Question:
Let $\lambda \in R$. The system of linear equations
$2 x_{1}-4 x_{2}+\lambda x_{3}=1$
$x_{1}-6 x_{2}+x_{3}=2$
$\lambda x_{1}-10 x_{2}+4 x_{3}=3$
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 $\lambda=-\frac{2}{3}$.
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