Question

If $a+b+c=0$; $x\ne$ 0 and $\left|\begin{array}{ccc}a-x& c& b\\ c& b-x& a\\ b& a& c-x\end{array}\right|=0$ , then $x$ is equal to

• A. $\pm \sqrt{\frac{3}{2}(a^2+b^2+c^2)}$
• B. $\pm (a^2+b^2+c^2)$
• C. $\pm (a+b+c)$
• D. $\pm (a^3+b^3+c^3)$

Answer 1

The correct option is A $\pm \sqrt{\frac{3}{2}(a^2+b^2+c^2)}$

$\left|\begin{array}{ccc}a-x& c& b\\ c& b-x& a\\ b& a& c-x\end{array}\right|$=0
$C_1{\to C}_1{+C}_2+C_3$
$\left|\begin{array}{ccc}a+b+c-x& c& b\\ a+b+c-x& b-x& a\\ a+b+c-x& a& c-x\end{array}\right|=0$

$(a+b+c-x)\left|\begin{array}{ccc}1& c& b\\ 1& b-x& a\\ 1& a& c-x\end{array}\right|=0$
$\because (a+b+c)=0$

$\therefore (-x)\left|\begin{array}{ccc}1& c& b\\ 1& b-x& a\\ 1& a& c-x\end{array}\right|=0$
$R_1{\to R}_1{-R}_2,R_2{\to R}_2{-R}_3$
$\Rightarrow \left|\begin{array}{ccc}0& c-b+x& b-a\\ 0& b-x-a& a-c+x\\ 1& a& c-x\end{array}\right|=0$
$\Rightarrow (c-b+x)(a-c+x)-(b-a)(b-x-a)=0$
$\Rightarrow (x^2-(a^2+b^2+c^2)+ab+bc+ac)=0$
$\Rightarrow x^2-(a^2+b^2+c^2)+\frac{{(a+b+c)}^2-(a^2+b^2+c^2)}{2}=0$
$\Rightarrow x=\pm \sqrt{\frac{3(a^2+b^2+c^2)}{2}}$