Question

Let `lambda` be a real number for which the system of linear equations
x + y +z =6, 4x + `lambday -lambdaz = lambda -2` and 3x + 2y-4z =-5
has infinitely many solutions. Then `lambda` is a root of the quadratic equation
A. `lambda^(2)-3lambda-4 = 0`
B. `lambda^(2) + 3lambda-4 =0`
C. `lambda^(2)-lambda-6=0`
D. `lambda^(2) + lambda-6 =0`

Answer 1

Given, system of linear equations
`x+ y+ z =6 " "… (i)`
`4x+ lambda y - lambda z = lambda-2 " "…(ii)`
`"and " 3x + 2y-4z = -5 " "... (iii)`
has infinitely many solutions, then `Delta = 0`
`rArr |{:(1, 1, 1), (4, lambda, -lambda), (3, 2, -4):}| = 0`
`rArr 1(-4 lambda + 2lambda) -1 (-16 + 3lambda) +1(8-3lambda) = 0`
`rArr -8lambda + 24 =0 rArr lambda = 3`
From, the option `lambda = 3`, satisfy the quadratic equation `lambda^(2) - lambda - 6 = 0`.