Question

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

Answer 1

consider the following question.

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

$c_1\to c_1+c_2+c_3$

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

$R_2\to R_2-R_1{R}_3\to R_3-R_1$

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

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

$(a+b+c-x)[(b-c-x)(c-b-x)-(a-b)(a-c)]=0$

$(a+b+c-x)\left((bc-bx-b^2-c^2+cx+bc-cx+x^2+bx-a^2+ac+ab-bc)\right)=0$

$(0-x)(x^2-b^2-c^2+bc+ca+ab)=0$

$x=0$

$(x^2-b^2-c^2+bc+ca+ab)=0$

$x=\sqrt{b^2-c^2+bc+ca+ab}$

Hence, this is the required answer.