If the system of equations \(2x + 3y = 10, x + ky = 4\) has a unique solution, then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(k \ne 3/2\)
A system AX = B has a unique solution when the coefficient determinant is non-zero:
\( \displaystyle |A| = \begin{vmatrix}2 & 3\\ 1 & k\end{vmatrix} = 2k - 3 \)
\(2k - 3 \ne 0\)
\( \displaystyle k \ne \frac{3}{2} \)
For \(k = \frac{3}{2}\) the two lines are parallel (the second equation times 2 is \(2x + 3y = 8\)), so there is no solution at all.