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. $\pm \frac{3}{2}(a^2+b^2+c^2)$
• 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

The correct option is C $0,\pm \sqrt{\frac{3}{2}(a^2+b^2+c^2)}$

$\left|\begin{array}{ccc}a-x& c& b\\ c& b-x& a\\ b& a& c-x\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}a+b+c-x& c& b\\ a+b+c-x& b-x& a\\ a+b+c-x& a& c-x\end{array}\right|=0\Rightarrow x=\sum a=0$ (by hypothesis)
or $1\left\{\right(b-x\left)\right(c-x)-a^2\}-c\{c-x-a\}+b\{a-b+x\}=0$ by expanding the determinant.
or $x^2-(a^2+b^2+c^2)+(ab+bc+ca)=0$
or $x^2-(\sum a^2)-\frac{1}{2}(\sum a^2)=0\{\begin{array}{c}\because a+b+c=0\Rightarrow (a+b+c{)}^2=0\end{array}\begin{array}{c}\Rightarrow \sum a^2+2\sum ab=0\sum ab=-\frac{1}{2}\sum a^2\end{array}\}orx=\pm \sqrt{\frac{3}{2}\sum a^2}$
$\therefore$ 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 $\left|\begin{array}{ccc}1-x& 0& -1\\ 0& -1-x& 1\\ -1& 1& -x\end{array}\right|=0\Rightarrow (1-x)\left\{x\right(1+x)-1\}+1(1+x)=0\Rightarrow (1-x)\{x^2+x-1\}+x+1=0\Rightarrow x(x^2-3)=0$
Now putting these in the options, we find that option (c) gives the same values i.e., 0, $\pm \sqrt{3}$.