For a real number `alpha,` if the system `[1alphaalpha^2alpha1alphaalpha^2alpha1][x y z]=[1-1 1]` of linear equations, has infinitely many solutions, then `1+alpha+alpha^2=`

Explanation

To solve the problem, we need to find the value of \(1 + \alpha + \alpha^2\) given that the system of linear equations has infinitely many solutions. This occurs when the determinant of the coefficient matrix is zero.

### Step-by-step Solution:

1. **Set up the system of equations:**
The system can be represented in matrix form as:
\[ \begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \]

2. **Find the determinant of the coefficient matrix:**
We need to calculate the determinant of the matrix:
\[ D = \begin{vmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{vmatrix} \]
The determinant can be expanded as follows:
\[ D = 1 \cdot \begin{vmatrix} 1 & \alpha \\ \alpha & 1 \end{vmatrix} - \alpha \cdot \begin{vmatrix} \alpha & \alpha \\ \alpha^2 & 1 \end{vmatrix} + \alpha^2 \cdot \begin{vmatrix} \alpha & 1 \\ \alpha^2 & \alpha \end{vmatrix} \]

3. **Calculate the 2x2 determinants:**
- For the first determinant:
\[ \begin{vmatrix} 1 & \alpha \\ \alpha & 1 \end{vmatrix} = 1 \cdot 1 - \alpha \cdot \alpha = 1 - \alpha^2 \]
- For the second determinant:
\[ \begin{vmatrix} \alpha & \alpha \\ \alpha^2 & 1 \end{vmatrix} = \alpha \cdot 1 - \alpha \cdot \alpha^2 = \alpha - \alpha^3 = \alpha(1 - \alpha^2) \]
- For the third determinant:
\[ \begin{vmatrix} \alpha & 1 \\ \alpha^2 & \alpha \end{vmatrix} = \alpha \cdot \alpha - 1 \cdot \alpha^2 = \alpha^2 - \alpha^2 = 0 \]

4. **Substituting back into the determinant:**
Now substituting these values back into the determinant \(D\):
\[ D = 1(1 - \alpha^2) - \alpha(\alpha(1 - \alpha^2)) + 0 \]
Simplifying this gives:
\[ D = 1 - \alpha^2 - \alpha^2 + \alpha^4 = 1 - 2\alpha^2 + \alpha^4 \]

5. **Set the determinant to zero for infinitely many solutions:**
For the system to have infinitely many solutions, we set the determinant equal to zero:
\[ 1 - 2\alpha^2 + \alpha^4 = 0 \]
This can be rearranged to:
\[ \alpha^4 - 2\alpha^2 + 1 = 0 \]
Let \(u = \alpha^2\), then the equation becomes:
\[ u^2 - 2u + 1 = 0 \]
This factors to:
\[ (u - 1)^2 = 0 \]
Thus, \(u = 1\) or \(\alpha^2 = 1\), which gives:
\[ \alpha = 1 \quad \text{or} \quad \alpha = -1 \]

6. **Calculate \(1 + \alpha + \alpha^2\):**
- For \(\alpha = 1\):
\[ 1 + 1 + 1^2 = 1 + 1 + 1 = 3 \]
- For \(\alpha = -1\):
\[ 1 - 1 + (-1)^2 = 1 - 1 + 1 = 1 \]

### Conclusion:
Thus, the possible values of \(1 + \alpha + \alpha^2\) are \(3\) and \(1\).