Question

If $\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=Xabc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$. Find the value of $X$.

Answer 1

$\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=\left|\begin{array}{ccc}a& 0& 1\\ -b& b& 1\\ 0& -c& 1+c\end{array}\right|$, apply $C_1\to C_1-C_2$ and $C_2\to C_2-C_3$

$=a\left[b\right(1+c)+c]-0+1[bc-0]$, expand along first row

$=abc+ab+bc+ca$

$=abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
The value of $X$ is $1$.