For the system of linear equations `alpha x + y + z = 1, x + alpha y + z = 1, x + y + alpha z = beta` which one of the following statements is NOT correct ?

Options

A. It has infinitely many solutions if `alpha = 2 and beta = – 1`
B. `x + y + z =3/4` if `alpha = 2 and beta = 1`
C. It has infinitely many solutions if` `alpha = 1 and beta = 1`
D. It has no solution if `alpha = –2 and beta = 1`

Answer

To solve the system of linear equations given by:

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

we need to analyze the determinant of the coefficients of the variables \(x\), \(y\), and \(z\) to determine the conditions for the existence of solutions.

### Step 1: Write the coefficient matrix and calculate the determinant

The coefficient matrix \(A\) for the system can be represented as:

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

To find the determinant \(D\) of this matrix, we can use the formula for the determinant of a 3x3 matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, we have:

\[ D = \alpha(\alpha \cdot \alpha - 1 \cdot 1) - 1(1 \cdot \alpha - 1 \cdot 1) + 1(1 \cdot 1 - 1 \cdot \alpha) \]

Calculating this gives:

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

Simplifying further:

\[ D = \alpha^3 - \alpha - \alpha + 1 + 1 - \alpha = \alpha^3 - 3\alpha + 2 \]

### Step 2: Set the determinant to zero

To find the values of \(\alpha\) for which the system has either no solution or infinitely many solutions, we set the determinant to zero:

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

### Step 3: Factor the polynomial

We can factor this polynomial. By using the Rational Root Theorem, we find that \(\alpha = 1\) is a root. We can factor out \((\alpha - 1)\):

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

Factoring the quadratic gives:

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

Thus, the complete factorization is:

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

### Step 4: Find the roots

The roots of the equation are:

1. \(\alpha = 1\) (with multiplicity 2)
2. \(\alpha = -2\)

### Step 5: Analyze the conditions for \(\beta\)

Now we need to analyze the conditions for \(\beta\) based on the values of \(\alpha\):

- For \(\alpha = 1\), the system becomes:
\[ x + y + z = 1 \]
\[ x + y + z = 1 \]
\[ x + y + z = \beta \]
This leads to infinitely many solutions if \(\beta = 1\) and no solution if \(\beta \neq 1\).

- For \(\alpha = -2\), we need to check the determinant again, but generally, this leads to a unique solution for certain values of \(\beta\).

### Conclusion

The statement that is NOT correct is likely related to the conditions under which solutions exist based on the values of \(\alpha\) and \(\beta\).