Question

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

Answer 1

Write the given determinant as product of two determinants

$\left|\begin{array}{ccc}b+c& c+a& a+b\\ c+a& a+b& b+c\\ a+b& b+c& c+a\end{array}\right|=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|$.

Now $\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|=0-1(0-1)+1(1-0)=2$, expand along first row

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