Question

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

• A. $4$
• B. $2$
• C. $0$
• D. $2a^2{b}^2{c}^2$

Answer 1

Answer: (C)

$0$

Let $△=\left|\begin{array}{ccc}{-a}^2& ab& ac\\ ab& {-b}^2& bc\\ ac& bc& {-c}^2\end{array}\right|+\left|\begin{array}{ccc}{a}^2& {-b}^2& {-c}^2\\ {-a}^2& b^2& {-c}^2\\ {-a}^2& {-b}^2& c^2\end{array}\right|$

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

$=\left|\begin{array}{ccc}0& b(a-b)& c(a-c)\\ a(b-a)& 0& c(b-c)\\ a(c-a)& b(c-b)& 0\end{array}\right|$

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

$\therefore △=abc\left|\begin{array}{ccc}0& a-b& a-c\\ b-a& 0& b-c\\ c-a& c-b& 0\end{array}\right|$

Expanding it along $R_1,$ we get

$△=abc\{0-(a-b)(0-(c-a)(b-c))+(a-c)((b-a)(c-b)-0)\}$

$=abc\{+(a-b)(a-c)(b-c)-(a-c)(b-a)(c-b)\}$

$=abc\left\{0\right\}=0$