The system of equations
\(x + y + z = 5\)
\(2x + y + 3z = 9\)
\(3x + 2y + \lambda z = \mu\)
has infinitely many solutions if
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\lambda = 4,\ \mu = 14\)
If the determinant of the coefficient matrix is not 0, the system has a unique solution. So infinitely many solutions need this determinant to be 0:
\( \displaystyle |A| = \begin{vmatrix}1 & 1 & 1\\ 2 & 1 & 3\\ 3 & 2 & \lambda\end{vmatrix} \)
Expanding along the first row:
\(|A| = 1(\lambda - 6) - 1(2\lambda - 9) + 1(4 - 3)\)
\(= \lambda - 6 - 2\lambda + 9 + 1\)
\(= 4 - \lambda\)
So \(|A| = 0\) only when \(\lambda = 4\).
With \(\lambda = 4\), adding the first two equations gives
\((x + y + z) + (2x + y + 3z) = 5 + 9\)
\(3x + 2y + 4z = 14\)
The third equation has the same left side, \(3x + 2y + 4z = \mu\). Hence:
if \(\mu = 14\), the third equation is the sum of the first two, so only two independent equations remain and the system is consistent with infinitely many solutions;
if \(\mu \ne 14\), the third equation contradicts the first two, so there is no solution.
Therefore \(\lambda = 4\) and \(\mu = 14\).
Why the other options are incorrect
Option (1) \(\lambda = 4,\ \mu \ne 14\) — Incorrect. Then \(3x + 2y + 4z\) would equal both 14 and \(\mu\), so the system is inconsistent (no solution).
Option (2) \(\lambda \ne 4,\ \mu = 14\) — Incorrect. Then \(|A| = 4 - \lambda \ne 0\), so the solution is unique.
Option (3) \(\lambda = -4,\ \mu = 14\) — Incorrect. Then \(|A| = 4 - (-4) = 8 \ne 0\), so the solution is unique.