Question

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 infinitely many solutions. Then \(\lambda\) is a root of the quadratic equation :

• A. \(\lambda^2+\lambda-6=0\)
• B. \(\lambda^2-\lambda-6=0\)
• C. \(\lambda^2-3\lambda-4=0\)
• D. \(\lambda^2+3\lambda-4=0\)

Hint:

Infinitely many solutions need the coefficient determinant to vanish; this fixes \(\lambda\), then check consistency.

Answer 1

The Correct Option is B

Coefficient determinant:
\(\Delta=\begin{vmatrix}1&1&1\\4&\lambda&-\lambda\\3&2&-4\end{vmatrix}=1(-4\lambda+2\lambda)-1(-16+3\lambda)+1(8-3\lambda)=24-8\lambda\)
For infinitely many solutions \(\Delta=0\Rightarrow \lambda=3\).
Check consistency at \(\lambda=3\): the equations are \(x+y+z=6\), \(4x+3y-3z=1\), \(3x+2y-4z=-5\). Subtracting the third from the second gives \(x+y+z=6\), the same as the first. So the system is consistent with infinitely many solutions.
Now \(\lambda=3\): \(\lambda^2-\lambda-6=9-3-6=0\) (option 2), while \(\lambda^2+\lambda-6=6\), \(\lambda^2-3\lambda-4=-4\), \(\lambda^2+3\lambda-4=14\).
Answer: (2) \(\lambda^2-\lambda-6=0\)