If the system of equation
\( \lambda x+3y-z=4 \)
\( x+2y+z=20 \)
\( -\lambda x+y+2z=-5 \)
does not possess a unique solution, then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \lambda=-\displaystyle\frac{7}{2} \)
A system AX = B has a unique solution exactly when |A| ≠ 0. So it does not have a unique solution when |A| = 0.
\( \displaystyle |A|=\begin{vmatrix}\lambda & 3 & -1\\1 & 2 & 1\\-\lambda & 1 & 2\end{vmatrix} \)
Expanding along the first row:
\( |A|=\lambda(4-1)-3(2+\lambda)+(-1)(1+2\lambda) \)
\( =3\lambda-6-3\lambda-1-2\lambda \)
\( =-2\lambda-7 \)
\( \displaystyle -2\lambda-7=0\ \Rightarrow\ \lambda=-\frac{7}{2} \)
Why the other options are incorrect
Option (1) \( \lambda=\displaystyle\frac{7}{2} \) — Incorrect. At \( \lambda=\frac{7}{2} \), \( |A|=-14\ne 0 \), so the system has a unique solution.
Option (3) \( \lambda\ne-\displaystyle\frac{7}{2} \) — Incorrect. \( \lambda\ne-\frac{7}{2} \) is exactly the condition for a unique solution, the opposite of what is asked.
Option (4) \( \lambda\ne\displaystyle\frac{7}{2} \) — Incorrect. \( \lambda\ne\frac{7}{2} \) includes values such as \( \lambda=0 \), where \( |A|=-7\ne 0 \) and the solution is unique.