Consider the system of equations \( x+y+z=2,2x+3y+2z=5,x+2y+\lambda z=\mu \) then which of the following statements are TRUE?
A. The system of equations has unique solution if \( \lambda=2 \) and \( \mu\in\mathrm{R} \) (set of real number)
B. The system of equations is inconsistent if \( \lambda=1 \) and \( \mu=1 \)
C. The system of equations is consistent if \( \lambda=1 \) and \( \mu=1 \)
D. The system of equations is consistent if \( \lambda=1 \) and \( \mu=3 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A, B and D only
The coefficient determinant is
\( \displaystyle |A|=\begin{vmatrix} 1 & 1 & 1\\ 2 & 3 & 2\\ 1 & 2 & \lambda\end{vmatrix} \)
\( \displaystyle =1(3\lambda-4)-1(2\lambda-2)+1(4-3) \)
\( \displaystyle =\lambda-1 \)
If \( \lambda\ne 1 \), \( |A|\ne 0 \) and the system has a unique solution for every \( \mu \).
If \( \lambda=1 \), subtracting the first equation from the second gives
\( \displaystyle x+2y+z=3 \)
The third equation is then \( x+2y+z=\mu \). The two agree only when \( \mu=3 \) (infinitely many solutions); for \( \mu\ne 3 \) they contradict each other (no solution).
So for \( \lambda\ne 1 \) the solution is unique, and for \( \lambda=1 \) the system is consistent only when \( \mu=3 \).
Checking the statements
(A) The system of equations has unique solution if \( \lambda=2 \) and \( \mu\in\mathrm{R} \) (set of real number) — Correct. With \( \lambda=2 \), \( |A|=1\ne 0 \).
(B) The system of equations is inconsistent if \( \lambda=1 \) and \( \mu=1 \) — Correct. x + 2y + z would have to equal both 3 and 1.
(C) The system of equations is consistent if \( \lambda=1 \) and \( \mu=1 \) — Incorrect. x + 2y + z would have to equal both 3 and 1, so there is no solution.
(D) The system of equations is consistent if \( \lambda=1 \) and \( \mu=3 \) — Correct. The third equation repeats x + 2y + z = 3.