A) \[1+abc+ab+bc+ca\]
B) \[abc\]
C) \[4abc\]
D) \[abc\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\]
E) \[abc\left( 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\]
Correct Answer: E
Solution :
\[\left| \begin{matrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \\ \end{matrix} \right|\] \[=abc\left| \begin{matrix} 1+\frac{1}{a} & \frac{1}{b} & \frac{1}{c} \\ \frac{1}{a} & 1+\frac{1}{b} & \frac{1}{c} \\ \frac{1}{a} & \frac{1}{b} & 1+\frac{1}{c} \\ \end{matrix} \right|\] \[=abc\left( 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\left| \begin{matrix} 1 & \frac{1}{b} & \frac{1}{c} \\ 1 & 1+\frac{1}{b} & \frac{1}{c} \\ 1 & \frac{1}{b} & 1+\frac{1}{c} \\ \end{matrix} \right|\] \[=abc\left( 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\left| \begin{matrix} 1 & \frac{1}{b} & \frac{1}{c} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right|\] \[=abc\left( 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\]
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