The value of k for which the system of equations \(2x + ky = 3\), \(4x + 6y = 7\) has no solution is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3
If the coefficient determinant is not zero, the system has a unique solution. So for no solution it must be zero.
\(\displaystyle \begin{vmatrix}2 & k\\ 4 & 6\end{vmatrix} = 12 - 4k\)
\(12 - 4k = 0\)
\(k = 3\)
Check that k = 3 really gives no solution: doubling \(2x + 3y = 3\) gives \(4x + 6y = 6\), but the second equation says \(4x + 6y = 7\). The two cannot both hold, so the system is inconsistent.
For every other value of k the determinant is non-zero and there is exactly one solution.