If `a , b , c` are non-zero, then the system of equations `(alpha+a)x+alphay+alphaz=0,` `alphax+(alpha+b)y+alphaz=0,` `alphax+alphay+(alpha+c)z=0` has a non-trivial solution if

Options

A. `2alpha=a+b+c`
B. `alpha^(-1)=a+b+c`
C. `alpha+a+b+c=1`
D. `alpha^(-1)=-(a^(-1)+b^(-1)+c^(-1))`

Explanation

To determine the conditions under which the given system of equations has a non-trivial solution, we need to analyze the determinant of the coefficient matrix. Let's break down the solution step by step.

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \((\alpha + a)x + \alpha y + \alpha z = 0\)
2. \(\alpha x + (\alpha + b)y + \alpha z = 0\)
3. \(\alpha x + \alpha y + (\alpha + c)z = 0\)

We can express this in matrix form as:
\[ \begin{pmatrix} \alpha + a & \alpha & \alpha \\ \alpha & \alpha + b & \alpha \\ \alpha & \alpha & \alpha + c \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]

### Step 2: Set up the determinant condition
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:
\[ \text{det}\begin{pmatrix} \alpha + a & \alpha & \alpha \\ \alpha & \alpha + b & \alpha \\ \alpha & \alpha & \alpha + c \end{pmatrix} = 0 \]

### Step 3: Calculate the determinant
Let \(D\) denote the determinant:
\[ D = \begin{vmatrix} \alpha + a & \alpha & \alpha \\ \alpha & \alpha + b & \alpha \\ \alpha & \alpha & \alpha + c \end{vmatrix} \]

Using the determinant formula for a \(3 \times 3\) matrix, we can expand \(D\):
\[ D = (\alpha + a)((\alpha + b)(\alpha + c) - \alpha^2) - \alpha(\alpha(\alpha + c) - \alpha^2) + \alpha(\alpha(\alpha + b) - \alpha^2) \]

### Step 4: Simplify the determinant expression
Expanding and simplifying the determinant:
1. The first term expands to:
\[ (\alpha + a)(\alpha^2 + b\alpha + c\alpha + bc - \alpha^2) = (\alpha + a)(b\alpha + c\alpha + bc) \]

2. The second term simplifies to:
\[ -\alpha(\alpha^2 + c\alpha - \alpha^2) = -\alpha c\alpha \]

3. The third term simplifies to:
\[ \alpha(\alpha^2 + b\alpha - \alpha^2) = \alpha b\alpha \]

Combining these terms gives us:
\[ D = (\alpha + a)(b\alpha + c\alpha + bc) - \alpha c\alpha + \alpha b\alpha \]

### Step 5: Set the determinant to zero
We need to set \(D = 0\) and solve for \(\alpha\):
\[ (\alpha + a)(b\alpha + c\alpha + bc) - \alpha c\alpha + \alpha b\alpha = 0 \]

### Step 6: Rearranging and solving
After rearranging, we can express this as:
\[ \alpha(b + c + \frac{bc}{\alpha + a}) + a(b + c + \frac{bc}{\alpha + a}) = 0 \]

### Step 7: Final condition
From the above expression, we can derive the condition:
\[ \frac{1}{\alpha} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \]

### Conclusion
Thus, the system of equations has a non-trivial solution if:
\[ \frac{1}{\alpha} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \]