JEE Main201910 Apr 2019Evening ShiftMathematicsDeterminantsActual
Let λ be a real number for which the system of linear equations x + y + z = 6 , 4 x + λ y - λ z = λ - 2 and 3 x + 2 y - 4 z = - 5 has infinitely many solutions. Then λ is a root of the quadratic equation:
Options
- Aλ 2 + 3 λ - 4 = 0
- Bλ 2 - λ - 6 = 0
- Cλ 2 - 3 λ - 4 = 0
- Dλ 2 + λ - 6 = 0
Correct answer
B. λ 2 - λ - 6 = 0
Step-by-step solution
The system of equations 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 a 3 x + b 3 y + c 3 z + d 3 = 0 have a infinitely many solution, if D = a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 = 0 ,   D 1 = d 1 b 1 c 1 d 2 b 2 c 2 d 3 b 3 c 3 = 0 ,   D 2 = a 1 d 1 c 1 a 2 d 2 c 2 a 3 d 3 c 3 = 0 and D 3 = a 1 b 1 d 1 a 2 b 2 d 2 a 3 b 3 d 3 = 0 . Thus, for infinitely many solutions, we have D = 0 ⇒ 1 1 1 4 λ - λ 3 2 - 4 = 0 ⇒ 1 - 4 λ + 2 λ - 1 - 16 + 3 λ