The system of equations
\(x + y + z = 4\)
\(x + 2y +3z= 12\)
\(x + 3y+\lambda z = \mu\) has a unique solution if
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\lambda\ne 5, \mu\) can be any real number
A system \(AX = B\) has a unique solution exactly when \(|A| \ne 0\). Expanding along the first row:
\( \displaystyle |A| = \begin{vmatrix}1 & 1 & 1\\ 1 & 2 & 3\\ 1 & 3 & \lambda\end{vmatrix} \)
\(= 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2)\)
\(= \lambda - 5\)
So the solution is unique when \(\lambda \ne 5\). The value of \(\mu\) does not enter \(|A|\), so it can be any real number.
Why the other options are incorrect
With \(\lambda = 5\), subtracting the first equation from the second, and the second from the third:
\(y + 2z = 8\)
\(y + 2z = \mu - 12\)
Option (1) \(\lambda=5, \mu\) can be any real number — Incorrect. With \(\lambda = 5\), \(|A| = 0\), so the solution is never unique.
Option (2) \(\lambda= 5,\mu \ne 20\) — Incorrect. Then \(\mu - 12 \ne 8\), the two equations clash, and there is no solution.
Option (4) \(\lambda=5,\mu = 20\) — Incorrect. Then both equations read \(y + 2z = 8\), and there are infinitely many solutions.