# The system of linear equations

Question:

Let $\lambda \in R$. The system of linear equations

$2 x_{1}-4 x_{2}+\lambda x_{3}=1$

$x_{1}-6 x_{2}+x_{3}=2$

$\lambda x_{1}-10 x_{2}+4 x_{3}=3$

1. (1) exactly one negative value of $\lambda$

2. (2) exactly one positive value of $\lambda$

3. (3) every value of $\lambda$

4. (4) exactly two value of $\lambda$

Correct Option: 1

Solution:

$\because\left|\begin{array}{ccc}2 & -4 & \lambda \\ 1 & -6 & 1 \\ \lambda & -10 & 4\end{array}\right|=0 \Rightarrow 3 \lambda^{2}-7 \lambda-12=0$

$\Rightarrow \lambda=3$ or $-\frac{2}{3}$

$D_{1}=\left|\begin{array}{lll}1 & -4 & \lambda \\ 2 & -6 & 1 \\ 3 & -10 & 4\end{array}\right|=2(3-\lambda)$

$\therefore$ When $\lambda=-\frac{2}{3}, D_{1} \neq 0$

Hence, equations will be inconsistent when $\lambda=-\frac{2}{3}$.