Question

Without expanding the determinants, prove that
$\left|\begin{array}{ccc}a& b& c\\ x& y& z\\ p& q& r\end{array}\right|=\left|\begin{array}{ccc}y& b& q\\ x& a& p\\ z& c& r\end{array}\right|=\left|\begin{array}{ccc}x& y& z\\ p& q& r\\ a& b& c\end{array}\right|$.

Answer 1

$\left|\begin{array}{ccc}a& b& c\\ x& y& z\\ p& q& r\end{array}\right|=\left|\begin{array}{ccc}y& b& q\\ x& a& p\\ z& c& r\end{array}\right|\left|\begin{array}{ccc}x& y& z\\ p& q& r\\ a& b& c\end{array}\right|\Rightarrow {\Delta }_1={\Delta }_2={\Delta }_3{\Delta }_2=\left|\begin{array}{ccc}y& b& q\\ x& a& p\\ z& c& r\end{array}\right|$

Taking transpose as $\left|A\right|=\left|A^T\right|$

$\Rightarrow {\Delta }_2=\left|\begin{array}{ccc}y& x& z\\ b& a& c\\ q& z& r\end{array}\right|$

Interchanging $C_1$ and $C_2$

$\Rightarrow {\Delta }_2=-\left|\begin{array}{ccc}x& y& z\\ a& b& c\\ p& q& r\end{array}\right|$

Interchanging $R_1$ and $R_2$

$\Rightarrow {\Delta }_2=\left|\begin{array}{ccc}a& b& c\\ x& y& z\\ p& q& r\end{array}\right|......\left(i\right)$

${\Delta }_3=\left|\begin{array}{ccc}x& y& z\\ p& q& r\\ a& b& c\end{array}\right|$

Interchanging $R_1$ and $R_3$

$\Rightarrow {\Delta }_3=-\left|\begin{array}{ccc}a& b& c\\ p& q& r\\ x& y& z\end{array}\right|$

Interchanging $R_2$ and $R_3$

$\Rightarrow {\Delta }_3=\left|\begin{array}{ccc}a& b& c\\ x& y& z\\ p& q& r\end{array}\right|.....\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$

${\Delta }_1={\Delta }_2={\Delta }_3$

Hence proved.