Question

lf $a\ne b\ne c$ Then one 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 given by

• A. x = a
• B. x = b
• C. x = c
• D. x = 0

Answer 1

The correct option is D x = 0

Suppose $x=0$

Then, the determinant becomes

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

$=0(0+c^2)+a(0+bc)-b(ac-0)$

$=abc-abc$

$=0$

So LHS $=$ RHS

Hence the answer is option D