Question

Solve $\left|\begin{array}{ccc}0& p-q& p-r\\ q-p& 0& q-r\\ r-p& r-q& 0\end{array}\right|=$

• A. $p+q+r$
• B. $pq+qr+rp$
• C. 0
• D. $p^2+q^2+r^2$

Answer 1

The correct option is C 0

$\left|\begin{array}{ccc}0& p-q& p-r\\ q-p& 0& q-r\\ r-p& r-q& 0\end{array}\right|=\left|\begin{array}{ccc}0& p& p\\ q& 0& q\\ r& r& 0\end{array}\right|-\left|\begin{array}{ccc}0& q& r\\ p& 0& r\\ p& q& 0\end{array}\right|$

Taking $p,q$ and $r$ common from $R_1,R_2$ and $R_3$ respectively from first matrix and $p,q$ and $r$ from $C_1,C_2$ and $C_3$ respectively from second matrix, we get

$=pqr\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|-pqr\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|=0$