Discussion Forum



1)

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 -($\alpha+\beta+\gamma$)

B) 0 and $\alpha+\beta+\gamma$

C) 1 and $\alpha-\beta-\gamma$

D) 0 and $\alpha^{2}+\beta^{2}+\gamma^{2}$

Answer:

Option A

Explanation:

Given $\begin{vmatrix}x+\alpha & \beta & \gamma \\\gamma & x+\beta & \alpha \\\alpha & \beta& x+\gamma \end{vmatrix} =0$

Operate C1 → C1 + C2 + C3

$\begin{vmatrix}x+\alpha+\beta+\gamma & \beta & \gamma \\x+\alpha+\beta+\gamma & x+\beta & \alpha \\x+\alpha+\beta+\gamma & \beta& x+\gamma \end{vmatrix} =0$

$(x+\alpha+\beta+\gamma )\begin{vmatrix}1& \beta & \gamma \\1 & x+\beta & \alpha \\1 & \beta& x+\gamma \end{vmatrix} =0$

x = -($\alpha+\beta+\gamma$)

Again if 

$\begin{vmatrix}1& \beta & \gamma \\1 & x+\beta & \alpha \\1 & \beta& \gamma \end{vmatrix} =0$→

$\begin{vmatrix}1& \beta & \gamma \\0 & x & \alpha-\gamma \\0 & 0& x \end{vmatrix}$=0

x2 = 0 → x = 0

.'. Solutions of the equation x = 0, -($\alpha+\beta+\gamma$)