Question

Using the properties of determinants, show that:

$\left|\begin{array}{ccc}{-a}^2& ab& ac\\ ba& {-b}^2& bc\\ ca& cb& {-c}^2\end{array}\right|={4a}^2{b}^2{c}^2$

Answer 1

$\left|\begin{array}{ccc}{-a}^2& ab& ac\\ ba& {-b}^2& bc\\ ca& cb& {-c}^2\end{array}\right|$

Taking $a,b,c$ common from $R_1,R_2,R_3$ respectively

$=abc\left|\begin{array}{ccc}-a& b& c\\ a& -b& c\\ a& b& -c\end{array}\right|$

Taking $a,b,c$ common from $C_1,C_2,C_3$ respectively

$a^2{b}^2{c}^2\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|$

$C_1\to C_1+C_2,c_2\to C_2+C_3$

$=a^2{b}^2{c}^2\left|\begin{array}{ccc}0& 2& 1\\ 0& 0& 1\\ 2& 0& -1\end{array}\right|$

Expanding along the first column,

$=a^2{b}^2{c}^2\left[2\right(2-0\left)\right]$

$=4a^2{b}^2{c}^2$