Question

Prove the following identities :

$\left|\begin{array}{ccc}a+b+2c& a& b\\ c& b+c+2a& b\\ c& a& c+a+2b\end{array}\right|=2(a+b+c{)}^3$

Answer 1

Let $\Delta =\left|\begin{array}{ccc}a+b+2c& a& b\\ c& b+c+2a& b\\ c& a& c+a+2b\end{array}\right|=2(a+b+c{)}^3$.

Applying $C_1\to C_1+C_2+C_3,$ we get

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

$\Delta =2(a+b+c)\left|\begin{array}{ccc}1& a& b\\ 1& b+c+2a& b\\ 1& a& c+a+2b\end{array}\right|$ [Taking $2(a+b+c)$ common from $C_1$]

$\Rightarrow \Delta =2(a+b+c)\left|\begin{array}{ccc}1& a& b\\ 0& b+c+a& 0\\ 0& 0& c+a+b\end{array}\right|$ [Applying $R_2\to R_2-R_1$ & $R_3\to R_3-R_1$]

$\Rightarrow \Delta =2(a+b+c{)}^3\left|\begin{array}{ccc}1& a& b\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$

$\Rightarrow \Delta =2(a+b+c{)}^3\times 1=2(a+b+c{)}^3$ [Expanding along $C_1$]