The system of equations x + y + z = 2 x + ky + 2z = 3 x + 2y + kz = 3…

The correct answer is Option (4) → 0

Let the system be \(AX = B\). If \(|A| \ne 0\), it has a unique solution, so a value of k that gives no solution must make \(|A| = 0\).

    \( \displaystyle |A| = \begin{vmatrix} 1 & 1 & 1 \ 1 & k & 2 \ 1 & 2 & k \end{vmatrix} \)

    \(|A| = 1(k^2 - 4) - 1(k - 2) + 1(2 - k)\)

    \(|A| = k^2 - 4 - k + 2 + 2 - k\)

    \(|A| = k^2 - 2k = k(k - 2)\)

So \(|A| = 0\) only for \(k = 0\) or \(k = 2\), and each of these values has to be checked.

For k = 2, the second and third equations both become \(x + 2y + 2z = 3\). Only two different equations in three unknowns remain, so the system has infinitely many solutions.

For k = 0:

    \( \displaystyle A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 0 & 2 \ 1 & 2 & 0 \end{bmatrix}, \quad \mathrm{adj}\,A = \begin{bmatrix} -4 & 2 & 2 \ 2 & -1 & -1 \ 2 & -1 & -1 \end{bmatrix} \)

    \( \displaystyle (\mathrm{adj}\,A)B = \begin{bmatrix} -4 & 2 & 2 \ 2 & -1 & -1 \ 2 & -1 & -1 \end{bmatrix} \begin{bmatrix} 2 \ 3 \ 3 \end{bmatrix} = \begin{bmatrix} 4 \ -2 \ -2 \end{bmatrix} \ne O \)

Since \(|A| = 0\) and \((\mathrm{adj}\,A)B \ne O\), the system is inconsistent. Indeed, adding the second and third equations gives \(2x + 2y + 2z = 6\), that is \(x + y + z = 3\), while the first equation says \(x + y + z = 2\).

Why the other options are incorrect

• Option (1) 2 — Incorrect. For k = 2 the second and third equations are the same, so the system has infinitely many solutions, not none.

• Option (2) 0 or 2 — Incorrect. Both values make \(|A| = 0\), but only k = 0 gives no solution; k = 2 gives infinitely many.

• Option (3) \(-2\) — Incorrect. \(|A| = (-2)(-4) = 8 \ne 0\), so the system has a unique solution.