Multiple choice

Let $\lambda$ be a real number for which the system of linear equations $x+y+z=6$ $4x+\lambda y-\lambda z=\lambda -2$ $3x+2y-4z=-5$ has many infinitely many solutions. Then $\lambda$ is a root of the quadratic equation

  1. ${ \lambda }^{ 2 }-3\lambda -4=0\quad $
  2. ${ \lambda }^{ 2 }-\lambda -6=0$
  3. ${ \lambda }^{ 2 }+3\lambda -4=0$
  4. ${ \lambda }^{ 2 }+\lambda -6=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For infinitely many solutions, the determinant of the coefficient matrix must be 0. Matrix: [[1,1,1],[4,L,-L],[3,2,-4]]. Det = 1(-4L + 2L) - 1(-16 + 3L) + 1(8 - 3L) = -2L + 16 - 3L + 8 - 3L = -8L + 24 = 0, so L=3. Checking the quadratic options, L^2 - L - 6 = 0 has roots 3 and -2. Since L=3 is a root, option B is correct.

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