Question

If a + b + c = 0 and $\left|\begin{array}{ccc}a-x& c& b\\ c& b-x& a\\ b& a& c-x\end{array}\right|=0$ then show that x = 0, $x=\sqrt{\frac{3}{2}(a^2+b^2+c^2)}$

Answer 1

$\left|\begin{array}{ccc}a-x& c& b\\ c& b-x& a\\ b& a& c-x\end{array}\right|=C_1=C_1+C_2+C_3\left|\begin{array}{ccc}a+b+c-x& c& b\\ a+b+c-x& b-x& a\\ a+b+c-x& a& c-x\end{array}\right|$
$=-x\left|\begin{array}{ccc}1& c& b\\ 1& b-x& a\\ 1& a& c-x\end{array}\right|$ ($\because )a+b+c=0$
$=-x\left\{\right(b-x\left)\right(c-x)-a^2-c[(c-x)-a]+b[a-b+x\left]\right\}$
$=-x\{bc+x^2-x(b+c)-a^2-c^2+cx+ac+ab-b^2+bx\}$
$=+x\{a^2+b^2+c^2-(ab+bc+ca-x^2\left)\right\}$
Now $(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)=0(\because a+b+c.a)$
$2(ab+bc+ca)=-(a^2+b^2+c^2)$
$Arab+bc+ca\frac{(a_2+b_2+c_2)}{2}$
$\Delta =x\{a^2+b^2+c^2+\frac{a^2+b^2+c^2}{2}-x^2]=0$
$=x\left\{\frac{3}{2}\right(a^2+b^2+c^2)-x^2\}=0$
$\Rightarrow x=0$ or $x=\frac{\sqrt{3}}{2}(a^2+b^2+c^2)$