For what values of \( \lambda \) and \( \mu \) , the following system of linear equations:
x + y + 2z = 1, 2x + y − z = 4, x + 2y + \( \lambda \)z = \( \mu \)
have infinitely many solution?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \lambda=7,\quad\mu=-1 \)
A system has infinitely many solutions only if the determinant of the coefficient matrix is zero and the equations are consistent.
\( \displaystyle \Delta=\begin{vmatrix}1 & 1 & 2\\2 & 1 & -1\\1 & 2 & \lambda\end{vmatrix} \)
\( \displaystyle =1(\lambda+2)-1(2\lambda+1)+2(4-1) \)
\( \displaystyle =7-\lambda \)
\( \displaystyle \Delta=0\ \Rightarrow\ \lambda=7 \)
With \( \lambda=7 \), the third equation must be a combination of the first two. Writing it as a(first) + b(second) and comparing the coefficients of x and y:
\( \displaystyle a+2b=1,\quad a+b=2 \)
\( \displaystyle b=-1,\quad a=3 \)
\( \displaystyle z:\ 3(2)+(-1)(-1)=7=\lambda\ \text{(agrees)} \)
\( \displaystyle \text{right side}:\ 3(1)+(-1)(4)=-1 \)
So the equations are consistent, with infinitely many solutions, only when \( \mu=-1 \).
Why the other options are incorrect
Option (2) \( \lambda\ne 7,\quad\mu \) may have any value — Incorrect. For \( \lambda\ne 7 \), \( \Delta\ne 0 \) and the system has a unique solution.
Option (3) \( \lambda=7,\quad\mu\ne -1 \) — Incorrect. With \( \lambda=7 \) and \( \mu\ne -1 \), the third equation contradicts the first two, so there is no solution.
Option (4) \( \lambda\ne 7,\quad\mu\ne -1 \) — Incorrect. \( \lambda\ne 7 \) gives \( \Delta\ne 0 \), so the solution is unique.