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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2019 · 10 April · Evening Shift · Mathematics · Determinants

Let λ be a real number for which the system of linear equations x +...

Let λ\lambda be a real number for which the system of linear equations

x+y+z=6,x + y + z = 6,

4x+\lambday−\lambdaz=λ−24x + \lambday - \lambdaz = \lambda - 2 and

3x+2y−4z=−53x + 2y - 4z = - 5

has infinitely many solutions. Then λ\lambda is a root of the quadratic equation:

Options

  1. Aλ2+3λ−4=0{\lambda}^{2} + 3\lambda - 4 = 0
  2. Bλ2−λ−6=0{\lambda}^{2} - \lambda - 6 = 0
  3. Cλ2−3λ−4=0{\lambda}^{2} - 3\lambda - 4 = 0
  4. Dλ2+λ−6=0{\lambda}^{2} + \lambda - 6 = 0

Correct answer

B. λ2−λ−6=0{\lambda}^{2} - \lambda - 6 = 0

Step-by-step solution

The system of equations a1x+b1y+c1z+d1=0,a2x+b2y+c2z+d2=0{a}_{1}x + {b}_{1}y + {c}_{1}z + {d}_{1} = 0, {a}_{2}x + {b}_{2}y + {c}_{2}z + {d}_{2} = 0 and a3x+b3y+c3z+d3=0{a}_{3}x + {b}_{3}y + {c}_{3}z + {d}_{3} = 0 have a infinitely many solution, if

D = \left|\begin{cases}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{cases}\right| = 0, {D}_{1} = \left|\begin{cases}{d}_{1} & {b}_{1} & {c}_{1} \\ {d}_{2} & {b}_{2} & {c}_{2} \\ {d}_{3} & {b}_{3} & {c}_{3}\end{cases}\right| = 0, {D}_{2} = \left|\begin{cases}{a}_{1} & {d}_{1} & {c}_{1} \\ {a}_{2} & {d}_{2} & {c}_{2} \\ {a}_{3} & {d}_{3} & {c}_{3}\end{cases}\right| = 0 and {D}_{3} = \left|\begin{cases}{a}_{1} & {b}_{1} & {d}_{1} \\ {a}_{2} & {b}_{2} & {d}_{2} \\ {a}_{3} & {b}_{3} & {d}_{3}\end{cases}\right| = 0 .

Thus, for infinitely many solutions, we have D=0D = 0

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

⇒1(−4λ+2λ)−1(−16+3λ)+1(8−3λ)=0⇒ 1\left(- 4\lambda + 2\lambda\right) - 1\left(- 16 + 3\lambda\right) + 1\left(8 - 3\lambda\right) = 0

⇒−2λ+16−3λ+8−3λ=0⇒ - 2\lambda + 16 - 3\lambda + 8 - 3\lambda = 0

⇒24−3λ=0⇒ 24 - 3\lambda = 0

⇒λ=3⇒ \lambda = 3

And, λ=3\lambda = 3 satisfies the equation λ2−λ−6=0.{\lambda}^{2} - \lambda - 6 = 0 .

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