Question

$\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

Answer 1

$\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|$

Multiplying and dividing $R_1,R_2,R_3$ by $a,b,c$ respectively, and taking out common $a,b,c$ from $R_1,R_2,R_3$ respectively

$=abc\left|\begin{array}{ccc}1+\frac{1}{a}& \frac{1}{a}& \frac{1}{a}\\ \frac{1}{b}& 1+\frac{1}{b}& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& 1+\frac{1}{c}\end{array}\right|$

$R_1\to R_1+R_2+R_3$

$=abc\left|\begin{array}{ccc}1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}& 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}& 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\\ \frac{1}{b}& 1+\frac{1}{b}& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& 1+\frac{1}{c}\end{array}\right|$

Taking out common $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ from $R_1$

$=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left|\begin{array}{ccc}1& 1& 1\\ \frac{1}{b}& 1+\frac{1}{b}& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& 1+\frac{1}{c}\end{array}\right|$

$C_2\to C_2-C_1$

$C_3\to C_3-C_1$

$=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left|\begin{array}{ccc}1& 0& 0\\ \frac{1}{b}& 1& 0\\ \frac{1}{c}& 0& 1\end{array}\right|$

Expanding along $R_1$, We get.

$=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$