Let λ be a real number for which the system of linear equations x+y+z=6, 4x+λy-λz=λ-2, 3x+2y-4z=-5 has infinitely many solutions. Then λ is a root of the quadratic equation
λ²−3λ−4=0
λ²−λ−6=0
λ²+λ−6=0
λ²+3λ−4=0
Solution:
Correct option is B. λ²−λ−6=0 D=0|1 1 1| = 0 |4 λ -λ| => λ = 3 |3 2 -4| Now,(3)²−3(3)−4=−4≠0 (3)²+3(3)−4=14≠0 (3)²+3−6=6≠0 (3)²−3−6=0 Therefore, Option B is correct; i.e λ=3 is the root of λ²−λ−6=0