Determinants Matrices Question 48
Question: If the system of equations $ \lambda x_1+x_2+x_3=1, $
$ x_1+\lambda x_2+x_3=1, $
$ x_1+x_2+\lambda x_3=1 $ is consistent, then $ \lambda $ can be
Options:
A) $ 5 $
B) $ -2/3 $
C) $ -3 $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] Let $ \Delta = \begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \\ \end{vmatrix}= \begin{vmatrix} \lambda +2 & 1 & 1 \\ \lambda +2 & \lambda & 1 \\ \lambda +2 & 1 & \lambda \\ \end{vmatrix} $
$ [C_1\to C_1+C_2+C_3] $
$ =(\lambda +2) \begin{vmatrix} 1 & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \\ \end{vmatrix}=(\lambda +2) \begin{vmatrix} 1 & 0 & 0 \\ 1 & \lambda -1 & 0 \\ 1 & 0 & \lambda -1 \\ \end{vmatrix} $
$ =(\lambda +2){{(\lambda -1)}^{2}} $ [using $ C_2\to C_2-C_1 $ and $ C_3\to C_3-C_1 $ ] If $ \Delta =0 $ , then $ \lambda =-2 $ or $ \lambda =1 $ . But when $ \lambda =1 $ , the system of equation becomes $ x_1+x_2+x_3=1 $ which has infinite number of solutions. When $ \lambda =-2 $ , by adding three equations, we obtain $ 0=3 $ and thus, the system of equations is inconsistent.
BETA
