Question

If $a+b+c=0$, then the solution of the equation $\left|\begin{array}{ccc}a-x & c & b \\ c & b-x & a \\ b & a & c-x\end{array}\right|=0$, is

(A) 0

(B) $\pm \frac{3}{2}\left(a^2+b^2+c^2\right)$

(C) $0, \pm \sqrt{\frac{3}{2}\left(a^2+b^2+c^2\right)}$

(D) $0, \pm \sqrt{a^2+b^2+c^2}$

Answer 1

Applying $R_1 \rightarrow R_1+R_2+R_3$ and put $a+b+c=0$, we get

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

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

[In $R_1$, taking common $-x$ and applying $C_2 \rightarrow C_2-C_1$ and $\left.C_3 \rightarrow C_3-C_1\right]$

$\begin{array}{l}\Rightarrow-x[(b-c-x)(c-b-x)-(a-b)(a-c)]=0 \\\Rightarrow-x\left[x^2-b^2-a^2-c^2+b c+c a+a b\right]=0 \\{\left[\begin{array}{l}\because+b+c=0 \\\Rightarrow a^2+b^2+c^2=-2(a b+b c+c a)\end{array}\right]} \\\Rightarrow \quad-x\left[x^2-\frac{3}{2}\left(a^2+b^2+c^2\right)\right]=0 \\\Rightarrow \quad x=0, \pm \sqrt{\frac{3}{2}\left(a^2+b^2+c^2\right)} \\\end{array}$

So, The correct option is (C).