The system of equations \(2x + \lambda y = 8,\ \lambda x+8y = 3\) has a unique solution if the value of \(\lambda\) is (are):
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\lambda \ne \pm 4\)
Two linear equations in two unknowns have a unique solution exactly when the determinant of the coefficients is non-zero:
\( \displaystyle \begin{vmatrix}2&\lambda\\ \lambda&8\end{vmatrix} \ne 0 \)
\(16 - \lambda^2 \ne 0\)
\(\lambda \ne \pm 4\)
For \(\lambda = 4\) or \(\lambda = -4\) the determinant is 0; the equations then represent parallel lines (for example \(x + 2y = 4\) and \(x + 2y = \frac{3}{4}\) when \(\lambda = 4\)), so there is no solution at all.