4y=02z + 4y = 0

⇒z=−2y⇒ z = - 2y

∴xzy2=(−2y)×(−5y)y2=10∴ \dfrac{xz}{{y}^{2}} = \dfrac{\left(- 2y\right) \times \left(- 5y\right)}{{y}^{2}} = 10
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