The system of equation \( x-y+z=4,x-2y-2z=9,\ 2x+y+\alpha z=1 \) has unique solution then value (/s) of \( \alpha \) is (/are) (where R is set of real numbers)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \alpha\in\mathrm{R}\sim\{11\} \)
A system AX = B of three equations in three unknowns has a unique solution exactly when |A| ≠ 0.
\( \displaystyle |A|=\begin{vmatrix}1 & -1 & 1\\1 & -2 & -2\\2 & 1 & \alpha\end{vmatrix} \)
Expanding along the first row:
\( \displaystyle |A|=1(-2\alpha+2)-(-1)(\alpha+4)+1(1+4) \)
\( \displaystyle =-2\alpha+2+\alpha+4+5 \)
\( \displaystyle =11-\alpha \)
\( \displaystyle |A|\ne 0\ \Rightarrow\ \alpha\ne 11 \)
So the system has a unique solution for every real \( \alpha \) except 11, that is, \( \alpha\in\mathrm{R}\sim\{11\} \).
Why the other options are incorrect
Option (1) \( \alpha\in\mathrm{R} \) — Incorrect. For \( \alpha=11 \), |A| = 0 and the solution is not unique (there is none or infinitely many), so not every real value works.
Option (2) \( \alpha\in\mathrm{R}\sim\{-11\} \) — Incorrect. The sign is wrong: |A| = 11 − α vanishes at α = 11, not −11.
Option (3) 11 — Incorrect. α = 11 is the only value for which the solution is not unique.