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 solve the given system of equations for the value of \(\alpha\) such that the system has no solution, we will follow these steps:

### Step 1: Write the system of equations
The system of equations is given as:
1. \(\alpha(x - 1) + y + z = -1\)
2. \(x + \alpha(y - 1) + z = -1\)
3. \(x + y + \alpha(z - 1) = -1\)

### Step 2: Rearrange the equations
Rearranging each equation to standard form \(Ax + By + Cz = D\):
1. \(\alpha x + y + z = \alpha - 1\)
2. \(x + \alpha y + z = \alpha - 1\)
3. \(x + y + \alpha z = \alpha - 1\)

### Step 3: Write the coefficient matrix and constant matrix
The coefficient matrix \(A\) and the constant matrix \(B\) can be represented as:
\[ A = \begin{bmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{bmatrix}, \quad B = \begin{bmatrix} \alpha - 1 \\ \alpha - 1 \\ \alpha - 1 \end{bmatrix} \]

### Step 4: Calculate the determinant of the coefficient matrix
To find the value of \(\alpha\) for which the system has no solution, we need to calculate the determinant of matrix \(A\) and set it to zero:
\[ \text{det}(A) = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} \]

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

### Step 5: Set the determinant to zero
We need to set the determinant to zero to find the values of \(\alpha\):
\[ \alpha^3 - 3\alpha + 2 = 0 \]

### Step 6: Factor the polynomial
To factor the polynomial, we can use the Rational Root Theorem or synthetic division. Testing possible rational roots, we find that \(\alpha = 1\) is a root:
\[ \alpha^3 - 3\alpha + 2 = (\alpha - 1)(\alpha^2 + \alpha - 2) \]
Factoring further:
\[ \alpha^2 + \alpha - 2 = (\alpha - 1)(\alpha + 2) \]
Thus, the complete factorization is:
\[ (\alpha - 1)^2(\alpha + 2) = 0 \]

### Step 7: Solve for \(\alpha\)
Setting each factor to zero gives:
1. \((\alpha - 1)^2 = 0 \Rightarrow \alpha = 1\)
2. \((\alpha + 2) = 0 \Rightarrow \alpha = -2\)

### Step 8: Determine conditions for no solution
For the system to have no solution, the determinant must be zero, and at least one of the determinants formed by replacing the columns must not be zero. We can check the conditions for \(\alpha = 1\) and \(\alpha = -2\).

### Conclusion
The system of equations has no solution if \(\alpha = 1\) or \(\alpha = -2\).