Question

Prove the following using properties of determinants:
$\left|\begin{array}{ccc}a+b+2c& a& b\\ c& b+c+2a& b\\ c& a& c+a+2b\end{array}\right|=2(a+b+c{)}^3$

Answer 1

$LHS=\left|\begin{array}{ccc}a+b+2c& a& b\\ c& b+c+2a& b\\ c& a& c+a+2b\end{array}\right|$
Applying $C_1$ to $C_1+C_2+C_3$
$\left|\begin{array}{ccc}2a+2b+2c& a& b\\ 2a+2b+2c& b+c+2a& b\\ 2a+2b+2c& a& c+a+2b\end{array}\right|$
$=(2a+2b+2c)\left|\begin{array}{ccc}1& a& b\\ 1& b+c+2a& b\\ 1& a& c+a+2b\end{array}\right|$
Applying $R_2\Rightarrow R_2-R_1$
$R_3\Rightarrow R_3-R_1$
$=(2a+2b+2c)\left|\begin{array}{ccc}1& a& b\\ 0& b+c+a& 0\\ 0& 0& c+a+b\end{array}\right|$
Expanding along third row, we get
$=2(a+b+c)\left[\right(b+c+a{)}^2-0]$
$=2(a+b+c{)}^3=$ RHS
Hence proved.