The system of equations kx + 2y = 5 3x + (k - 1)y = 7 has no solution…

The correct answer is Option (2) → \(k = 3\) or \(k = -2\)

If the coefficient determinant is not zero, the system has exactly one solution. So a system with no solution needs

    \( \displaystyle \begin{vmatrix}k & 2\\ 3 & k - 1\end{vmatrix} = 0 \)

    \( k(k - 1) - 6 = 0 \)
    \( k^2 - k - 6 = 0 \)
    \( (k - 3)(k + 2) = 0 \)

Each of these values must still be tested, because a zero determinant can also give infinitely many solutions.

For k = 3 the equations are

    \( 3x + 2y = 5 \)
    \( 3x + 2y = 7 \)

The same left side cannot equal both 5 and 7, so there is no solution.

For k = −2 the equations are

    \( \displaystyle -2x + 2y = 5,\ \text{that is } x - y = -\frac{5}{2} \)

    \( \displaystyle 3x - 3y = 7,\ \text{that is } x - y = \frac{7}{3} \)

Again x − y cannot take two different values, so there is no solution.

For every other k the determinant is non-zero and there is exactly one solution. Hence the system has no solution exactly when k = 3 or k = −2.