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

$\text{L.H.S}$
$△=\left|\begin{array}{ccc}b+c& c+a& a+b\\ q+r& r+p& p+q\\ y+z& z+x& x+y\end{array}\right|C_3\to C_3-C_1-C_2$

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

$=-2.\left|\begin{array}{ccc}b+c& c+a& c\\ q+r& r+p& r\\ y+z& z+x& z\end{array}\right|\begin{array}{c}{C}_1\to C_1-C_3\\ C_2\to C_2-C_3\end{array}$

$=-2.\left|\begin{array}{ccc}b& a& c\\ q& p& r\\ y& x& z\end{array}\right|C_1\to C_2$

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