Q. The value of $x$ obtained from the equation $\left| \begin{matrix} x + \alpha & \beta & \gamma \\ \gamma & x + \beta & \alpha \\ \alpha & \beta & x + \gamma \end{matrix} \right| = 0$ will be

A) 0 and $–(\alpha + \beta + \gamma)$
B) 0 and $(\alpha + \beta + \gamma)$
C) 1 and $(\alpha - \beta - \gamma)$
D) 0 and $(\alpha^{2} + \beta^{2} + \gamma^{2})$
Explanation

(a)
We have, $\left| \begin{matrix}
x + \alpha & \beta & \gamma \\
\gamma & x + \beta & \alpha \\
\alpha & \beta & x + \gamma
\end{matrix} \right| = 0$
Applying $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$
$\Rightarrow \ \left| \begin{matrix}
x + \alpha + \beta + \gamma & \beta & \gamma \\
x + \alpha + \beta + \gamma & x + \beta & \alpha \\
x + \alpha + \beta + \gamma & \beta & x + \gamma
\end{matrix} \right| = 0$
$\Rightarrow (x + \alpha + \beta + \gamma)\left| \begin{matrix}
1 & \beta & \gamma \\
1 & x + \beta & \alpha \\
1 & \beta & x + \gamma
\end{matrix} \right| = 0$
Applying $R_{2} \rightarrow R_{2} - R_{1}\ and\ R_{3} \rightarrow R_{3} - R_{1}$
$\Rightarrow \ (x + \alpha + \beta + \gamma)\left| \begin{matrix}
1 & \beta & \gamma \\
0 & x & \alpha - \gamma \\
0 & 0 & x
\end{matrix} \right| = 0$
$\Rightarrow \ (x + \alpha + \beta + \gamma)\left( x^{2} - 0 \right) = 0$
$\Rightarrow x = 0\ \ or\ x = - (\alpha + \beta + \gamma)$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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