Question

show that $\left|\begin{array}{ccc}-bc& ca+ab& ca+ab\\ ab+bc& -ca& ab+bc\\ bc+ca& bc+ca& -ab\end{array}\right|=(\sum ab{)}^3$

Answer 1

Solution :- given

$=\left|\begin{array}{ccc}-bc& ab+ac& ac+ab\\ ab+bc& -ac& bc+ab\\ ca+bc& bc+ac& -ab\end{array}\right|$

$=\left|\begin{array}{ccc}ab+bc+ca& ab+bc+ca& ab+bc+ca\\ ab+bc& -ca& bc+ab\\ ca+bc& bc+ac& -ab\end{array}\right|$ [Here $R_1^1=R_1+R_2+R_3]$

$=(ab+bc+ca)\left|\begin{array}{ccc}1& 1& 1\\ ab+bc& -ca& bc+ab\\ ca+bc& bc+ca& -ab\end{array}\right|$ [taking $(ab+bc+ca)$ common from first Row]

$=(ab+bc+ca)\left|\begin{array}{ccc}0& 0& 1\\ ab+bc+ca& -(ab+bc+ca)& bc+ab\\ 0& ab+bc+ca& -ab\end{array}\right|$ [Here $C_1=C_1-C_2,C_2=C_2-C_3]$

$=(ab+bc+ca)\left|\begin{array}{cc}ab+bc+ca& -(ab+bc+ca)\\ 0& ab+bc+ca\end{array}\right|$ [ expanding by 1st Row]

$=(ab+bc+ca)(ab+bc+ca{)}^2$

$=(ab+bc+ca{)}^3$

$=(\sum ab{)}^3$ Proved