Question

Using properties of determinants, prove that: $\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=abc+bc+ca+ab.$

Answer 1

LHS $=\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|$
Applying $R_1\to R_1-R_2$
$R_2\to R_2-R_3$

and row column transformation,
$=\left|\begin{array}{ccc}a& 0& 1\\ -b& b& 1\\ 0& -c& 1+c\end{array}\right|$
$=a(b+bc+c)+1(bc-0)$
$=ab+abc+ac+bc$
$=abc+ab+bc+ca=$ RHS
Hence proved.