Question

If $a+b+c=0$, solve for x: $\left|\begin{array}{ccc}a-x& c& b\\ c& b-x& a\\ b& a& c-x\end{array}\right|=0$

• A. $x=0$ or $x=\pm \sqrt{(a^2+b^2+c^2-ab-bc-ca)}$
• B. $x=0$ or $x=\sqrt{(a^2+b^2+c^2-ab-bc-ca)}$
• C. $x=0$ or $x=-\sqrt{(a^2+b^2+c^2-ab-bc-ca)}$
• D. $x=0$ or $x=\pm \sqrt{(a^2+b^2+c^2+ab+bc+ca)}$

Answer 1

The correct option is A $x=0$ or $x=\pm \sqrt{(a^2+b^2+c^2-ab-bc-ca)}$

$\left|\begin{array}{ccc}a-x& c& b\\ c& b-x& a\\ b& a& c-x\end{array}\right|=0$
$R_1\to R_1+R_2+R_3$
$\Rightarrow \left|\begin{array}{ccc}a+b+c-x& c& b\\ a+b+c-x& b-x& a\\ a+b+c-x& a& c-x\end{array}\right|=\left|\begin{array}{ccc}-x& c& b\\ -x& b-x& a\\ -x& a& c-x\end{array}\right|=0$
$\Rightarrow -x\left|\begin{array}{ccc}1& c& b\\ 1& b-x& a\\ 1& a& c-x\end{array}\right|=0$
$\Rightarrow x=0$ or $\left|\begin{array}{ccc}1& c& b\\ 1& b-x& a\\ 1& a& c-x\end{array}\right|=0$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$\Rightarrow \left|\begin{array}{ccc}0& c-b+x& b-a\\ 0& b-a-x& a-c+x\\ 1& a& c-x\end{array}\right|=0$
$\Rightarrow (c-b+x)(a-c+x)-(b-a)(b-a-x)=0$
$\Rightarrow (c-b)(a-c)+x^2+x(a-b)-(b-a{)}^2+x(b-a)=0$
$\Rightarrow x^2=a^2+b^2+c^2-ab-bc-ca$
$\Rightarrow x=\pm \sqrt{(a^2+b^2+c^2-ab-bc-ca)}$