The system of equations `alpha `x + y + z = `alpha` - 1
x + `alpha` y + = `alpha` - 1 , x + y + `alpha`z = `alpha` - 1
has infinite solutions, if `alpha` is

Options

A. `-2`
B. either -2 or 1
C. not - 2
D. 1

Answer

To solve the problem, we need to determine the values of \( \alpha \) for which the system of equations has infinite solutions. The system of equations is given as follows:

1. \( \alpha x + y + z = \alpha - 1 \)
2. \( x + \alpha y + z = \alpha - 1 \)
3. \( x + y + \alpha z = \alpha - 1 \)

### Step 1: Write the system in matrix form
We can represent the system of equations in matrix form as \( AX = B \), where:

\[ A = \begin{bmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} \alpha - 1 \\ \alpha - 1 \\ \alpha - 1 \end{bmatrix} \]

### Step 2: Find the determinant of matrix \( A \)
For the system to have infinite solutions, the determinant of matrix \( A \) must be zero. We calculate the determinant \( |A| \):

\[ |A| = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix, we have:

\[ |A| = \alpha \begin{vmatrix} \alpha & 1 \\ 1 & \alpha \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \alpha \end{vmatrix} + 1 \begin{vmatrix} 1 & \alpha \\ 1 & 1 \end{vmatrix} \]

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:

\[ |A| = \alpha(\alpha^2 - 1) - (\alpha - 1) + (1 - \alpha) \]

Simplifying this gives:

\[ |A| = \alpha^3 - \alpha - \alpha + 1 + 1 - \alpha = \alpha^3 - 3\alpha + 2 \]

### Step 3: Set the determinant to zero
For infinite solutions, we set the determinant equal to zero:

\[ \alpha^3 - 3\alpha + 2 = 0 \]

### Step 4: Factor the polynomial
We can factor the polynomial \( \alpha^3 - 3\alpha + 2 \):

\[ (\alpha - 1)^2(\alpha + 2) = 0 \]

### Step 5: Solve for \( \alpha \)
Setting each factor to zero gives us:

1. \( \alpha - 1 = 0 \) → \( \alpha = 1 \)
2. \( \alpha + 2 = 0 \) → \( \alpha = -2 \)

### Step 6: Check conditions for infinite solutions
We need to check the conditions for infinite solutions. For \( \alpha = 1 \):

Substituting \( \alpha = 1 \) into the determinant \( D_1 \):

\[ D_1 = \begin{vmatrix} 0 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]

This determinant is zero, indicating infinite solutions.

For \( \alpha = -2 \):

Substituting \( \alpha = -2 \) into \( D_1 \):

\[ D_1 = \begin{vmatrix} -3 & -3 & -3 \\ -2 & -2 & 1 \\ -2 & 1 & -2 \end{vmatrix} \]

Calculating this determinant gives a non-zero value, indicating no solutions.

### Conclusion
Thus, the value of \( \alpha \) for which the system has infinite solutions is:

\[ \alpha = 1 \]