The system of equations
`alphax+y+z=alpha-1,`
`x+alphay+z=alpha-1`
`x+y+alphaz=alpha-1`
and has no solution if `alpha` is

Options

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

Answer

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

### Given System of Equations:
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 express the system in the form \(AX = B\), where:
\[ A = \begin{pmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} \alpha - 1 \\ \alpha - 1 \\ \alpha - 1 \end{pmatrix} \]

### Step 2: Find the determinant of the coefficient matrix \(A\)
To find the values of \(\alpha\) for which the system has no solution, we need to compute the determinant of matrix \(A\) and set it to zero.

\[ \text{det}(A) = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the formula for the determinant of a 3x3 matrix, we expand:

\[ \text{det}(A) = \alpha \begin{vmatrix} \alpha & 1 \\ 1 & \alpha \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \alpha \end{vmatrix} + 1 \begin{vmatrix} 1 & \alpha \\ 1 & 1 \end{vmatrix} \]

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 expression:

\[ \text{det}(A) = \alpha(\alpha^2 - 1) - (\alpha - 1) + (1 - \alpha) \]
\[ = \alpha^3 - \alpha - \alpha + 1 + 1 - \alpha \]
\[ = \alpha^3 - 3\alpha + 2 \]

### Step 4: Set the determinant to zero
To find when the system has no solution, we set the determinant equal to zero:

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

### Step 5: Factor the polynomial
We can try to find the roots of the polynomial. By testing possible rational roots, we find:

1. \(\alpha = 1\) is a root.
2. Factoring out \((\alpha - 1)\):

Using synthetic division or polynomial long division, we can factor the polynomial:

\[ \alpha^3 - 3\alpha + 2 = (\alpha - 1)(\alpha^2 + \alpha - 2) \]

### Step 6: Factor the quadratic
Now we factor the quadratic:

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

Thus, the complete factorization is:

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

### Step 7: Solve for \(\alpha\)
Setting each factor to zero gives us:

1. \(\alpha - 1 = 0 \Rightarrow \alpha = 1\) (this leads to infinite solutions)
2. \(\alpha + 2 = 0 \Rightarrow \alpha = -2\) (this leads to no solution)

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