The value/s of 'λ' for which the system of equations
\( \lambda x+3y-z=1 \)
\( x+2y+z=2 \)
\( -\lambda x+y+2z=-1 \)
has unique solution is/are:
(where R is set of real numbers)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( R-\left\{-\displaystyle\frac{7}{2}\right\} \)
A system AX = B of three equations has a unique solution exactly when \( |A|\ne 0 \).
\( \displaystyle |A|=\begin{vmatrix}\lambda & 3 & -1\\1 & 2 & 1\\-\lambda & 1 & 2\end{vmatrix} \)
\( \displaystyle =\lambda(4-1)-3(2+\lambda)+(-1)(1+2\lambda) \)
\( \displaystyle =3\lambda-6-3\lambda-1-2\lambda \)
\( \displaystyle =-2\lambda-7 \)
\( \displaystyle |A|=0\ \Rightarrow\ \lambda=-\frac{7}{2} \)
So the system has a unique solution for every real λ except \( -\frac{7}{2} \), that is, for \( \lambda\in R-\left\{-\frac{7}{2}\right\} \).
Why the other options are incorrect
Option (1) \( \displaystyle\frac{7}{2} \) — Incorrect. \( \lambda=\frac{7}{2} \) does give a unique solution, but so does every other value except \( -\frac{7}{2} \); a single value is not the complete answer.
Option (2) \( -\displaystyle\frac{7}{2} \) — Incorrect. \( \lambda=-\frac{7}{2} \) is the only value for which |A| = 0, so it is exactly the value that does not give a unique solution.
Option (3) \( R-\left\{\displaystyle\frac{7}{2}\right\} \) — Incorrect. This excludes \( \frac{7}{2} \) instead of \( -\frac{7}{2} \), a sign slip in solving \( -2\lambda-7=0 \).