The value of x obtained from the equation lef | vedclass

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Question

(A) Given the determinant equation: $\left| \begin{array}{ccc} x + \alpha & \beta & \gamma \ \gamma & x + \beta & \alpha \ \alpha & \beta & x + \gam

Vedclass · Accepted Answer

$0$ and $-(\alpha + \beta + \gamma)$

Vedclass · Answer

(A) Given the determinant equation: $\left| \begin{array}{ccc} x + \alpha & \beta & \gamma \ \gamma & x + \beta & \alpha \ \alpha & \beta & x + \gamma \end{array} ight| = 0$. Applying the column operation $C_1 	o C_1 + C_2 + C_3$: $\left| \begin{array}{ccc} x + \alpha + \beta + \gamma & \beta & \gamma \ x + \alpha + \beta + \gamma & x + \beta & \alpha \ x + \alpha + \beta + \gamma & \beta & x + \gamma \end{array} ight| = 0$. Taking $(x + \alpha + \beta + \gamma)$ common from $C_1$: $(x + \alpha + \beta + \gamma) \left| \begin{array}{ccc} 1 & \beta & \gamma \ 1 & x + \beta & \alpha \ 1 & \beta & x + \gamma \end{array} ight| = 0$. Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$: $(x + \alpha + \beta + \gamma) \left| \begin{array}{ccc} 1 & \beta & \gamma \ 0 & x & \alpha - \gamma \ 0 & 0 & x \end{array} ight| = 0$. Expanding the determinant along $C_1$: $(x + \alpha + \beta + \gamma) [1(x^2 - 0)] = 0$. $x^2(x + \alpha + \beta + \gamma) = 0$. Therefore,the values of $x$ are $x = 0$ and $x = -(\alpha + \beta + \gamma)$.

The value of $x$ obtained from the equation $\left| \begin{array}{ccc} x + \alpha & \beta & \gamma \\ \gamma & x + \beta & \alpha \\ \alpha & \beta & x + \gamma \end{array} \right| = 0$ is:

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