The values of \(\lambda\) for which the system of equation \(x + 2y + z = 14, -x + y + z = 10, x + \lambda y + z = 2\) has unique solution is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\mathrm{R}-\{2\}\); where R is set of real numbers.
A system of three linear equations in three unknowns has a unique solution exactly when the determinant of the coefficients is not zero.
\(\displaystyle D = \begin{vmatrix}1&2&1\\-1&1&1\\1&\lambda&1\end{vmatrix}\)
Expanding along the first row:
\(D = 1(1 - \lambda) - 2(-1 - 1) + 1(-\lambda - 1)\)
\(= 1 - \lambda + 4 - \lambda - 1\)
\(= 4 - 2\lambda\)
\(4 - 2\lambda = 0\) only when \(\lambda = 2\)
So the system has a unique solution for every real \(\lambda\) except 2, that is, \(\lambda \in \mathrm{R} - \{2\}\).