Question

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

Answer 1

First take LHS: $\Delta =\left|\begin{array}{ccc}b+c& q+r& y+z\\ c+a& r+p& z+x\\ a+b& p+q& x+y\end{array}\right|$
Applying, $R_1\leftrightarrow R_3$ and $R_3\leftrightarrow R_2$, we get
$=\left|\begin{array}{ccc}a+b& p+q& x+y\\ b+c& q+r& y+z\\ c+a& r+p& z+x\end{array}\right|$
Applying, $R_1\to R_1+R_2+R_3$, we get
$=\left|\begin{array}{ccc}2(a+b+c)& 2(p+q+r)& 2(x+y+z\\ b+c& q+r& y+z\\ c+a& r+p& z+x\end{array}\right|$
$\Rightarrow 2\left|\begin{array}{ccc}a+b+c& p+q+r& x+y+z\\ b+c& q+r& y+z\\ c+a& r+p& z+x\end{array}\right|$
Applying, $R_1\to R_1-R_2$
$=2\left|\begin{array}{ccc}a& p& x\\ b+c& q+r& y+z\\ c+a& r+p& z+x\end{array}\right|$
Applying, $R_3\to R_3-R_1$
$=2\left|\begin{array}{ccc}a& p& x\\ b+c& q+r& y+z\\ c& r& z\end{array}\right|$
Applying, $R_2\to R_2-R_3$
$\Delta =2\left|\begin{array}{ccc}a& p& x\\ b& q& y\\ c& r& z\end{array}\right|=RHS$
Hence proved.