Question

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

• A. 0
• B. $\pm \frac{3}{2}(a^{2} + b^{2} + c^{2})$
• C. $0, \pm \sqrt{\frac{3}{2}(a^{2} + b^{2} + c^{2})}$
• D. $0, \pm \sqrt{a^{2} + b^{2} + c^{2}}$

Answer 1

Answer: (C)

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

Answer 2

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

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

1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{matrix} \right| = 0$$ $\Rightarrow x = \sum_{}^{}{a = 0}$ (by hypothesis) Or 1$\{(b - x)(c - x) - a^{2}\} - c\{ c - x - a\} + b\{ a - b + x\} = 0$ by expanding the determinant. or $x^{2} - (a^{2} + b^{2} + c^{2}) + (ab + bc + ca) = 0$ or $x^{2} - \left( \sum_{}^{}a^{2} \right) - \frac{1}{2}\left( \sum_{}^{}a^{2} \right) = 0$ $$\left\{ \because a + b + c = 0 \Rightarrow (a + b + c)^{2} = 0 \right.\ $$ $\Rightarrow$ $\left. \ \sum_{}^{}a^{2} + 2\sum_{}^{}{ab} = 0 \Rightarrow \sum_{}^{}{ab} = - \frac{1}{2}\sum_{}^{}a^{2} \right\}$ or $x = \pm \sqrt{\frac{3}{2}\sum_{}^{}a^{2}}$ $\therefore$The solution is $x = 0$ or $\pm \sqrt{\frac{3}{2}\sum_{}^{}a^{2}}$. **Trick:** Put $a = 1,b = - 1$ and $c = 0$ so that they satisfy the condition $a + b + c = 0$. Now the determinant becomes $$\left| \begin{matrix} 1 - x & 0 & - 1 \\ 0 & - 1 - x & 1 \\ - 1 & 1 & - x \end{matrix} \right| = 0$$ $$\mathbf{\Rightarrow}\mathbf{(1}\mathbf{-}\mathbf{x)\{ x(1 + x)}\mathbf{-}\mathbf{1\} + 1(1 + x) = 0}$$ $\mathbf{\Rightarrow (1 - x)\{}\mathbf{x}^{\mathbf{2}}\mathbf{+ x - 1\} + x + 1 = 0}$ $\mathbf{\Rightarrow}x(x^{2} - 3) = 0$ Now putting these in the options, we find that option (3) gives the same values i.e., 0, $\pm \sqrt{3}$.