Question

Using properties of determinants, prove that $\left|\begin{array}{ccc}b+c& c+a& a+b\\ q+r& r+p& p+q\\ y+z& z+x& x+y\end{array}\right|=2\left|\begin{array}{ccc}a& b& c\\ p& q& r\\ x& y& z\end{array}\right|.$

Answer 1

Let $C_1,C_2,C_3$ be first,second and third columns respectively

$\left|\begin{array}{ccc}b+c& c+a& a+b\\ q+r& r+p& p+q\\ y+z& z+x& x+y\end{array}\right|$

Applying $C_1=C_1+C_2+C_3,$ we get,

$=2\left|\begin{array}{ccc}a+b+c& c+a& a+b\\ p+q+r& r+p& p+q\\ x+y+z& z+x& x+y\end{array}\right|$

Applying $C_2=C_2-C_1$ and $C_3=C_3-C_1,$ we get,

$=2\left|\begin{array}{ccc}a+b+c& b& c\\ p+q+r& q& r\\ x+y+z& y& z\end{array}\right|$

Finally, Applying $C_1=C_1-(C_2+C_3),$ we get,

$=2\left|\begin{array}{ccc}a& b& c\\ p& q& r\\ x& y& z\end{array}\right|$

Hence, proved!