A) \[xyz\left( 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\]
B) \[xyz\]
C) \[1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\]
D) \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\]
Correct Answer: A
Solution :
\[\{\therefore {{C}_{1}}\equiv {{C}_{2}}\}\] =\[xyz\left( 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\] \[\,\left| \,\begin{matrix} 1 & 1 & 1 \\ \frac{1}{y} & 1+\frac{1}{y} & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & 1+\frac{1}{z} \\ \end{matrix}\, \right|\], by \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\] =\[xyz\left( 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\]\[\,\left| \,\begin{matrix} 1 & 0 & 0 \\ 1/y & 1 & 0 \\ 1/z & 0 & 1 \\ \end{matrix}\, \right|\], by \[\begin{align} & {{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \\ & {{C}_{3}}\to {{C}_{3}}-{{C}_{1}} \\ \end{align}\] = \[xyz\left( 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\] \[\left| \,\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix}\, \right|=xyz\left( 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\]. Trick: Put \[x=1,\,y=2\] and \[z=3\], then \[\left| \,\begin{matrix} 2 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 4 \\ \end{matrix}\, \right|=2(11)-1(3)+1(1-3)=17\] Option (a) gives, \[1\times 2\times 3\,\left( 1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3} \right)=17\].
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