Consider the system of equations `alphax+y+z = p, x+alphay+z=q and x+y+alphaz=r`, then the sum of all possible distinct value(s) of `alpha` for which system does not possess a unique solution is

Options

A. `-2`
B. 1
C. `-1`
D. 0

Answer

To solve the given system of equations for the values of \( \alpha \) for which the system does not possess a unique solution, we need to analyze the determinant of the coefficient matrix.

The system of equations is:
1. \( \alpha x + y + z = p \)
2. \( x + \alpha y + z = q \)
3. \( x + y + \alpha z = r \)

### Step 1: Write the coefficient matrix
The coefficient matrix \( A \) for the system can be represented as:
\[ A = \begin{bmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{bmatrix} \]

### Step 2: Calculate the determinant of the matrix
To find the values of \( \alpha \) for which the system does not have a unique solution, we need to set the determinant of matrix \( A \) to zero:
\[ \text{det}(A) = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we can calculate:
\[ \text{det}(A) = \alpha \begin{vmatrix} \alpha & 1 \\ 1 & \alpha \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \alpha \end{vmatrix} + 1 \begin{vmatrix} 1 & \alpha \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} \alpha & 1 \\ 1 & \alpha \end{vmatrix} = \alpha^2 - 1 \)
2. \( \begin{vmatrix} 1 & 1 \\ 1 & \alpha \end{vmatrix} = \alpha - 1 \)
3. \( \begin{vmatrix} 1 & \alpha \\ 1 & 1 \end{vmatrix} = 1 - \alpha \)

Substituting these back into the determinant:
\[ \text{det}(A) = \alpha (\alpha^2 - 1) - (\alpha - 1) + (1 - \alpha) \]
\[ = \alpha^3 - \alpha - \alpha + 1 + 1 - \alpha \]
\[ = \alpha^3 - 3\alpha + 2 \]

### Step 3: Set the determinant to zero
Now, we set the determinant to zero to find the values of \( \alpha \):
\[ \alpha^3 - 3\alpha + 2 = 0 \]

### Step 4: Factor the polynomial
We can factor this polynomial. By trial and error, we can find that \( \alpha = 1 \) is a root. We can perform polynomial long division or synthetic division to factor out \( (\alpha - 1) \):
\[ \alpha^3 - 3\alpha + 2 = (\alpha - 1)(\alpha^2 + \alpha - 2) \]

Next, we can factor \( \alpha^2 + \alpha - 2 \):
\[ \alpha^2 + \alpha - 2 = (\alpha - 1)(\alpha + 2) \]

Thus, the complete factorization is:
\[ (\alpha - 1)^2(\alpha + 2) = 0 \]

### Step 5: Find the roots
The roots of the equation are:
1. \( \alpha = 1 \) (with multiplicity 2)
2. \( \alpha = -2 \)

### Step 6: Sum the distinct values
The distinct values of \( \alpha \) are \( 1 \) and \( -2 \). Therefore, the sum of all possible distinct values of \( \alpha \) is:
\[ 1 + (-2) = -1 \]

### Final Answer
The sum of all possible distinct values of \( \alpha \) for which the system does not possess a unique solution is \( -1 \).

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