Question

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

Answer 1

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

$=(1+a)\left[\right(1+b\left)\right(1+c)-1]-(1+c-1)+(-1-b+1)$

$=(1+a)[1+b+c+bc-1]-c-b$

$=(1+a)[b+c+bc]-c-b$

$=b+c+bc+ab+ac+abc-c-b$

$=bc+ab+ac+abc$

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

Hence, proved.