Loading [Contrib]/a11y/accessibility-menu.js

Discussion Forum



1)

Consider  the following system of equations in matrix form 

$\begin{bmatrix}1  \\2 \\\lambda \end{bmatrix}$  (1    2    $\lambda$) $\begin{bmatrix}x  \\y \\z\end{bmatrix}  =0 $

Then which one of the following statements  is ture?


A) $\forall \lambda\epsilon(-\infty,\infty)$ , the given system has non trivial solution

B) $\forall \lambda\epsilon(-\infty,\infty)$ , the given system has only trivial solution

C) For $\lambda\neq0$ , the given system does not have any solution

D) For $\lambda =0$ , the given system is inconsistent

Answer:

Option A

Explanation:

We have ,

$\begin{bmatrix}1  \\2 \\\lambda \end{bmatrix}$  (1    2    $\lambda$) $\begin{bmatrix}x  \\y \\z\end{bmatrix}  =0 $

   $\begin{bmatrix}1 & 2 &\lambda \\2 & 4 & 2\lambda \\\lambda &2\lambda &\lambda^{2} \end{bmatrix}\begin{bmatrix}x  \\y \\z \end{bmatrix}=0$

 Let        A=   $\begin{bmatrix}1 & 2 &\lambda \\2 & 4 & 2\lambda \\\lambda &2\lambda &\lambda^{2} \end{bmatrix}$

 |A|  = $1(4 \lambda^{2}-4 \lambda^{2})-2(2 \lambda^{2}-2 \lambda^{2})+\lambda (4 \lambda-4 \lambda)$

|A|= 0, $\forall \lambda \epsilon R$

$\therefore$ the given syste m has non trivial solution.