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

Answer 1

$x=0$ (double root) or $x=-(\alpha+\beta+\gamma)$

Answer 2

Step 1. Let $A$ be the matrix

$$A = \begin{pmatrix} \alpha & \beta & \gamma\\ \gamma & \beta & \alpha\\ \alpha & \beta & \gamma \end{pmatrix}.$$

We need $\det(xI + A)=0$.

Step 2. Note that row 1 and row 3 of $A$ are identical, so $\rank(A)=2$. Hence one eigenvalue of $A$ is $0$. Also

$$\tr(A)=\alpha+\beta+\gamma,\quad \sum \text{(principal minors of order 2)} = 0,\quad \det(A)=0.$$

The characteristic polynomial of $A$ is

$$\lambda^3 - (\alpha+\beta+\gamma)\,\lambda^2 + 0\cdot\lambda - 0 = \lambda^2(\lambda - (\alpha+\beta+\gamma)).$$

Thus the eigenvalues of $A$ are $0$ (double) and $\alpha+\beta+\gamma$.

Step 3. The eigenvalues of $xI+A$ are $x+0$ (twice) and $x+(\alpha+\beta+\gamma)$. Hence

$$\det(xI + A) = x^2\bigl(x + (\alpha+\beta+\gamma)\bigr).$$

Setting this to zero gives

$$x^2\bigl(x + (\alpha+\beta+\gamma)\bigr)=0 \quad\Longrightarrow\quad x=0\;\text{(double)}\quad\text{or}\quad x=-(\alpha+\beta+\gamma).$$