Question

The value of x obtained from the equation $\begin{vmatrix}x+\alpha&\beta&\gamma\\ \gamma&x+\beta&\alpha\\ \alpha&\beta&x+\gamma\end{vmatrix}=0$ will be

• A. 0 and $-\left(\alpha+\beta+\gamma\right)$
• B. 0 and $\alpha+\beta+\gamma$
• C. 1 and $\left(\alpha-\beta-\gamma\right)$
• D. $0 and \alpha^{2}+\beta+\gamma^{2}$

Answer 1

The Correct Option is A

$Given \begin{vmatrix}x+\alpha&\beta&\gamma\\ \gamma&x+\beta&\alpha\\ \alpha&\beta&x+\gamma\end{vmatrix}=0$ Operate $C_{1} \to C_{1}+C_{2}+C_{3}$ $\begin{vmatrix}x+\alpha+\beta+\gamma&\beta&\gamma\\ x+\alpha+\beta+\gamma&x+\beta&\alpha\\ x+\alpha+\beta+\gamma&\beta&x+\gamma\end{vmatrix}=0$ $=\left(x+\alpha+\beta+\gamma\right) \begin{vmatrix}1&\beta&\gamma\\ 1&x+\beta&\alpha\\ 1&\beta&x+\gamma\end{vmatrix}=0$ $\Rightarrow\quad x+\alpha+\beta+\gamma=0 \Rightarrow\, x=-\left(\alpha+\beta+\gamma\right)$ Again if $\begin{vmatrix}1&\beta&\gamma\\ 1&x+\beta&\alpha\\ 1&\beta&\gamma\end{vmatrix}=0 \Rightarrow \begin{vmatrix}1&\beta&\gamma\\ 0&x&\alpha-\gamma\\ 0&0&x\end{vmatrix}=0$ $\Rightarrow\quad x^{2}=0 \Rightarrow x=0$ $\therefore\quad$ Solutions of the equation are x = 0, $-\left(\alpha+\beta+\gamma\right)$