A) \[abc\]
B) \[ab+bc+ca+abc\]
C) \[ab+bc+ca\]
D) \[3(a+b+c)\]
Correct Answer: B
Solution :
\[\left| \begin{matrix} a+1 & 1 & 1 \\ 1 & b+1 & 1 \\ 1 & 1 & c+1 \\ \end{matrix} \right|=abc\left| \begin{matrix} \frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \\ \end{matrix} \right|\] \[=abc\left| \begin{matrix} 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \\ \end{matrix} \right|\] \[[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}]\] \[=abc\left( 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\left| \begin{matrix} 1 & 1 & 1 \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \\ \end{matrix} \right|\] \[=abc\left( \frac{abc+bc+ac+ab}{abc} \right)\left| \begin{matrix} 1 & 0 & 0 \\ 1/b & 1 & 0 \\ 1/c & 0 & 1 \\ \end{matrix} \right|\] \[[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\,and\,{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}],\] \[=(abc+bc+ac+ab)[1(1-0)]\] \[=abc+ab+bc+ca\]
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