Question

Using the property of determinants and without expanding, 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

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

$R_1\to R_1+R_2+R_3$

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

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

$R_1\to R_1-R_2$

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

$R_3\to R_3-R_1$

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

$R_2\to R_2-R_3$

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

$R_1\leftrightarrow R_2$

$=-2\left|\begin{array}{ccc}c& r& z\\ b& q& y\\ a& p& x\end{array}\right|$ (When two rows are interchanged, the value of determinant differs by a negative sign)

$R_1\leftrightarrow R_3$

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