Question

Using properties of determinates, prove that:
$\left|\begin{array}{ccc}a& b& b+c\\ c& a& c+a\\ b& c& a+b\end{array}\right|=(a+b+c)(a-c{)}^2$

Answer 1

$L.H.S=$

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

$R_1=R_1+R_2+R_3$

$=\left|\begin{array}{ccc}a+b+c& a+b+c& 2a+2b+2c\\ c& a& c+a\\ b& c& a+b\end{array}\right|$

$=(a+b+c)\left|\begin{array}{ccc}1& 1& 2\\ c& a& c+a\\ b& c& a+b\end{array}\right|$

$C_1=C_1+C_2$

$=(a+b+c)\left|\begin{array}{ccc}2& 1& 2\\ a+c& a& c+a\\ b+c& c& a+b\end{array}\right|$

$C_1=C_3-C_1$

$=(a+b+c)\left|\begin{array}{ccc}0& 1& 2\\ 0& a& c+a\\ a-c& c& a+b\end{array}\right|$

$=(a+b+c)(a-c)\left|\begin{array}{ccc}0& 1& 2\\ 0& a& c+a\\ 1& c& a+b\end{array}\right|$

$=(a+b+c)(a-c)(-1(a+c-2a))$

$=(a+b+c){(a-c)}^2$

$=R.H.S$