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 will follow these steps:

### Step 1: Write the system of equations in matrix form
The system of equations is:
1. \(\alpha x + y + z = p\)
2. \(x + \alpha y + z = q\)
3. \(x + y + \alpha z = r\)

We can represent this system in matrix form as:
\[ \begin{bmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} p \\ q \\ r \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
To determine when the system does not have a unique solution, we need to find the determinant of the coefficient matrix and set it to zero:
\[ D = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the formula for the determinant of a \(3 \times 3\) matrix, we have:
\[ D = \alpha \begin{vmatrix} \alpha & 1 \\ 1 & \alpha \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \alpha \end{vmatrix} + 1 \begin{vmatrix} 1 & \alpha \\ 1 & 1 \end{vmatrix} \]

Calculating the \(2 \times 2\) 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. \(\begin{vmatrix} 1 & \alpha \\ 1 & 1 \end{vmatrix} = 1 - \alpha\)

Putting it all together:
\[ D = \alpha(\alpha^2 - 1) - (\alpha - 1) + (1 - \alpha) \]
\[ D = \alpha^3 - \alpha - \alpha + 1 + 1 - \alpha \]
\[ D = \alpha^3 - 3\alpha + 2 \]

### Step 4: Set the determinant to zero
To find the values of \(\alpha\) for which the system does not have a unique solution, we set the determinant to zero:
\[ \alpha^3 - 3\alpha + 2 = 0 \]

### Step 5: Factor the polynomial
We can factor this polynomial:
\[ (\alpha - 1)(\alpha^2 + \alpha - 2) = 0 \]

Now, we can factor the quadratic:
\[ \alpha^2 + \alpha - 2 = (\alpha - 1)(\alpha + 2) = 0 \]

### Step 6: Find the roots
Setting each factor to zero gives us:
1. \(\alpha - 1 = 0 \Rightarrow \alpha = 1\)
2. \(\alpha - 1 = 0 \Rightarrow \alpha = 1\) (duplicate root)
3. \(\alpha + 2 = 0 \Rightarrow \alpha = -2\)

Thus, the distinct values of \(\alpha\) for which the system does not possess a unique solution are \(\alpha = 1\) and \(\alpha = -2\).

### Step 7: Calculate the sum of distinct values
Now, we find the sum of all distinct values:
\[ 1 + (-2) = -1 \]

### 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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