A) \[\frac{3}{4}\]
B) \[-4\]
C) \[\frac{1}{2}\]
D) \[-\frac{1}{4}\]
Correct Answer: C
Solution :
\[\left| \begin{matrix} 2 & 3 & -1 \\ 1 & K & -2 \\ 2 & -1 & 1 \\ \end{matrix} \right|=0\] By solving\[K=\frac{9}{2}\] \[2x+3y2y=0\] ?(1) \[x+\frac{9}{2}y-2z=0\] ?(2) \[2x-y+z=0\] ?(3) \[(1)-(3)\Rightarrow 4y-2z=0\] \[2y=z\] ?(4) \[\] ?(5) Put z form eqn. (4) into (1) \[2x+3y-2y=0\] \[2x+y=0\] \[\] ?(6) \[\frac{(6)}{(5)}\] \[\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+K=\frac{1}{2}\]You need to login to perform this action.
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