For a real number, `alpha` if the system
`[{:(,1,alpha,alpha^(2)),(,alpha,1,alpha),(,alpha^(2),alpha,1):}] [{:(,x),(,y),(,z):}]=[{:(,1),(,-1),(,1):}]`
of linear equations has infinitely many solutions then `1+alpha+alpha^(2)`=

Answer

To solve the problem, we need to analyze the given system of linear equations represented in matrix form. The system is given by:

\[ \begin{pmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \]

We need to find the condition under which this system has infinitely many solutions. This occurs when the determinant of the coefficient matrix is zero.

### Step 1: Calculate the Determinant of the Coefficient Matrix

The determinant of the matrix

\[ A = \begin{pmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{pmatrix} \]

can be calculated using the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows respectively.

Substituting the values:

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

This simplifies to:

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

\[ = 1 - \alpha^2 - \alpha^2 + \alpha^4 \]

\[ = \alpha^4 - 2\alpha^2 + 1 \]

### Step 2: Set the Determinant to Zero

For the system to have infinitely many solutions, we set the determinant equal to zero:

\[ \alpha^4 - 2\alpha^2 + 1 = 0 \]

### Step 3: Factor the Polynomial

This can be factored as:

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

### Step 4: Solve for Alpha

From this, we find:

\[ \alpha^2 - 1 = 0 \implies \alpha^2 = 1 \implies \alpha = \pm 1 \]

### Step 5: Calculate \( 1 + \alpha + \alpha^2 \)

Now, we need to evaluate \( 1 + \alpha + \alpha^2 \) for both values of \( \alpha \):

1. For \( \alpha = 1 \):

\[ 1 + 1 + 1^2 = 1 + 1 + 1 = 3 \]

2. For \( \alpha = -1 \):

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

### Conclusion

Thus, the value of \( 1 + \alpha + \alpha^2 \) can be either 3 or 1 depending on the value of \( \alpha \).

### Final Answer

The possible values of \( 1 + \alpha + \alpha^2 \) are \( 3 \) or \( 1 \).

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