Question

If$a \neq b \neq c$, the value of x which satisfies the equation

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

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

Answer 1

Answer: (A)

$x = 0$

Answer 2

Expanding determinant, we get,

$\Delta = - (x - a)\lbrack - (x + b)(x - c)\rbrack + (x + b)\lbrack(x + a)(x + c)\rbrack = 0$

$\Rightarrow 2x^{3} - (2\Sigma ab)x = 0$

$\Rightarrow$ Either $x = 0$ or $x^{2} = \Sigma ab$.

Since $x = 0$ satisfies the given given equation.

Trick : On putting $x = 0$, we observe that the determinant becomes

0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{matrix} \right| = 0$$ $\therefore x = 0$ is a root of the given equation.