Question

The system of equations $6x+5y+\lambda z=0$ ,$3x-y+4z=0$ ,$x+2y-3z=0$ has

• A. Only a trivial solution for $\lambda ϵR$
• B. exactly one nontrivial solution for some real $\lambda$
• C. infinite number of nontrivial solutions for one value of $\lambda$
• D. only one solution for$\lambda \ne -5$

Answer 1

Answer: (C), (D)

The correct options are
C infinite number of nontrivial solutions for one value of $\lambda$
D only one solution for$\lambda \ne -5$

The determinant will be,
$\left|\begin{array}{ccc}6& 5& \lambda \\ 3& -1& 4\\ 1& 2& -3\end{array}\right|$
By, $R_1=R_1-2R_2$ and$R_2=R_2-3R_1$
$=\left|\begin{array}{ccc}0& 7& \lambda -8\\ 0& -7& 13\\ 1& 2& -3\end{array}\right|$$=91+7(\lambda -8)$
$=7\lambda +91-56$$=7\lambda +35$
$=7(\lambda +5)$
So, if this is $0$,
i.e,$\lambda =-5$ it will have infinitely many solutions
and
if $\lambda \ne -5$, it will have one solution.