The system of equations x + y + z = 7 x + 2y + 3z = 5 x + 3y + λ z =…

The correct answer is Option (3) → \(\lambda \ne 5\) and \(\mu\) is any real number

A system of three linear equations in three unknowns has a unique solution exactly when the determinant of the coefficient matrix is non-zero; the right-hand sides do not matter.

    \( \displaystyle \Delta = \begin{vmatrix}1 & 1 & 1\\1 & 2 & 3\\1 & 3 & \lambda\end{vmatrix} \)

Expanding along the first row:

    \(\Delta = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2)\)
    \(= 2\lambda - 9 - \lambda + 3 + 1\)
    \(= \lambda - 5\)

So the solution is unique if and only if \(\lambda \ne 5\), whatever the value of \(\mu\).

Why the other options are incorrect

• Option (1) \(\lambda = 5\) and \(\mu = 7\) — Incorrect. With \(\lambda = 5\), \(\Delta = 0\), so the solution cannot be unique. (Subtracting, the equations give \(y + 2z = -2\) and \(2y + 4z = \mu - 7\), which agree only for \(\mu = 3\); with \(\mu = 7\) there is no solution at all.)

• Option (2) \(\lambda \ne 1\) and \(\mu = 5\) — Incorrect. \(\lambda \ne 1\) still allows \(\lambda = 5\), where \(\Delta = 0\).

• Option (4) \(\lambda \ne 1\) and \(\mu \ne 5\) — Incorrect. \(\lambda \ne 1\) still allows \(\lambda = 5\), where \(\Delta = 0\); the condition on \(\mu\) is irrelevant.