The system of equations 2x + ky = 5 kx + 8y = 10 has no solution for

The correct answer is Option (1) → \(k = -4\)

In matrix form the system is \(AX = B\) with

    \( \displaystyle A = \begin{bmatrix} 2 & k \\ k & 8 \end{bmatrix},\quad B = \begin{bmatrix} 5 \\ 10 \end{bmatrix} \)

If \(|A| \ne 0\), there is a unique solution. So a system with no solution needs

    \(|A| = 16 - k^2 = 0\)

    \(k = 4\) or \(k = -4\)

Each value is now checked with \((\mathrm{adj}\ A)B\), where \(\mathrm{adj}\ A = \begin{bmatrix} 8 & -k \\ -k & 2 \end{bmatrix}\).

For \(k = 4\):

    \( \displaystyle (\mathrm{adj}\ A)B = \begin{bmatrix} 40 - 40 \\ -20 + 20 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \)

So for \(k = 4\) the system is consistent with infinitely many solutions (the second equation is just twice the first).

For \(k = -4\):

    \( \displaystyle (\mathrm{adj}\ A)B = \begin{bmatrix} 40 + 40 \\ 20 + 20 \end{bmatrix} = \begin{bmatrix} 80 \\ 40 \end{bmatrix} \ne O \)

A singular A with \((\mathrm{adj}\ A)B \ne O\) means the system is inconsistent. So there is no solution only for \(k = -4\).

Why the other options are incorrect

• Option (2) \(k = 4\) — Incorrect. The equations become \(2x + 4y = 5\) and \(4x + 8y = 10\), the same line, so there are infinitely many solutions.

• Option (3) \(k = \pm 4\) — Incorrect. \(|A| = 0\) for both values, but \(k = 4\) gives infinitely many solutions, not none.

• Option (4) \(k = 0\) — Incorrect. \(|A| = 16 \ne 0\), so the system has the unique solution \(x = \frac{5}{2}\), \(y = \frac{5}{4}\).