The system of equations `alpha(x-1)+y+z=-1, x+alpha(y-1)+z=-1 and x+y+alpha(z-1)=-1` has no solution, if `alpha` is equal to

Options

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

Answer

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

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

### Step 1: Rearranging the equations
We can rearrange each equation to express them in standard form \(Ax + By + Cz = D\).

1. Rearranging the first equation:
\[ \alpha x - \alpha + y + z = -1 \implies \alpha x + y + z = \alpha - 1 \]

2. Rearranging the second equation:
\[ x + \alpha y - \alpha + z = -1 \implies x + \alpha y + z = \alpha - 1 \]

3. Rearranging the third equation:
\[ x + y + \alpha z - \alpha = -1 \implies x + y + \alpha z = \alpha - 1 \]

Now we have the system:
1. \(\alpha x + y + z = \alpha - 1\)
2. \(x + \alpha y + z = \alpha - 1\)
3. \(x + y + \alpha z = \alpha - 1\)

### Step 2: Writing the system in matrix form
We can express the system in the form of a matrix equation \(A\mathbf{x} = \mathbf{b}\):
\[ \begin{bmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \alpha - 1 \\ \alpha - 1 \\ \alpha - 1 \end{bmatrix} \]

### Step 3: Finding the determinant of the coefficient matrix
For the system to have no solution, the determinant of the coefficient matrix must be zero:
\[ \text{det}(A) = 0 \]

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

Using the determinant formula for a \(3 \times 3\) matrix:
\[ \text{det}(A) = \alpha(\alpha^2 - 1) - 1(\alpha - 1) + 1(1 - \alpha) \]
\[ = \alpha^3 - \alpha - \alpha + 1 + 1 - \alpha \]
\[ = \alpha^3 - 3\alpha + 2 \]

Setting the determinant to zero for no solution:
\[ \alpha^3 - 3\alpha + 2 = 0 \]

### Step 4: Solving the cubic equation
To find the roots of the cubic equation, we can factor or use the Rational Root Theorem. Testing possible rational roots, we find:
\[ \alpha = 1 \quad \text{and} \quad \alpha = -2 \]

### Step 5: Conditions for no solution
For the system to have no solution, we need at least one of the determinants of the modified matrices (denoted as \(\Delta_1, \Delta_2, \Delta_3\)) to be non-zero.

From the previous steps, we conclude:
- \(\alpha = 1\) leads to a determinant of zero, which gives us a dependent system.
- \(\alpha = -2\) does not lead to a dependent system.

Thus, the value of \(\alpha\) for which the system has no solution is:
\[ \alpha = 1 \]

### Final Answer
The system of equations has no solution if \(\alpha = 1\).