Question

Prove the following :
$\left|\begin{array}{ccc}bc& b{c}^{\prime }+b^{\prime }c& b^{\prime }{c}^{\prime }\\ ca& c{a}^{\prime }+c^{\prime }a& c^{\prime }{a}^{\prime }\\ ab& a{b}^{\prime }+d^{\prime }b& d^{\prime }{b}^{\prime }\end{array}\right|=(b{c}^{\prime }-b^{\prime }c)(c{a}^{\prime }-c^{\prime }a)(a{b}^{\prime }-db)$

Answer 1

Hint : Multiply $R_1,R_2,R_3$ by $a,b,c$ respectively and divide the det. by $abc$.
Now operate $R_2-R_1$ and $R_3-R_1$. This will give
$\Delta =\cfrac { 1 }{ abc } \begin{vmatrix} abc & abc'+ab'c & ab'c' \\ 0 & c\left( db-b'a \right) & c'\left( bd-ab' \right) \\
0 & b\left( dc-c'a \right) & b'\left( c

d-ac' \right) \end{vmatrix}=\left( bd-ab' \right) \left( dc-c'a \right) $$\left|\begin{array}{cc}c& c^{\prime }\\ b& b^{\prime }\end{array}\right|$$=\left( bd-ab' \right) \left( dc-c'a \right) \left( cb'-bc' \right)=\left( bc'-cb' \right) \left( c'a-ac' \right) \left( ab'-bd \right) $.