Question

Suppose $a,b,cϵR,a+b+c>0$, $A=bc-a^2$, $B=ca-b^2$ and $C=ab-c^2$ and $\left|\begin{array}{ccc}A& B& C\\ B& C& A\\ C& A& B\end{array}\right|$ = 49 then$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ equals ?

• A. $-7$
• B. $2401$
• C. $-2401$
• D. $7$

Answer 1

The correct option is A $-7$

We have,
$\left|\begin{array}{ccc}A& B& C\\ B& C& A\\ C& A& B\end{array}\right|=(A+B+C)(AC+BC+AC-A^2-B^2-C^2)$
Substituting the values of $a,b$ and $c$,
$\left|\begin{array}{ccc}A& B& C\\ B& C& A\\ C& A& B\end{array}\right|$
$=(ac+bc+ac-a^2-b^2-c^2)(3a^{2}bc+3a{b}^{2}c+3ab{c}^2-b^{3}c-a^{3}c-a^4-a{c}^3-a{b}^3-b{c}^3-b^4-a^{3}b-c^4)$
$=(ac+bc+ac-a^2-b^2-c^2)(a+b+c)(3abc-a^3-b^3-c^3)$
$=(ac+bc+ac-a^2-b^2-c^2{)}^2(a+b+c{)}^2$
$={\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|}^2$
Hence,
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=\pm 7$
However, $(a+b+c)>0$ and $(ab+bc+ac-a^2-b^2-c^2)<0$
Hence,
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=-7$