Consider the system of equations
\(x + y + z = 3\)
\(x + 2y + 3z = 6\)
\(x + 3y + \lambda z = \mu\)
and the following statements:
(A) The system has a unique solution if \(\lambda \ne 5\).
(B) If \(\lambda = 5\) and \(\mu = 9\), the system has infinitely many solutions.
(C) If \(\lambda = 5\) and \(\mu = 10\), the system has infinitely many solutions.
(D) If \(\lambda = 4\), the system has no solution.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
The determinant of the coefficient matrix is
\( \displaystyle |A| = \begin{vmatrix}1 & 1 & 1\\ 1 & 2 & 3\\ 1 & 3 & \lambda\end{vmatrix} \)
\(|A| = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2)\)
\(|A| = \lambda - 5\)
When \(\lambda = 5\): twice the second equation minus the first gives
\(2(x + 2y + 3z) - (x + y + z) = 12 - 3\)
\(x + 3y + 5z = 9\)
So with \(\lambda = 5\) the third equation repeats the first two if \(\mu = 9\), and contradicts them if \(\mu \ne 9\).
So the solution is unique when \(\lambda \ne 5\); for \(\lambda = 5\) there are infinitely many solutions when \(\mu = 9\) and none when \(\mu \ne 9\).
Checking the statements
(A) Unique solution if \(\lambda \ne 5\) — Correct. Then \(|A| = \lambda - 5 \ne 0\).
(B) \(\lambda = 5,\ \mu = 9\): infinitely many solutions — Correct. The third equation is then \(2 \times\) (second) − (first).
(C) \(\lambda = 5,\ \mu = 10\): infinitely many solutions — Incorrect. \(x + 3y + 5z\) would equal both 9 and 10, so there is no solution.
(D) \(\lambda = 4\): no solution — Incorrect. \(|A| = 4 - 5 = -1 \ne 0\), so the solution is unique.