Question

If $a\ne b\ne c$, 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

Answer: (D)

$x=0$

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

On putting $x=a$, we get

$\Delta =\left|\begin{array}{ccc}0& x-a& a-b\\ 2a& 0& a-c\\ a+b& a+c& 0\end{array}\right|=(a+c)(a+b)(a-c)$

Clearly $\Delta \ne 0$ on expansion along second column, so that $x=a$ does not satisfy the equation $\Delta =0$.

Similarly $x=b$ and $x=c$ also do not satisfy.

Now, put $x=0$, we get

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

Hence, $x=0$ satisfies the equation $\Delta =0$.