Question

Using the property of determinants and without expanding.

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

Answer 1

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

$=\left|\begin{array}{ccc}b+c& c+a& a+b\\ q+r& r+p& p+q\\ y+z& z+x& x+y\end{array}\right|$ (interchange row and column)

$=\left|\begin{array}{ccc}b+c& c+a& -2c\\ q+r& r+p& -2r\\ y+z& z+x& -2z\end{array}\right|$ [using $C_3\to C_3-(C_1+C_2)$]

$=-2\left|\begin{array}{ccc}b+c& c+a& c\\ q+r& r+p& r\\ y+z& z+x& z\end{array}\right|$ (taking '-2' common from $C_3$)

$=-2\left|\begin{array}{ccc}b& a& c\\ q& p& r\\ y& x& z\end{array}\right|$ (using $C_1\to C_1-C_3$ and $C_2\to C_2-C_3$)

$=2\left|\begin{array}{ccc}a& b& c\\ p& q& r\\ x& y& z\end{array}\right|$ ($using{C}_1\leftrightarrow C_2$)

$=2\left|\begin{array}{ccc}a& p& x\\ b& q& y\\ c& r& z\end{array}\right|=RHS$ (interchange column and row)