The system of equations `alphax+y+z=alpha-1, x+alphay+z=alpha-1, x+y+alphaz=alpha-1` has no solution if alpha is (A) 1 (B) not -2 (C) either -2 or 1 (D) -2

Options

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

Explanation

To determine the value of \(\alpha\) for which the system of equations has no solution, we will analyze the given equations using determinants.

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

We can represent this system in matrix form as \(AX = B\), where:
\[ A = \begin{bmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} \alpha - 1 \\ \alpha - 1 \\ \alpha - 1 \end{bmatrix} \]

To find when the system has no solution, we need to check the determinant of matrix \(A\) (denoted as \(D\)). The system has no solution if \(D = 0\) and at least one of the determinants of the modified matrices (denoted as \(D_1\), \(D_2\), \(D_3\)) is non-zero.

### Step 1: Calculate the Determinant \(D\)
The determinant \(D\) of matrix \(A\) can be calculated using the formula for the determinant of a \(3 \times 3\) matrix:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \(A\):
\[ D = \alpha \begin{vmatrix} \alpha & 1 \\ 1 & \alpha \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \alpha \end{vmatrix} + 1 \begin{vmatrix} 1 & \alpha \\ 1 & 1 \end{vmatrix} \]

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

Substituting these back into the determinant \(D\):
\[ D = \alpha(\alpha^2 - 1) - 1(\alpha - 1) + (1 - \alpha) \]
\[ D = \alpha^3 - \alpha - \alpha + 1 + 1 - \alpha \]
\[ D = \alpha^3 - 3\alpha + 2 \]

### Step 2: Set the Determinant to Zero
To find when the system has no solution, we set \(D = 0\):
\[ \alpha^3 - 3\alpha + 2 = 0 \]

### Step 3: Factor the Polynomial
We can factor this polynomial to find the roots. Testing possible rational roots, we find:
\[ (\alpha - 1)(\alpha^2 + \alpha - 2) = 0 \]
Factoring further:
\[ (\alpha - 1)(\alpha - 1)(\alpha + 2) = 0 \]
Thus, the roots are:
\[ \alpha = 1 \quad \text{and} \quad \alpha = -2 \]

### Step 4: Determine Conditions for No Solution
For the system to have no solution, we need \(D = 0\) and at least one of \(D_1\), \(D_2\), or \(D_3\) must be non-zero.

1. If \(\alpha = 1\), we can check the modified determinants, and they will also be zero, leading to infinite solutions.
2. If \(\alpha = -2\), we can check that at least one of the modified determinants is non-zero.

### Conclusion
The system of equations has no solution if \(\alpha = -2\).

Thus, the answer is:
**(D) -2**