Question

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

• A. x=0
• B. x=a
• C. x=b
• D. x=c

Answer 1

The correct option is A x=0

Obviously, on putting x=0, we observe that the determinant becomes
${\Delta }_{x=0}=\left|\begin{array}{ccc}0& -a& -b\\ a& 0& -c\\ b& c& 0\end{array}\right|=a\left(bc\right)-b\left(ac\right)=0\therefore$ x = 0 is a root of the given equation.
Aliter : Expanding $\Delta$, we get
$\Delta \equiv -(x-a)[-(x+b\left)\right(x-c\left)\right]+(x-b)\left[\right(x+a\left)\right(x+c\left)\right]=0\Rightarrow 2x^3-(2\sum ab)x=0\Rightarrow Eitherx=0or{x}^2=\sum ab(i.e.,x=\pm \sum ab)$
Again x = 0satisfies the given equation.